Continuity and strict positivity of the multi-layer extension of the stochastic heat equation

We prove the continuity and strict positivity of the multi-layer extension to the stochastic heat equation introduced in [OW11] which form a hierarchy of partition functions for the continuum directed random polymer. This shows that the corresponding free energy (logarithm of the partition function) is well defined. This is also a step towards proving the conjecture stated at the end of the above paper that an array of such partition functions has the Markov property.


Introduction
In [OW11] O'Connell and Warren introduced the following: for each n = 1, 2, . . ., t > 0 and x, y ∈ R define where ∆ k (t) = {0 < s 1 < s 2 < · · · < s k < t}, s = (s 1 , . . . , s k ), y ′ = (y ′ 1 , . . . , y ′ k ) and R k (s, y ′ ; t, x, y) is the k-point correlation function for a collection of n non-intersecting Brownian bridges each of which starts at x at time 0 and ends at y at time t. p t (x − y) is the heat kernel (2πt) −1/2 e −(x−y) 2 /2t . The integral is a k-fold stochastic integral with respect to space-time white noise, see Section 2 for the definition of such integrals. It was shown in [OW11] by considering local times of non-intersecting Brownian bridges that the infinite sum in the definition is convergent in L 2 with respect to the white noise.
(2) By a solution to the above we mean a random field u which satisfies almost surely the mild form Iterating equation (3) multiple times gives the chaos expansion (1) for n = 1. One can express the solution u(t, x, y) in a more suggestive notation: where b is a Brownian bridge that starts at x at time 0 and ends at y at time t and E b x,y;t denotes the corresponding expectation. E xp is the Wick exponential defined by for a martingale M . The Feynman-Kac formula (4) is not rigorous as it is unclear how one would define the integral of the white noise along a Brownian path and moreover to exponentiate such an expression. However, Taylor expanding the exponential, then switching the expectation with the infinite sum and evaluating the expectation, one obtains the chaos expansion of u. With this in mind, (4) can be thought of as a short hand for the chaos expansion (1) in the case n = 1. On the other hand, one can obtain an rigorous expression by replacing W in (4) with a smoothed version of the space-time white noise. Indeed, Bertini and Cancrini showed in [BC95] that such expression has a meaningful limit as one takes away the smoothing and that the limit solves the SHE. With this Feynman-Kac interpretation, one can think of the solution to the stochastic heat equation as the partition function (up to a multiplication by the heat kernel) of the continuum directed random polymer [AKQ14a]. Analogously, we write where (X 1 s , . . . , X n s , 0 ≤ s ≤ t) denotes the trajectories of the above mentioned collection of n non-intersecting Brownian bridges and E X x,y;t is the corresponding expectation. In the same manner as in the n = 1 case, (5) should be thought of as the short hand for the chaos expansion (1). Therefore, in view of (5) one can interpret Z n as the partition function (up to a factor of the heat kernel) of a natural extension of the continuum directed random polymer involving multiple non-intersecting Brownian paths.
Since the work of Bertini and Giacomin [BG97], it is widely accepted that the logarithm of u is the Cole-Hopf solution to the KPZ equation [KPZ86], with narrow wedge initial condition. This solution arises as the scaling limit of the corner growth model under weak asymmetry. The Cole-Hopf solution to the KPZ equation via the Feynman-Kac formula (4) can be seen as the free energy of the continuum directed random polymer. With this interpretation the Cole-Hopf solution can be regarded as the continuum analogue of the longest increasing subsequence of a random permutation, length of the first row of a random Young diagram, directed last passage percolation and free energy of a discrete/semi-discrete polymer in random media etc., see [BDJ99a], [BDJ99b], [BOO00], [Joh99], [Joh01a], [PS02], [Joh03], [COSZ14] and the references therein. In each of these discrete models, there is further structure provided either by multiple non-intersecting up-right paths on lattices, multi-layer growth dynamics or Young diagrams constructed from the RSK correspondence. The work in the above mentioned references have shown that in some cases, utilisation of this additional structure have lead to derivations of exact formulae for the distribution of quantities of interest. The above mentioned discrete models provide examples of what is called integrability or exact solvability. The motivation for introducing the partition functions Z n , which are the continuum analogue of the structures mentioned above, is that they should provide insight to the integrable structure in the continuum setting.
The main result of this paper is that the continuum partition functions possess some nice regularity properties.
Theorem 1.1. For all n ≥ 1, the function (t, x, y) → Z n (t, x, y) has a version that is continuous over (0, ∞) × R × R. Moreover, P[Z n (t, x, y) > 0 for all t > 0 and x, y ∈ R] = 1. Now define for n = 1, 2, . . . h n (t, x) = log Z n (t, 0, x) Z n−1 (t, 0, x) , with the convention that Z 0 ≡ 1, then h 1 (t, x) is the Cole-Hopf solution to the KPZ equation with narrow wedge initial data. An immediate corollary to the above theorem is Corollary 1.2. For all n ≥ 1, h n is well defined and it is a continuous function of (t, x) over (0, ∞) × R.
The collection {h n , n ≥ 1} represents a multi-layer extension to the free energy of the continuum directed random polymer. It is the analogue in the setting of the KPZ of the multi-layer PNG or its discrete counterpart studied in [PS02] and [Joh03] respectively.
We mention here the work of [CH13]. The authors showed the existence of a collection of random continuous curves such that the lowest indexed curve is distributed as the time t Cole-Hopf solution to the KPZ with narrow wedge initial data. It is believed (see [CH13,Conjecture 2.17]) that for each t > 0 fixed, their collection of curves is equal to {h n (t, x) : n ≥ 1, x ∈ R} defined by (7). Proving this will give an alternative proof of the continuity and strict positivity of Z n at a fixed time t. In this paper, we provide a direct proof of this and furthermore our proof gives a stronger result since t can vary over (0, ∞).
The continuity and strict positivity of u = Z 1 was proved by considering its mild form which suggests that to prove Theorem 1.1 one could consider the evolution equation satisfied by Z n . By considering a smooth space-time potential, the authors in [OW11] showed that Z n should satisfy a certain SPDE, see [OW11, Proposition 3.3 and 3.7], however unfortunately it is not immediately obvious that this SPDE makes sense in the white noise setting. Instead, we shall show that a natural extension of Z n does satisfy a rigorous evolution equation which can be regarded as a multi-dimensional stochastic heat equation. This allows us to derive the continuity and strict positivity of the extension and from which Theorem 1.1 follows as a corollary.
Denote by W n the Weyl chamber {x ∈ R n : x 1 ≥ x 2 · · · ≥ x n }, then for n = 1, 2, . . ., t > 0 and x, y ∈ W n define where R k is the k-point correlation function of a collection of n non-intersection Brownian bridges which starts at x at time 0 and ends at y at time t. p * n (t, x, y) = det[p t (x i − y j )] n i,j=1 is by the Karlin-McGregor formula [KM59] the transition density of Brownian motion killed at the boundary of W n . It was proved in [OW11, Proposition 3.2] that K n also satisfies a Karlin-McGregor type formula: where each term in the determinant are solutions to (2) each driven by the same white noise. Now, define for t > 0, where ∆(x) = 1≤i<j≤n (x i − x j ) is the Vandermonde determinant. It follows from (8) that M n has chaos expansion By (9) and the continuity of the solution to the stochastic heat equation, it is easy to see that K n (t, x, y) is almost surely continuous on (0, t) × W n × W n and is zero on the boundary of W n × W n . It follows that M n (t, x, y) is continuous in the interior W • n × W • n . By [BBO09,Lemma 5.11], p * n (t, x, y)/∆(x)∆(y) is a smooth function of (x, y) over R n × R n and since the k-point correlation function R k extends continuously to the boundary of the Weyl chamber, see Section 2.4, we see from its chaos expansion (11) that M n (t, x, y) is defined for x, y ∈ ∂W n . This also suggests that M n (t, x, y) is a continuous function on W n × W n . Furthermore, from (9) we see that M n being a ratio of determinants is a permutation symmetric function of its spatial variables, that is for any permutations π, σ of {1, . . . , n}, M n (t, πx, σy) = M n (t, x, y). Hence, we can extend M n by symmetry to a function on R n × R n and we will show that there exists a version of M n that is almost surely strictly positive and continuous on the whole of R n × R n and for all t > 0. Moreover, when all the x coordinates are equal and likewise for y, M n agrees up to a multiplicative constant with Z n , that is where c n,t := n−1 i=1 i! −1 t −n(n−1)/2 and 1 = (1, . . . , 1). Equation (12) was shown to hold in [OW11] but there the continuity of M n on the boundary of W n was only established in an L 2 sense; here we extend it to almost sure continuity. Note that (9) suggests that K n (t, x, y) and M n (t, x, y) can be regarded as the stochastic analogue of p * n (t, x, y) and p * n (t, x, y)/∆(x)∆(y) respectively where the latter has limit at the boundary equal to c n,t p t (a − b) n .
