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The Harnack inequality and the Jordan-Kinderlehrer-Otto scheme

We establish a version of the Harnack inequality for the Jordan-Kinderlehrer-Otto scheme of the heat equation on the flat torus.


Introduction
The Harnack inequality is one the most important inequalities for elliptic and parabolic equations. In the linear elliptic case, the inequality was first proved in [21]. It was extended to more general elliptic and parabolic equations in [16,17] (see also [18,9]). In this paper, we prove a version of the Harnack inequality for the Jordan-Kinderlehrer-Otto (JKO) scheme of the heat equation on the flat torus.
The JKO scheme is a time discrete scheme introduced in [12] using the theory of optimal transportation. It produces time discrete approximations to various gradient flows on the Wasserstein space, the space of probability measures. For the infinite dimensional geometry behind this gradient flow, see [19,1]. Since its discovery, the JKO scheme has been an intensive area of research. Here is a list which is far from exhaustive [1,2,3,5,7,10] (see also [22] for an overview).
In what follows, we will discuss the detail construction of this scheme in the case of the heat equation in order to give a precise statement of the main result.
Let µ 0 and µ 1 be two Borel probability measures on the flat T n and let d be the distance function on T n . The theory of optimal transportation begins with the following problem which searches for a map ϕ transporting the first mass µ 0 to the second one µ 1 with the minimal amount of total cost: where the infimum is taken over all Borel maps ϕ : T n → T n pushing µ 0 forward to µ 1 .
The above problem defines a distance function d, called L 2 Wasserstein distance, on the Wasserstein space. More precisely, Assuming the first measure is absolutely continuous with respect to the Lebesgue measure, the existence and uniqueness of minimizer to the above problem (1.1) was established in [4,6,15]. This minimizer is called the optimal map that pushes µ 0 forward to µ 1 .
For each fixed number K > 0, each positive integer N, and each probability measure ρ 0 dx n , we consider the following family of minimization problems parametrized by k = 1, ..., N: where ρ N 0 = ρ 0 and the infimum is taken over the set of L 1 functions ρ : T n → [0, ∞) satisfying ρdµ = 1.
Assuming that the function ρ 0 is in L 1 , the existence and uniqueness of minimizer to the problem (1.2) was established in [12]. Moreover, if K and ρ 0 are fixed, the initial condition ρ 0 is in C 2,α , and N is large enough, then it was shown in [13] is an optimal map pushing ρ N k−1 dx n forward to ρ N k dx n . It also satisfies the following Monge-Ampere type equation The above minimization problem (1.2) gives a sequence of functions It was shown in [12] that u N t converges in L 1 to the solution of the heat equation with the initial condition u 0 = ρ 0 as N goes to infinity. This discretization scheme is the so-called Jordan-Kinderlehrer-Otto (JKO) scheme. It was also shown in [13] that there is a uniform C 1 bound for the family Γ and so the convergence is improved to C 0,α in space (the C 0 bound was also observed in [20]). In this paper, we establish a uniform lower bound for the second derivatives of the elements in Γ. More precisely, There is a constant 1 ≥ C ≥ 1 2 and, for each fixed K > 0 and each ρ 0 , there is an integer N 0 > 0 such that By letting N goes to ∞ in (1.4), we recover the matrix differential Harnack inequality proved in [11] for the heat equation. This matrix differential Harnack inequality is the matrix analogue of a scalar version proved in [14]. They are called differential Harnack inequalities because one can recover the Harnack inequality by integrating the differential ones along geodesics. Theorem 1.1 shows that the matrix differential Harnack inequality holds at the level of the JKO scheme. The following is the Harnack inequality for the JKO scheme mentioned at the beginning of the introduction.
Theorem 1.2. For each fixed K > 0 and each ρ 0 , there is an integer N 0 > 0 such that By letting N goes to ∞, we recover a version of the Harnack inequality for the heat equation.
The rest of the paper is devoted to the proof of Theorem 1.1 and 1.2.
2. Proof of Theorem 1. 1 By definition, the function f N k is locally semi-concave. In particular, it is differentiable Lebesgue almost everywhere and the map φ N k (x) := for Lebesgue almost all x and for all x and y in T n .
Therefore, for Lebesgue almost all x, the function y → 1 2 d 2 (x, y) + K N log ρ N k (y) achieves its minimum at φ N k (x). Therefore, Therefore, by (1.3) and the uniform bound of ρ N k , dϕ N k ≥ cI > 0, where c is independent of k and N. It also follows that x → x + tK N ∇ log ρ N k (x) =: ψ N k (t, x) is a diffeomorphism for each t in [0, 1]. Hence, by the method of characteristics (see [8]), the Hamilton-Jacobi equationġ N k + 1 2 |∇g N k | 2 = 0 with initial condition g N k (0, x) = K N log ρ N k (x) has a smooth solution. Moreover, ψ N k is the flow of the vector field ∇g N k . It follows that x) is length minimising between its endpoints x and ϕ N k (x). It follows from integrating (2.6) that The last equality follows from (2.1) and (2.4).
On the other hand, sinceψ N k = ∇g N k (ψ N k ), it follows that Therefore, by (2.7), (2.8) By [15,Theorem 11], φ N k is the optimal map pushing forward ρ N k−1 dx n to ρ N k dx n and so it satisfies By (2.4), f N k is in C 4,α . By differentiating the above equation twice, we obtain By using the same arguments in the proof of (2.5), we obtain I − , ξ achieves its minimum λ N k among all points in T n × S n , then where a N k = inf (x,ξ)∈T n ×S n ∇ 2 log ρ N k (ξ, ξ). It follows from this and (2.8) that for any unit vector v.
Therefore, by letting (x, v) be a minimizer of ∇ 2 log ρ N k (ξ, ξ) in T n × S n , we obtain We claim that there is an integer N 0 > 0 depending on ρ 0 and a constant 1 2 ≤ C ≤ 1 such that The constant C will be chosen such that (1−C) 2 2C−1 ≤ 1. The inequality (2.10) is clearly true for k = 0 if we choose N 0 large enough. Assume that a N k−1 ≥ − C K N k , we show (2.10) holds. Indeed, we have By (2.5), 1 + K N a N k > 0. Therefore, the above equation is equivalent to k .
An elementary calculation shows that Since we chose C close enough to 1, the above inequality holds for all k ≥ 1. This proves the claim and therefore the theorem.

Proof of Theorem 1.2
It follows from (2.8) and Theorem 1.1 that

It follows from this and (1.3) that
On the other hand, by (2.1) and (2.2), By combining this with (3.1), we obtain It follows from this and (3.2) that Therefore, if t 1 and t 2 are contained in (k 1 −1)K N , k 1 K N and k 2 K N , (k 2 +1)K N , respectively, then