In Section 4, we will show that for all (t, x, y) ∈ (0, ∞)× R n × R n , M n (t, x, y) satisfies almost surely the mild equation where A n = 1/(n − 1)! is a combinatorial constant, dy ′ * = dy 2 . . . dy n and is the transition density of Dyson's Brownian motion starting from x ∈ W n and ending at y ∈ W n and it satisfies We can extend Q t by symmetry to a function on R n × R n and so the integral over R n in the mild equation (13) is defined. Consider also the following integral equation for (t, y) ∈ (0, ∞) × R n , where g : R n → R is permutation symmetric and may be random but independent of the white noise. The function g is the initial condition for equation (15) in the sense that lim t→0 1 n! R n g(y ′ )Q t (y, y ′ ) dy ′ = lim t→0 Wn g(y ′ )Q t (y, y ′ ) dy ′ = g(y).
On the other hand, we say that M n (t, x, y) is the solution started from a delta initial data at x even though strictly speaking it is the ratio of K n (t, x, y), which can be shown to satisfy an integral equation similar to (15) with delta initial condition, and the product of Vandermonde determinants ∆(x)∆(y). To emphasise the initial data we sometimes write M g n (t, y) instead of M n (t, y). We now state the main results regarding the solutions of (13) and (15) from which Theorem 1.1 follows as a corollary by (12). Let B b (R) be the collection of Borel measurable subsets of R with finite Lebesgue measure and let W = W t (A), t ≥ 0, A ∈ B b (R) be space-time white noise on a complete probability space (Ω, F , P) endowed with a right-continuous filtration . From now on we fix this filtered probability space (Ω, F , (F t ) t≥0 , P). We use E to denote the expectation with repect to P and for p ≥ 1, 1/p denotes the L p (Ω) norm. Throughout this paper, c p ≤ 2 √ p is the constant appearing in the Burkholder-Davis-Gundy inequality.
(a) Suppose that g is F 0 -measurable and symmetric and satisfies for all p ≥ 2, sup y∈R n g(y) p ≤ K p,g < ∞, then there exists a solution M n (t, y), (t, y) ∈ [0, ∞)×R n to the integral equation (15) that is unique (in the sense of versions) in the class of all random fields v(t, y), for a constant A > 0 depending on n. Moreover, M n has a version such that (t, y) → M n (t, y) is locally Hölder continuous on (0, ∞) × R n with indices α < 1/2 in space and α < 1/4 in time.
(b) There exists a unique solution M n (t, x, y) ∈ (0, ∞) × R n × R n given by the chaos expansion (11) to the integral equation (13) such that for all p ≥ 2 and t > 0 sup for some constant C n,p > 0. Moreover, M n has a version such that (t, x, y) → M n (t, x, y) is locally Hölder continuous on (0, ∞) × R n × R n with indices α < 1/2 in space and α < 1/4 in time.
Theorem 1.4. Let g be as in Theorem 1.3(a) with the additional property that g is non-negative almost surely and P[g(y) > 0 for some y ∈ R n ] = 1. Then the solution M g n to (15) satisfies P[M g n (t, y) > 0 for all t > 0 and y ∈ R n ] = 1. Let M n be the random field defined by (11) then P[M n (t, x, y) > 0 for all t > 0 and x, y ∈ R n ] = 1.
Comparing (13) and (15) with (3), we see that they have a similar form to the mild equation of the SHE which has been well studied. It has been shown for various initial data that the solution is Hölder continuous with indices up to 1/2 in space and up to 1/4 in time. For example, the case with a bounded initial data was studied by Walsh in [Wal86]. Bertini and Cancrini stated the Hölder continuity in [BC95] for a class of initial data which includes a delta function. More recently, Chen and Dalang [CD14a] proved the Hölder continuity for a non-linear SHE with initial data µ being a signed Borel measure over R such that (|µ| * p t )(x) < ∞ for all t > 0 and x ∈ R. For other variants of the SHE see for example [CJKS14], [Shi94], [SSS02] and the references therein.
In each case the tool used to prove the continuity of the solution is Kolmogorov's continuity criterion. Denote the stochastic integral term of (3) by I(t, y) then the key is to show that for p large enough. This in turn requires showing some continuity estimate for the heat kernel and in our case, estimates for the kernel Q t , see Theorem 3.1 below. These estimates get increasingly involved for increasingly less regular initial data due to the pth moments E[|u(t, y)| p ] of the solution being unbounded as t ↓ 0 or as y → ∞ or both. However for certain initial data such as a delta function, even though the pth moments blow up as time t ↓ 0, they are for any fixed positive times uniformly bounded in space and thus one can in effect isolate the effects of the initial data by solving the equation for a small time and then start afresh with the current solution as the new initial condition. This is the case with M n (t, x, y). We will show that for all positive times t, E[|M n (t, x, y)| p ] is bounded uniformly in space for all p which puts us in the situation of (15) with g having uniformly bounded pth moments for which continuity is easier to obtain.
The strict positivity of the solution to the stochastic heat equation was first proved by Mueller in [Mue91]. He showed that if the initial data f is non-negative, continuous with compact support with f (x) > 0 for some x ∈ R, then for all t > 0 P[u(t, x) > 0 for every x ∈ R] = 1.
Bertini and Cancrini proved a weak comparison principle using the Feynman-Kac formula and used it to extend Mueller's result to a delta type initial data. Shiga in [Shi94] proved the stronger statement P[u(t, x) > 0 for every x ∈ R and every t > 0] = 1, for initial data being continuous function such that the tails grow no faster than e λ|x| for all λ > 0. More recently, Moreno Flores in [Flo14] proved the strict positivity of the solution for delta initial conditions, using a convergence result of a discrete polymer model to the SHE, see [AKQ14b]. Chen and Kim [CK14] further generalised the strict positivity result to the fractional SHE, which includes as a special case the SHE considered here, for measure-valued initial data by adapting Shiga's method.
In all of the proofs above (except for the polymer proof) a key result is a large deviation estimate on the stochastic integral term of the solution. Mueller proved such result using the fact that integrals of the type t 0 R f (s, y) W (ds, dy) can be considered as a time-changed Brownian motion. Chen and Kim using a method of [CJK12] derived a similar estimate for the fractional SHE using Kolmogorov's continuity criterion. We will adapt the approach of [CK14] since we will first derive the necessary estimates in order to prove Hölder continuity anyway.
The outline of the paper is as follows. In Section 2.1 we first briefly recall integration with respect to space-time white noise and multiple stochastic integrals. In Section 2.2 we derive an upper bound on the L p (Ω) norm of stochastic integrals which will be used repeatedly in this paper and we discuss briefly non-intersecting Brownian bridges in Section 2.4. We then prove some estimates on the transition density Q t in Section 3 which are central to the proof of existence and continuity. The existence, uniqueness and moment estimates part of Theorem 1.3 will be proved in Section 4. The proof of Hölder continuity is in Section 5. Finally, in Section 6 we prove a strong comparison principle for the integral equation (15) of which Theorem 1.4 is a corollary.
Acknowledgements. C.H.L would like to thank Roger Tribe for numerous helpful discussions on SPDEs. The research of C.H.L was supported by EPSRC grant number EP/H023364/1 through the MASDOC DTC.

Preliminaries
2.1. White Noise and Stochastic Integration. In this section we briefly recall the Walsh stochastic integral with respect to white noise, see for example [Wal86], [Kho09] and [Dal99] for details. Let B b (R d ) be the collection of Borel measurable subsets of R d with finite Lebesgue measure. A white noise on R d is a mean zero Gaussian random where |·| denotes the Lebesgue measure on R d . We will only consider the case d = 2 and we interpret one of the dimensions as time. More precisely, we define a space-time white where X is bounded and F a -measurable and A ∈ B(R). A simple function is a finite linear combination of elementary functions. We say that a random field f is predictable if it is measurable with respect to the σ-algebra generated by the simple functions and we say that f ∈ P 2 if it is predictable and f ∈ L 2 (Ω × [0, ∞) × R). According to Walsh's theory, [Wal86], {W t (A)} belongs to a suitable class of integrators called worthy martingale measures and the integral is defined for all f ∈ P 2 . Now we turn our attention to multiple stochastic integrals which appear in the chaos series in the introduction. Let k > 1 and let f ∈ L 2 S ([0, t] k × R k ) such that f (πs, πy) = f (s, y) for all (s, y) ∈ [0, t] k × R k and π ∈ S k where S k is the set of permutations of {1, . . . , k} and πs = (s π1 , . . . , s πk ). Let A 1 , . . . , A k be disjoint subsets of [0, t] × R. An For such f we define the k-fold integral by It can be shown that linear combinations of functions of the form (18) are dense in L 2 S ([0, t] k × R k ) and that for an elementary f , the integral (f · W ) k satisfies an Itô isometry, hence for a general f ∈ L 2 The resulting integral is a mean zero random variable with covariance given by For f ∈ L 2 ([0, t] k × R k ) that are not symmetric, we define its integral by first symmetrising f viaf (s, y) := 1 k! π∈S k f (πs, πy), and then define Finally, for functions f defined on ∆ k (t) × R k , for example the k-point correlation function R k appearing in (1) and (8), we first extend it to a function on [0, t] k by setting it to be zero for s / ∈ ∆ k (t) and then define . We will need the following result for the proof of continuity in the initial data.
Proof. The result in the case when f is an elementary function of the form (18) follows from the definition of the integral and the definition ofW .
be a sequence of elementary functions converging to f . The result of the lemma holds for (f n · W ) k (t) for all n and by taking limits we see that the result also holds for (f · W ) k (t).
2.2. L p Bounds on Stochastic Integrals. The following estimate is a useful bound on the L p (Ω) norm of stochastic integrals; it can be considered as a version of [CK12, Lemma 2.2] or [FK09, Lemma 3.3] adapted to the present setting. Recall that for brevity we denote y ′ * = dy ′ 2 . . . dy ′ n and c p ≤ 2 √ p is the constant appearing in the Burkhoider-Davis-Gundy inequality.
for a suitable random field w and Γ t (y, y ′ ) is a non-random measurable function on Proof. Fix t and y, then by the Burkholder-Davis-Gundy inequality applied to the martingale Proof. Since multiple stochastic integrals on ∆ k (t) coincides with iterated stochastic integrals, applying Burkholder-Davis-Gundy inequality and Minkowski's integral inequality k times gives the desired upper bound.

Predictability of Random Fields.
Recall that the Walsh integral is defined for random fields in P 2 , see Section 2.1 above, therefore it is convenient to have a set of conditions to verify the predictability of a random field. The following result is from [CD14b, Proposition 3.1] which is an extension of [DF98, Proposition 2] to space-time white noise.
is a well-defined Walsh integral.
In the sequel we will need to integrate functions of the form: for some random field M , let f (s, y ′ 1 ) = R n−1 Q t−s (y, y ′ )M (s, y ′ ) dy ′ * . (Note that we have suppressed the dependency of f on t and y to keep the notation simple). The following proposition provides convenient conditions to verify the integrability of such a random field.
is a well-defined Walsh integral.
Proof. We will show that f satisfies the three assumptions of Proposition 2.4. Since Q t−s (y, y ′ ) is continuous and deterministic, Q t−s (y, y ′ )M (s, y ′ ) is adapted by (i) and so the integral R n−1 Q t−s (y, y ′ )M (s, y ′ ) dy ′ * is also adapted. Assumption (iii) of Proposition 2.4 follows from (iii) above since by Lemma 2.2 and Lemma 3.8 below, we have It remains to show the L 2 (Ω)-continuity of f . We wish to show that for each (s, y) Then by the Harish-Chandra formula (22) and equation (23) below, we have The last line is integrable with respect to dz * = dz 2 . . . dz n and so by the dominated convergence theorem, the continuity of Q t and assumption (ii), the right hand side of converges to zero as (u, z 1 ) → (s, y ′ 1 ). Finally, an application of Proposition 2.4 completes the proof. [Dys62] can be realised as the eigenvalues of Hermitian Brownian motion, an n × n Hermitian matrix whose entries are (up to the Hermitian condition) independent standard complex Brownian motions. The eigenvalues of such a matrix is a Markov process with state space W n with transition density Q t (x, y). It also arises as the Doob h-transform of Brownian motion killed at the boundary ∂W n with h(x) = ∆(x) (see for example [Gra99] and [KT07]).

Non-intersecting Brownian Motions. Dyson Brownian motion introduced in
One can construct bridges of Dyson Brownian motion, which we will call Dyson Brownian bridge or non-intersecting Brownian bridges, using the framework of [FPY93]. For x, y ∈ W n , a collection of non-intersecting Brownian bridges X t = (X 1 t , . . . , X n t ) starting at x at time 0 and ending at y in time t is a process whose law is absolutely continuous to that of Dyson Brownian motion started at x with Radon-Nikodym derivative equal In particular, for 0 < s 1 < . . . < s k < t, the law of (X s1 , . . . , X s k ) is given by the density The above is well defined at the boundary of the Weyl chamber by (14); in particular, taking limits as x → a1, y → b1 where 1 = (1, . . . , 1) one obtains Notice that the integrand above is symmetric in the permutation of its arguments (y l 1 , . . . , y l n ) for all 1 ≤ l ≤ k and so we can rewrite each integral over W n−1 as integrals over R n−1 multiplied by a factor of 1/n!. Moreover, by symmetry each term in the sum over i 1 , . . . , i k gives the same contribution. There are in total n k of such k-tuples and hence we can rewrite the correlation function R k (s 1 , y 1 1 ), . . . , (s k , y k 1 ); t, x, y := R k (s, y 1 ; t, x, y), y 1 = (y 1 1 , . . . , y k 1 ) as where A n := 1/(n − 1)!. For each k we have chosen to leave the first coordinate of y k and integrated out the rest but this choice is arbitrary by symmetry. Note that this is also the reason for the form of the stochastic integral term in (13). In the sequel we will need to bound integrals of the square of the k-point correlation function R k . Correlation functions of densities given by a product of determinants have been studied extensively in the context of determinantal point processes, see for example [Joh06] and [Bor11]. They can be expressed as a determinant of a matrix whose entries are given by some kernel function. However for general start and end points x and y this kernel function is difficult to compute, but since all we need is the integral of the square of R k it is not necessary to compute R k explicitly and so we will not pursue this. Instead, the next two results proved in [OW11] which expresses the integral of R 2 k in terms of intersection local times of Brownian bridges will be used. Let X = (X 1 , . . . , X n ) and Y = (Y 1 , . . . , Y n ) be two independent copies of a collection of n non-intersecting Brownian bridges which start at x at time 0 and end at y at time t and let E X,Y x,y;t denote the corresponding expectation of the joint law of the bridges. Let L t (X i − Y j ) be the local time at 0 of the difference X i − X j . Then we have Lemma 2.6. Fix n ≥ 1. For all integers k ≥ 1 and all t > 0, x, y ∈ W n the following holds

CONTINUITY AND STRICT POSITIVITY OF THE MULTI-LAYER EXTENSION OF THE SHE 12
The following is used to bound the above moments of local times.
Lemma 2.7. For all a ≥ 1 and t > 0, there exists a constant C > 0 such that The above two lemmata shows that for each t > 0, Z n (t, x, y) 2 < ∞ uniformly in x and y and thus the chaos series (1) is convergent in L 2 (Ω). The same is also true for (8). We point out here that the bound on the pth moments of M n (t, x, y) can in fact be written as The bound (17) in Theorem 1.3(b) then follows from the above by Lemma 2.7.

Estimates on Q t
From now on we drop the bold typeface for vectors in R n or W n since we will only be working with functions of multi-dimensional spatial variables so there is no longer any risk of confusion.
Before proving Theorem 1.3 we need estimates on various quantities involving the kernel Q t . The following known as the Harish-Chandra/Itzykson-Zuber formula [IZ80] provides a useful alternate expression for Q t : for Hermitian matrices X and Y with eigenvalues x 1 , . . . , x n and y 1 , . . . , y n respectively. c n = n−1 i=1 i! −1 and the integral is with respect to the normalised Haar measure on the unitary group U(n). Furthermore, the integrand above is bounded uniformly in U as the following bound from [MRTZ06, Lemma 1] shows As mentioned in the introduction, Q t (x, y) is well defined on the boundary of the Weyl chamber and since it is a product and ratio of determinants, it is permutation invariant and so we can extend Q t to a function on R n × R n by symmetry. Denote K t (x, y 1 ) := R n−1 Q t (x, y) n i=2 dy i and K := K 1 . The following result strongly indicates the continuity of M n ; in fact it is a key estimate in its proof in Section 5.
(a) There is a constant C 1 > 0 depending only on n such that for all t > 0 and x, z ∈ R n we have (b) for all α < 1/2 and T > 0 there are positive constants C 2 := C 2 (α, n, T ) and The theorem is a consequence of the series of results below. First observe that Q t has the following scaling property: The left hand side of the inequality in Theorem 3.1(a) is bounded above by where we have changed the integration region to [0, ∞) in the time integral which results in an upper bound due to the positivity of the integrand. The equality follows from the scaling property (24) and a change of variables. Theorem 3.1(a) follows from (25) and Lemma 3.2 below.
for any x, z ∈ R n , then Thus, we need to show that K(x, y) satisfies the hypothesis of Lemma 3.2. Using the inequality (a + b) 2 ≤ 2(a 2 + b 2 ), the left hand side of (26) with K in place of R, can be bounded by On the other hand, let r(ρ) : [0, 1] → R n , r(ρ) = (1 − ρ)x + ρz be a parameterisation of the straight line from x to z, then where the gradient is with respect to the first variable of K(·, ·) and u · v denotes the usual inner product of two vectors in R n . Then by Minkowski's integral inequality and Cauchy-Schwarz inequality we have Therefore, in order to verify the hypothesis of Lemma 3.2 we need to show that and sup x∈R n R |∇K(x, y)| 2 dy < ∞.
We first concentrate on (28). It suffices to show that for all j = 1, . . . , n. Clearly, Proposition 3.3. For each j = 1, . . . n, Proof. We first assume (and prove later) that we can differentiate under the integral sign, that is ∂K ∂x j (x, y 1 ) = By the Harish-Chandra formula (22), Q 1 (x, y) can be written as where D x , D y are diagonal matrices with the eigenvalues of X and Y as its entries respectively. The second equality follows from the first due to the invariance of Haar measure on U(n). Observe that by the cyclic property of the trace and the fact that U is unitary, Expanding the trace inside the exponential we have where c ′ n = (2π) −n/2 c n . For a Hermitian matrix H, one can check that Tr H 2 = n i=1 h 2 ii + 2 i<j |h ij | 2 and therefore Tr ( where C = 2 sup x∈R xe −x 2 = 2/e. Hence, We can make a standard change of variables to the space of n × n Hermitian matrices H(n) by the rule dY = Z n ∆(y) 2 dydU where Z n = c n π n(n−1)/2 and dY is the product of Lebesgue measures i≤j dy ij i<j dy ji . The right hand side of the previous display is then equal to 2 −n/2 π −n 2 /2 It remains to justify the swapping of the derivative and the integral in (30) and (31). For this we shall use the following result from [Bil95, Theorem 16.8].
Proposition 3.4. Let (Y, µ) be a measure space. Suppose that f (x, y) is a continuous and integrable function of y for each x ∈ I, where I can be taken to be R and that for each y ∈ Y , ∂f ∂x (x, y) exists. If for each x * there exists a function g(x * , y) integrable in y such that ∂f ∂x (x, y) ≤ g(x * , y) for all y and all x in some neighbourhood of x * , then Thus, we need to show that Q 1 (x, y) satisfies the hypothesis of the above proposition. Since the function x → p * n (t, x, y)/∆(x)∆(y) is smooth on R n , the same is true for Q t (x, y) so it remains to find a dominating function g.
Therefore, for such x j , we have by the bounds (32) and (23) that i =j e −(yi−xi) 2 /4 e −(yj−(x * j +h)) 2 /4 e (x * j +h) 2 /4 =: g(x * , y), and g is integrable over R n−1 with respect to y 2 , . . . , y n due to the Gaussian factor. Considering y 1 , x i , i = j fixed and applying Proposition 3.4 with the above g proves (30) and hence completes the proof.
Proposition 3.5. For all j = 1, . . . , n To prove this we shall use the following formula for the one point correlation function K. For 1 ≤ N ≤ n it was shown in [Joh01b, Proposition 2.3] that the N -point correlation function of Q t is given by a determinant: where γ is a closed contour around the x i 's and Γ L : t → L + it, t ∈ R with L ∈ R large enough so that γ and Γ L do not intersect. Then K(x, y) is simply (n−1)! n!K 1 (x, y, y). It is sometimes convenient to use the following alternate expression forK t , see the equation below (2.18) in [Joh01b]: with the same contours as before. Observe that the integral formulas (33) and (34) make clear the symmetry ofK t with respect to the ordering of x 1 , . . . , x n and that there are no issues if any of the x i 's coincide.
Lemma 3.6. For all x ∈ R n and y ∈ R Proof. Since the derivative with respect to x j of the integrand in the formula for K(x, y) is equal to The rest of the proof is devoted to justifying the exchange of integral and derivative. Consider a bounded set B in the complex plane and let x = (x 1 , . . . , x n ) with the x i 's all lie on the real line in B. Let γ be a closed contour containing B and therefore also contains x, then there exist constants d > 0, C > 0 such that for all z ∈ γ, |z − x i | ≥ d for all i and |z| ≤ C. Moreover, Therefore, for all x ∈ B there is a constant b n such that The function g is integrable along the contours γ and Γ L . Indeed, where in the last line we have shifted the contour Γ L to Γ y : t → y + it by Cauchy's theorem. The integral with respect to w is just a Gaussian integral and integrates to a constant. The other term is treated in a similar fashion but the dw integral is instead equal to for each fixed y ∈ R. Thus, by Proposition 3.4, we can differentiate under the integral to see that the derivative of K(x, y) is given by Finally, by Cauchy's theorem we can shift the contour Γ L so that L = 0 since there is no longer a pole at z = w.
Proof of Proposition 3.5. It is clear from the contour integral (35) that ∂K ∂xj (x, y) is translation invariant in the sense that ∂K ∂xj (x + h1, y + h) = ∂K ∂xj (x, y) for all h ∈ R. Hence, sup (x,y)∈R n ×R ∂K ∂xj (x, y) is equivalent to sup x∈R n ∂K ∂xj (x, 0) so we only need to bound the latter. Fix a constant d > 0. By Cauchy's theorem, we can take γ to be the closed (rectangular) contour around x 1 , . . . , x n composed of four parts γ t , γ b , γ r and γ l , where is chosen so that the minimum distance between the contour γ and the x i 's is at least d. We shall consider each parts of the contour separately. Denote the integral along the contour γ t by I(γ t ) and likewise for the others.
Since |z − x i | ≥ d for all i and z ∈ γ, we have by (36) that On γ r , |z| = |R + vi| = (R 2 + v 2 ) 1/2 ≤ (R 2 + d 2 ) 1/2 and where length(γ r ) = 2d and Due to the exponential term e −R 2 /2 we see that the two terms on the right hand side of (37) vanishes as R → ∞ and hence the same is true for I(γ r ). By symmetry, the same argument shows that I(γ l ) also vanishes as R → ∞. Thus, we can deform the contour γ to the two horizontal lines, γ + : u → −u + di and γ − : u → u − di, u ∈ R.
On γ + , |z| = (u 2 + d 2 ) 1/2 and |e −z 2 /2 | = |e −(−u+di) 2 /2 | ≤ e −u 2 /2 e d 2 /2 . Hence, in a similar fashion as above, we have 2π R d n−1 e −u 2 /2 + 2 (n−3)/2 (u n−1 + d n−1 )e −u 2 /2 du = 2 n−2 d n−1 (1 + 2 (n−3)/2 ) + 2 n−2 2 (n−3)/2 √ 2π R u n−1 e −u 2 /2 du, and the integral on the last line is equal to zero if n is even and equal to (n − 2)!! if n is odd. By symmetry, the same bound applies for I(γ − ) and hence we have shown that there exists a constant C depending only on n and d and is independent of x such that since Wn Q 1 (x, y) dy = 1 for all x. So it suffices to show that sup x,y K(x, y) is bounded or equivalently by the translation invariance of K which follows from the translation invariance of Q t that sup x∈R n K(x, 0) is bounded.
Proof. It is convenient to use the contour integral formula (34) instead. Notice that there is no longer a pole at w = z and so we can deform the contour Γ L so that L = 0. Let γ be the contour in the proof of Proposition 3.5 comprising of four parts, γ r , γ l , γ t and γ b . It can be shown in the same manner as in the proof of Proposition 3.5 that the contributions from γ r and γ l vanishes at infinity in the direction of the real axis and so we can deform the contour γ to the two horizontal lines, γ + : u → −u + di and γ − : u → u − di, u ∈ R for a fixed d > 0. We then have Denote the contribution from γ + by I j (γ + ), j = 1, 2 and likewise for γ − . Note that on γ + , |z| 2 = (u 2 + d 2 ), |e −z 2 /2 | ≤ e −u 2 /2 e d 2 /2 and |z − x j | ≥ d for all j and z ∈ γ + . Hence, by (36) we have in a similar manner to the proof of Proposition 3.5 that for some constant C d,n . By symmetry I 1 (γ − ) is bounded by the same constant. It remains to bound I 2 . Observe that Thus in the same way as above, both |I 2 (γ + )| and |I 2 (γ − )| are bounded by some constant C ′ d,n . Combining this with (40) and (41) shows that there exists a constant C independent of x and depending only on n and d such that Lemma 3.8. For all t > 0 and x ∈ R n there exists a constant C 4 > 0 depending only on n such that R K t (x, y) 2 dy ≤ C 4 t −1/2 .

Proof. By the scaling property of Q t and a change of variables
By Lemma 3.7 and (39), the latter integral for each fixed n is bounded uniformly in x which gives the desired result.
Proof of Theorem 3.1(b). Let t = u + h where h > 0, then we need to estimate Assume for now that one can differentiate under the integral in formula (33). The time derivative of K t is then equal to ∂ ∂r K r (x, y) where x ′ j = (x j − y)/ √ r and γ ′ , Γ ′ L are the contours γ, Γ L translated by y and scaled by 1/ √ r. We can rewrite the derivative as where in the last line we have shifted the contour Γ ′ L so that L = 0 as there is no longer a pole at w = z.
Note that |(w + z)e w 2 /4 e −z 2 /4 | is uniformly bounded on the chosen contours and as in the proof of Proposition 3.5, we can deform γ to the two horizontal contours γ + and γ − . Thus, there exists a C ′ := C ′ (n, d) such that Essentially the same calculation as for Proposition 3.5 shows that sup x∈R n |I(x)| < ∞ and together with an application of Lemma 3.7 gives for some constant C independent of x. Now, rewrite the integrand in (42) as |K s+h (x, y) − K s (x, y)| 2 = |K s+h (x, y) − K s (x, y)| 2−α |K s+h (x, y) − K s (x, y)| α for α < 1/2. We estimate the latter factor by s+h s ∂ ∂r K r (x, y) dr For the other term we have by time scaling Therefore, for u ≤ t ≤ T , the right hand side of (42) is bounded above by (the constant C := C(n, α) may change from line to line) It remains to justify the differentiation under the integral sign in ∂ ∂r K r (x, y). We once again appeal to Proposition 3.4, which means finding a dominating function g for the derivative. Let f (r; w, z) denote the integrand in (33) for u = v = y for fixed y ∈ R and x ∈ R n (note that we have suppresed the dependency on x and y in the notation), then differentiating with respect to r we have w − x j z − x j e (w−y) 2 /2r e −(z−y) 2 /2r (w + z − 2y) =: I 1 + I 2 Let z R * = sup z∈γ Re(z) and z I * = sup z∈γ Im(z). Fix r * > 0 then for all r ∈ [r * /2, 2r * ] and z ∈ γ, we have that |e −(z−y) 2 /2r | ≤ e −(z R * −y) 2 /4r * e (z I * ) 2 /r * , and for all w ∈ Γ L : t → L + it, we have Hence, for all r ∈ [r * /2, 2r * ] Let g(r * ; w, z) be the sum of the upper bounds of |I 1 | and |I 2 |, then it can be shown in a similar fashion as in Lemma 3.6 that g(r * ; w, z) is integrable on the contours γ and Γ L and so an application of Proposition 3.4 completes the argument.
Finally, by Lemma 3.8 we have This completes the whole proof of the theorem.

Existence, Uniqueness and Moment Estimates
4.1. Bounded Initial Data. We now prove the existence, uniqueness and moment estimates part of Theorem 1.3(a). The proof of continuity will be delayed to Section 5. In the sequel constants will generally be denoted by c, C or K and possibly adorned with primes, tildes, subscripts or superscripts. They may differ from line to line and their dependence if any will always be specified. However, C i , 1 ≤ i ≤ 4 will always mean the constants in Theorem 3.1 and Lemma 3.8. T ≥ 0 will always denote the finite time horizon.
Proof of existence, uniqueness and moment estimates of Theorem 1.3(a). The proof is by a Picard iteration argument. Throughout the proof, we fix an arbitrary integer p ≥ 2. For (t, y) ∈ (0, ∞) × R n define m 0 (t, y) := J n (t, y) where J n was defined in (15) and for k ≥ 1, let We first show that each of the stochastic integrals above are well defined, that is for all (t, y) ∈ (0, ∞) × R n , the random field f k (s, x), (s, x) ∈ (0, t) × R defined by f k (s, y ′ 1 ) := R n−1 Q t−s (y, y ′ )m k (s, y ′ ) dy ′ * is in P 2 for all k ≥ 0. Fix (t, y) ∈ (0, ∞) × R n and consider f 0 (s, y ′ 1 ) = R n−1 Q t−s (y, y ′ )m 0 (s, y ′ ) dy ′ * . We need to show that m 0 satisfies the three assumptions of Proposition 2.5. Since the initial data g is F 0 -measurable, m 0 is adapted to the filtration (F t ) t≥0 . By assumption on g, sup y∈R n g(y) p ≤ K p,g < ∞ and hence by Minkowski's integral inequality Therefore, m 0 (s, y) 2 p is bounded by K 2 p,g uniformly for (s, y) ∈ [0, ∞)×R n . By Lemma 5.2 below, (s, y ′ ) → m 0 (s, y ′ ) is continuous in L 2 (Ω) on (0, t) × R n and so Proposition 2.5 implies that f 0 ∈ P 2 and is a well-defined Walsh integral. Consequently, the random field m 1 (t, y) = m 0 (t, y) + I 1 (t, y), (t, y) ∈ (0, ∞) × R n is well defined. We wish to show that the sequence {m k (t, y)} k≥0 is Cauchy in L p (Ω). To this end, let d k (t, y) := m k+1 (t, y) − m k (t, y) p . By Lemma 2.2, Lemma 3.8 and (44), we have for all (t, y) ∈ (0, ∞) × R n , where C 4 is the constant in Lemma 3.8 and Γ(3/2) = √ π/2. Now assume that for all 0 ≤ l ≤ k, m l (t, y), (t, y) ∈ (0, ∞) × R n is well defined and satisfies We want to show that the same is true for m k+1 and d k . 1 (t, y), and so to bound the pth moments of m k it suffices to bound each of the d l 's, 0 ≤ l ≤ k − 1. Indeed, by property (iii) and (44), we have which shows that sup (s,y)∈[0,t]×R n m k (s, y) 2 < ∞. This and the induction hypothesis shows that m k satisfies all three assumptions of Proposition 2.5 and so f k ∈ P 2 and is a well-defined Walsh integral for all (t, y) ∈ (0, ∞) × R n . Moreover, it is adapted and so m k+1 = m 0 + I k+1 is also adapted. We need to check the L 2 (Ω)-continuity of I k+1 . Fix α < 1/2, then for all 0 < r < u ≤ t and y, z ∈ R n by Theorem 3.1 which proves the L 2 (Ω)-continuity of m k+1 on (0, t) × R n .
For the bound on d k , we use Lemmata 2.2 and 3.8 and the induction hypothesis to obtain where we have used the Euler Beta integral [OLBC10, equation 5.12.1]: and the fact that Γ(1/2) = √ π to evaluate the time integral. It follows that the bound (45) holds with k replaced with k + 1 and that sup (s,y)∈[0,t]×R n m k+1 (s, y) 2 < ∞. Hence, m k+1 satisfies all the assumptions of Proposition 2.5 and therefore f k+1 ∈ P 2 .
Thus, by induction we conclude that for all integers k, the random field m k (t, y) = m 0 (t, y) + I k (t, y), (t, y) ∈ (0, ∞) × R n is well defined and satisfies properties (i), (ii) and (iii) listed above.
We now show that the sequence {m k (t, y)} k≥0 is Cauchy in L p (Ω). This follows from the fact that for any T > 0 which is a consequence of property (iii), the ratio test and the following asymptotic: [OLBC10,equation 5.11.12]. We conclude that there exist a random field which we denote by M n (t, y) such that m k (t, y) → M n (t, y) as k → ∞ in L p (Ω) and almost surely for a subsequence uniformly in y ∈ R n and t ∈ [0.T ].
Since each m k is adapted, M n is also adapted. The L 2 (Ω)-continuity of M n is inherited from that of m k since the convergence is uniform on [0, T ] × R n for all T > 0. Now take k → ∞ on both sides of (45). By [CD14b, Proposition 2.2], we know that for all (48) Using this with x = 2C 4 A 2 n c 2 p √ πt 1/2 gives the bound (16) in the statement of the theorem. Thus, by Proposition 2.5, for all (t, y) ∈ (0, ∞) × R n the random field f defined by f (s, y ′ 1 ) = R n−1 Q t−s (y, y ′ )M n (s, y ′ ) dy ′ * for (s, y ′ 1 ) ∈ (0, t) × R is in P 2 and the stochastic integral is well defined. It remains to show that the limit M n (t, y) solves (15). Fix (t, y) ∈ (0, ∞) × R n . By definition, m k (t, y) = m 0 (t, y) + I k (t, y) where the left hand side converges in L p (Ω) and almost surely for a subsequence to M n (t, y). For the right hand side we have by the uniform convergence L p (Ω) of m k that Therefore, we have L p (Ω) convergence of I k (t, y) to I n (t, y) and hence almost sure convergence for a subsequence to the same limit. The limit of both sides of m k (t, y) = m 0 (t, y) + I k (t, y) must be equal almost surely and so we have shown that for all (t, y) ∈ (0, ∞) × R n , M n (t, y) satisfies (15) almost surely. This proves existence. For uniqueness, suppose that M (t, y) and N (t, y) are both solutions to (15) with the same initial data g and let d(t, y) = M (t, y) − N (t, y) p then by a similar calculation as for existence we have which converges to 0 as n → ∞ since the expression on the right hand side is summable in n. Therefore, d ≡ 0 and so for all (t, y), M (t, y) = N (t, y) almost surely i.e. M and N are versions of each other. This proves uniqueness.

Delta Initial Data.
Proof of existence, uniqueness and moment estimates of Theorem 1.3(b). Fix an integer p ≥ 2. We first show that if solutions to (13) exists then it must be unique. Suppose M (t, x, y) and N (t, x, y) are two solutions to (13) and let d(t, x, y) = M (t, x, y) − N (t, x, y) p . By linearity of the equation (13), M (t, x, y) − N (t, x, y) is a solution to (15) with zero initial condition i.e. M (t, x, y) − N (t, x, y) = M g n (t, y) with g ≡ 0. Then by (16), sup x,y∈R n d(t, x, y) 2 is a bounded function of t ∈ [0, T ] for any T > 0. The bound (49) applies to d(t, x, y) 2 which shows that M (t, x, y) = N (t, x, y) almost surely for all (t, x, y). This proves uniqueness.
We now prove existence. We shall show that M n (t, x, y) defined by (11) satisfies equation (13) for all (t, x, y) ∈ (0, ∞) × R n × R n . Recall that M n (t, x, y) is well defined on the boundary of the Weyl chamber and it is symmetric under permutations of both its space variables, hence we can extend it to a function on R n × R n . Similarly we also extend Q t−s (x, y) to the whole of R n × R n . Substituting the chaos expansion of M n into the stochastic integral term of (13), using the expression for the correlation function R k (20) and the stochastic Fubini's theorem [Kho09,Theorem 5.30], we have bearing in mind that we can interchange the summation and integral because the series is convergent in L 2 (Ω) that R n Q t−s1 (y, y 1 )M n (s 1 , x, y 1 ) dy 1 * W (ds 1 , dy 1 1 ) = A n t 0 R n Q t−s1 (y, y 1 )p * n (s 1 , x, y 1 ) ∆(x)∆(y 1 ) dy 1 * W (ds 1 , dy 1 1 ) R 1 (s 1 , y 1 1 ; t, x, y) W (ds 1 , dy 1 1 ) where the last equality follows by a relabelling of the indices. Thus, the right hand side of (13) after the substitution is equal to which is the definition of M n (t, x, y) as required. It remains to estimate the pth moments of M n (t, x, y). The approach is to construct an approximating sequence to M n and estimate the moments of each term of the sequence and take limits. The natural candidate for the approximating sequence is the following: for each (t, x, y) ∈ (0, ∞) × R n × R n , let m 0 (t, x, y) := J n (t, x, y) where J n was defined in (13) and for k ≥ 1 define In other words, m k (t, x, y) is the kth partial sum of the chaos expansion for M n (t, x, y). Let d k−1 (t, x, y) := m k (t, x, y) − m k−1 (t, x, y) for k ≥ 1, then clearly It is easy to see that m k (t, x, y) = m 0 (t, x, y) + Each term in the sum above is equal to (2c 2 p ) l E X,Y x,y;t n i,j=1 L t (X i −Y j ) l /l! by Lemma 2.6 where X = (X 1 , . . . , X n ), Y = (Y 1 , . . . , Y n ) are independent copies of a collection of n non-intersecting Brownian bridges which start at x in time 0 and end at y in time t. Letting k → ∞ we have for all (t, x, y) For each t > 0, Lemma 2.7 shows that the right hand side of the above is bounded uniformly in x, y ∈ R n for any p ≥ 2. By Cauchy-Schwarz inequality x, y) p−1 2(p−1) , which converges to 0 as k, k ′ → ∞ by the L 2 (Ω) convergence of m k and the moment bound (51). Therefore, m k (t, x, y) also converges to M n (t, x, y) in L p (Ω) and we can replace the left hand side of (51) with M n (t, x, y) 2 p . This completes the proof of existence, uniqueness and moment estimates.

Continuity
We shall use the following version of Kolmogorov's continuity criterion which is due to Chen and Dalang, see [CD14a,Proposition 4.2].
Theorem 5.1. Consider a random field {f (t, y) : (t, y) ∈ R + × R d }. Suppose there are constants α 0 , . . . , α d ∈ (0, 1] such that for all p > 2(d + 1) and all M > 1, there exist a constant C := C(p, M ) depending on p and M such that 5.1. Bounded Initial Data. We now prove the Hölder continuity of the solution to (15) by verifying the assumptions of Kolmogorov's continuity criterion. We first estimate the increments of J n (t, y) = 1 n! R n g(y ′ )Q t (y, y ′ ) dy ′ where g satisfies the bound sup y∈R n g(y) p ≤ K p,g .
Proof. By the assumptions on g and Minkowsky integral inequality, we have For t ≥ 1/M , Q t has bounded derivatives in both time and space and the result follows by a direct calculation.
We now turn our attention to the stochastic integral term I n (t, y). Then by Lemma 2.2 and Theorem 3.1(a) For the temporal increment we have two terms (assuming without loss of generality that 0 ≤ u < t ≤ M ) I n (t, y) − I n (u, y) 2 p ≤ 2I + 2II, where by Theorem 3.1(b), for any α < 1/2 there exists a C 2 such that By the subadditivity of the function x → |x| β , for β ∈ (0, 1] we have Lemma 5.2 and Proposition 5.3 together shows that for all M > 1, α < 1/2 and p ≥ 2, there is a constant C := C(α, g, M, n, p) such that for all (t, y) and (t ′ , y ′ ) in Taking p large enough and applying Theorem 5.1 shows that M n has a version that is locally Hölder continuous on (0, ∞) × R n with indices up to 1/4 in time and up to 1/2 in space.
5.2. Delta Initial Data. We now turn our attention to M n (t, x, y). Observe that in this case we cannot apply the method used in Proposition 5.3 directly since the pth moments of M n (t, x, y) are not bounded uniformly in time, for instance if x = y then which converges to infinity as t ↓ 0. However, for any t > 0 fixed we have by (51) and Lemma 2.7 that there is a constant C := C(n, p) such that uniformly for x, y ∈ R n . Thus, for all positive times, M n belongs to the class of initial data in Theorem 1.3(a). It is clear that at any given time we can restart the equation taking the current solution as the new initial data. More precisely, let τ > 0 and consider the shifted white noiseẆ τ (s, y) :=Ẇ (τ + s, y). Define M τ n (t, x, y) := M n (τ + t, x, y) then it is easy to check by using the semigroup property of Q t that M τ n satisfies the integral equation In other words, M τ n is the solution to (15) driven by the shifted noiseẆ τ with initial condition M τ n (0, x, y) = M n (τ, x, y). Now definê Clearly,M n (t, x, y) solves (13) and by uniqueness, M τ n is a modification of the chaos series (11). Let M > 1, α < 1/2 and p ≥ 2 then since sup x,y∈R n M n (τ, x, y) p < ∞, Lemma 5.2 and Proposition 5.3 applies to show that there is a constant C := C(α, M, n, p, τ ) such that for all t, t ′ ∈ [τ, M ] and y, y ′ ∈ [−M, M ] n and x ∈ R n M τ n (t, x, y) − M τ n (t ′ , x, y ′ ) p ≤ C |t − t ′ | α/2 + |y − y ′ | 1/2 .
5.2.1. Continuity in the Initial Condition. We study the continuity of x → M n (t, x, y); in fact we show that (t, x, y) → M n (t, x, y) is jointly continuous. Recall the chaos expansion of M n (t, x, y): where for 0 < s 1 < . . . < s k < t, y = (y 1 1 , y 2 1 , . . . , y k 1 ) R k (s, y; t, x, y) It is easy to see that J n (t, x, y) = J n (t, y, x) and from the expression of R k one can see that for all k ≥ 1 R k (s, y; t, x, y) = R k (t − s,ỹ; t, y, x), where t−s := (t−s k , . . . , t−s 1 ), 0 < t−s k < . . . < t−s 1 < t andỹ := (y k 1 , y k−1 1 , . . . , y 1 1 ). Therefore, it is reasonable to think that each term in the sum above is symmetric in x and y provided one can reverse time in the multiple stochastic integral. This motivates the following proposition.
Proof. Fix k ≥ 1 and (t, x, y) ∈ (0, ∞) × R n × R n . Recall the time reversed white noisẽ W defined byW ([0, s] × A) =Ẇ ([t − s, t] × A) for s ≤ t and A ∈ B b (R). Extend R k (s, z; t, x, y) to a function on L 2 ([0, t] k × R k ) by setting it to be zero for s / ∈ ∆ k (t). LetR k be the symmetrisation of R k given bỹ R k (s, y; t, x, y) = 1 k! π∈S k R k (πs, πy; t, x, y), where πs = (s π(1) , . . . , s π(k) ) and likewise for πy. Clearly, we haveR k (s,ỹ; t, x, y) = R k (s, y; t, x, y). Therefore by Lemma 2.1 and (54), (recall the definition of the multiple stochastic integral in Section 2.1) Thus, applying the above to each term of the sum in (53) we see that for all (t, x, y) ∈ (0, ∞) × R n × R n and the result follows.
Finally, we return to proving the joint continuity of the solution to (13). We bound M n (t, x, y)−M n (t ′ , x ′ , y ′ ) 2 p by considering the increments in each variables separately. SinceM n (t, x, y) = M τ n (t − τ, x, y) for t ≥ 2τ , we have by Proposition 5.4 and (52) that for all M > 1, p ≥ 2 and α < 1/2 there is a constant C := C(α, M, n, p, τ ) such that for all (t, x, y) and (t ′ , Since τ > 0 is arbitrary, we can take 2τ = 1/M and thus we have shown that there exists a constantC =C(α, M, n, p) such that for all (t, x, y) and (t ′ , x ′ , y ′ ) ∈ [1/M, M ] × [−M, M ] 2n the above inequality holds withC in place of C. Finally, using the subadditivity of x → |x| β for β ∈ (0, 1] and applying Theorem 5.1 proves the existence of a Hölder continuous version. This concludes the entire proof of Theorem 1.3. 6. Strict Positivity 6.1. A Weak Comparision Principle. Recall that K n (t, x, y) can be expressed as K n (t, x, y) = det[u(t, x i , y j )] n i,j=1 where u(t, x, y) is the solution to (2) with initial data δ x . Bertini-Cancrini [BC95] proved that u(t, x, y) is the limit in L p (Ω) for all p ≥ 2 of u ε (t, x, y) as ε → ∞, where u ε (t, x, y) is the solution to the stochastic heat equation subject to a mollified white noise W ε in place of the space-time white noise. Its solution is given by the following Feymann-Kac formula which is well defined for the noise W ε : where the expectation is with respect to a Brownian bridge b starting from x at time 0 and ending in y at time t. By the above Feymann-Kac formula it is then clear that for all (t, x, y) ∈ (0, ∞) × R × R, with probability 1, u(t, x, y) ≥ 0. Using this and the determinant formula for K n , the authors in [OW11, Proposition 5.5] proved by a path switching argument that K n (t, x, y) ≥ 0 almost surely, for all (t, x, y) ∈ (0, ∞)×W n ×W n . In fact, a stronger result is true since the above implies that K n (t, x, y) ≥ 0 for all rational points (t, x, y) almost surely. It is well known that (t, x, y) → u(t, x, y) has a jointly continuous version and hence the same is true for K n as it is just a sum of products of the u's. Therefore, by continuity P[K n (t, x, y) ≥ 0 for all t > 0 and x, y ∈ W n ] = 1.
Since the Vandermonde determinant is non-negative on W n , we see that the same is true for M n in the interior W • n . By the continuity of M n proved in the previous section, this non-negativity extends to the boundary of the Weyl chamber and by symmetry to the whole of R n . That is, P[M n (t, x, y) ≥ 0 for all t > 0 and x, y ∈ R n ] = 1.
By the linearity of the equation (13), the non-negativity property above is equivalent to a weak comparison principle. The next lemma extends this to solutions M g n (t, y) of equation (15) with initial data g.
Lemma 6.1 (Weak comparison principle). Let M 1 n (t, y) and M 2 n (t, y) be the solution to (15) with symmetric initial data g 1 and g 2 respectively. If g 1 ≥ g 2 , then P[M 1 n (t, y) ≥ M 2 n (t, y) for all t > 0 and y ∈ R n ] = 1.
Proof. By linearity of the equation (15), it suffices to prove the lemma in the case g ≥ 0.
A direct calculation shows that v g satisfies (15) and so by uniqueness v g (t, y) = M g n (t, y) almost surely for all (t, y) ∈ [0, ∞) × R n . Now by (55) and the non-negativity of g and the Vandermonde determinant it is clear that for all (t, y) ∈ [0, ∞) × R n , v g (t, y) ≥ 0 almost surely. This and the continuity of (t, y) → M g n (t, y) shows that P[M g n (t, y) ≥ 0 for all t ≥ 0 and y ∈ R n ] = 1 as required.
6.2. A Strong Comparison Principle. We now prove a strong comparision principle of which Theorem 1.4 is an easy corollary.
(a) Let M 1 n (t, y) and M 2 n (t, y) be two solutions to (15) with initial data g 1 and g 2 respectively where g 1 and g 2 are as in Theorem 1.3(a). If furthermore g 1 ≥ g 2 and g 1 (y) > g 2 (y) for some y ∈ R n almost surely, then n (t, y) > M 2 n (t, y) for all t > 0 and y ∈ R n ] = 1. (b) Let M n (t, x, y) be the solution to (13), then P[M n (t, x, y) > 0 for all t > 0 and x, y ∈ R n ] = 1.
We begin with a lemma which provides a lower bound for the deterministic term J n (t, y) in (15). Proof. Since Dyson Brownian motion is realised as the eigenvalues of Brownian motion on the space of n × n Hermitian matrices H(n), we have that where P s (A − B) = 2 −n/2 (πs) −n 2 /2 e −Tr(A−B) 2 /2s for A, B ∈ H(n) is the transition density of Brownian motion on the space of Hermitian matrices and φ : H(n) → W n is such that φ(Y ) = y = (y 1 , . . . , y n ) = (φ 1 (Y ), . . . , φ n (Y )) is the vector of ordered eigenvalues of Y . D x is a diagonal matrix with entries x = (x 1 , . . . , x n ). Weyl's eigenvalue inequality [Bha97, Theorem III.2.1] implies that for two Hermitian matrices A, B with eigenvalues φ i (A) and φ i (B), 1 ≤ i ≤ n respectively, the following hold φ 1 (A + B) ≤ φ 1 (A) + φ(B) and φ n (A) + φ n (B) ≤ φ n (A + B). Therefore

and hence
Wn Q s (x, y)1 (−h,h) n (y) dy Let β > 0 be the constant in the statement of the lemma then for −h−M/m ≤ x i ≤ 0, 1 ≤ i ≤ n and t/2m ≤ s ≤ t/m, we have Wn Q s (x, y)1 (−h,h) n (y) dy Similarly, for 0 ≤ x i ≤ h + M/m, 1 ≤ i ≤ n and s in the same range as above, we have Wn Q s (x, y)1 (−h,h) n (y) dy 1{φ i (Y ) ∈ (−h(m/t) 1/2 , − √ 2M (tm) −1/2 )} dY. (57) Taking m large enough and noting that P 1 (Y ) is the probability density of a GUE matrix Y , we see that both (56) and (57) can be made greater than β and hence completes the proof.
We are now ready to prove the main result of this section.
Proof of Theorem 6.2. By linearity M 1 n − M 2 n is the solution to (15) with initial data g 1 − g 2 and so it suffices to prove that P[M g n (t, y) > 0 for all t > 0 and y ∈ R n ] = 1, for g such that g ≥ 0 and g(y) > 0 for some y ∈ R n almost surely.
We first consider the case when g is a continuous function such that g ≥ 0 and g(y) > 0 for some y ∈ R n so that one can find constants c > 0, d > 0 small enough such that g(x) ≥ c n i=1 1 (yi−d,yi+d) (x) for all x ∈ R n . Without loss of generality, we can assume c = 1 and take y to be the origin for convenience. By the weak comparision principle (Lemma 6.1), we can therefore assume that the initial data is g(·) = 1 (−d,d) n (·). From now on we drop the superscript g and just write M n (t, y). We consider first the sets A k . By Lemma 6.4, there is an m 0 such that for all m ≥ m 0 there is a δ(m) such that P[A 1 ] ≥ 1 − δ(m).
Now assume that A 1 ∩· · ·∩A k−1 occurs. On the event A k−1 we have M n ((k−1)t/m, y) ≥ γ k−1 1 (−d−M(k−1)/m,d+M(k−1)/m) n (y) for all y ∈ R n almost surely. Define a time shifted white noise byẆ k (s, y) =Ẇ ((k − 1)t/m + s, y). Let M k n (s, y) be the solution driven by the noiseẆ k with initial data given by γ k−1 1 (−d−M(k−1)/m,d+M(k−1)/m) n (y). On the event A k−1 , by the weak comparision principle, M n ((k − 1)t/m + s, y) ≥ M k n (s, y) for all s ≥ 0 and y ∈ R n almost surely. It is easy to see thatM k n (s, y) := γ −(k−1) M k n (s, y) is the solution to (15) with initial data 1 (−d−M(k−1)/m,d+M(k−1)/m) n (y). Lemma 6.4 applied toM k n with h = d + M (k − 1)/m shows that with the same m 0 and δ(·) as above that for all m ≥ m 0 , t m and y ∈ R n ≥ 1 − δ(m).

CONTINUITY AND STRICT POSITIVITY OF THE MULTI-LAYER EXTENSION OF THE SHE 36
Finally, by the union bound Since (1 − δ(m)) m → 1 as m → ∞, we conclude that Since t > 0 and M > 0 are arbitrary, this completes the proof in the case when the initial data g is a continuous function. We now prove the result for g satisfying the assumptions in Theorem 6.2(a). The idea is that after a small time τ > 0, we are back in the situation above. We shall prove that for all τ > 0, P[M n (t, y) > 0 for all t > τ and y ∈ R n ] = 1.
and since τ is arbitrary this would imply the desired result. LetẆ τ (s, y) =Ẇ (τ + s, y) be the time shifted white noise and let M τ n be the solution to (15) driven by the noiseẆ τ and with initial data M n (τ, ·). The weak comparison principle shows that P[M n (t, y) ≥ 0 for all t ≥ 0 and y ∈ R n ] = 1. We claim that P[M n (τ, y) > 0 for some y] = 1 then since y → M n (τ, y) is continuous, the strong comparison principle for continuous initial data proved above applied to the solution M τ n shows that P[M τ n (s, y) > 0 for all s > 0 all y ∈ R n ] = 1 which proves (62).
Therefore, it remains to prove the claim. Suppose the opposite is true, that is P[M n (τ, y) = 0 for all y] > 0 and consider the solution M n (s, ·) at time s ≤ τ . If M n (s, y) > 0 for some y almost surely then the strong comparison principle for continuous initial data applies to show that M n (τ, y) > 0 for all y almost surely. Hence, P[M n (s, y) = 0 for all y] > 0 for all 0 ≤ s ≤ τ which implies that M n (0, ·) ≡ 0 with strictly positive probability which is a contradiction. Thus, we must have that P[M n (τ, y) = 0 for all y] = 0 which proves the claim.
We now prove part (b) of the theorem; the everywhere strict positivity of M n (t, x, y). Fix τ > 0 then the same argument as above together with Proposition 5.4 shows that P[M n (τ, x, 0) > 0 for all x] = 1. By the joint continuity of M n , there exist random c = c(x) and d = d(x) strictly positive such that M n (τ, x, y) ≥ c1 (−d,d) n (y) for all x, y ∈ R n almost surely. For N ≥ 1 define the random set B N := {x ∈ R n : c(x) ≥ 1/N, d(x) ≥ 1/N }. Then M n (τ, x, y) ≥ (1/N )1 (−1/N,1/N ) n (y) for all y and all x ∈ B N almost surely. The strict positivity result proved above applied to the solution with initial data (1/N )1 (−1/N,1/N ) n (y) together with the weak comparision principle implies that