Elementary proof of logarithmic Sobolev inequalities for Gaussian convolutions on $\mathbb{R}$

In a 2013 paper, the author showed that the convolution of a compactly supported measure on the real line with a Gaussian measure satisfies a logarithmic Sobolev inequality (LSI). In a 2014 paper, the author gave bounds for the optimal constants in these LSIs. In this paper, we give a simpler, elementary proof of this result.


Introduction
A probability measure µ on R n is said to satisfy a logarithmic Sobolev inequality (LSI) with constant c ∈ R if Ent µ (f 2 ) ≤ c E (f, f ) for all locally Lipschitz functions f : R n → R + , where Ent µ , called the entropy functional, is defined as and E (f, f ), the energy of f , is defined as with |∇f | defined as |∇f |(x) := lim sup y→x |f (x) − f (y)| |x − y| so that |∇f | is defined everywhere and coincides with the usual notion of gradient where f is differentiable. The smallest c for which a LSI with constant c holds is called the optimal log-Sobolev constant for µ.
LSIs are a useful tool that have been applied in various areas of mathematics, cf. [1,2,4,5,7,8,9,10,11,12,13,14,15,16,18,19,20]. In [21], the present author showed that the convolution of a compactly supported measure on R with a Gaussian measure satisfies a LSI, and an application of this fact to random matrix theory was given. In [22,Thms. 2,3], bounds for the optimal constants in these LSIs were given, and the results were extended to R n . Those results are stated as Theorems 1 and 2 below. (See [17] for statements about LSIs for convolutions with more general measures). Theorem 1. Let µ be a probability measure on R whose support is contained in an interval of length 2R, and let γ δ be the centered Gaussian of variance δ > 0, i.e., dγ δ (t) = (2πδ) −1/2 exp(− t 2 2δ )dt. Then for some absolute constants K i , the optimal log-Sobolev constant c(δ) for µ * γ δ satisfies In particular, if δ ≤ R 2 , then The K i can be taken in the above inequalities to be K 1 = 6905, K 2 = 4989, K 3 = 7803.
K can be taken above to be 289. 1 Theorem 1 was proved in [22] using the following theorem due to Bobkov and Götze [3, p.
(2) There exists a W ∈ C 2 (R n ) with W ≥ 1 and constants b, c > 0 such that for all x ∈ R n . Then µ satisfies a LSI.
The goal of the present paper is to provide an elementary proof of Theorem 1. The result proved is the following: Theorem 5. Let µ be a probability measure on R whose support is contained in an interval of length 2R, and let γ δ be the centered Gaussian of variance In particular, if δ ≤ 16R 2 , we have The bound in Theorem 5 is worse than the bound in Theorem 1 for small δ, but still has an order of magnitude that is exponential in R 2 /δ. (It is shown in [22,Example 21] that one cannot do better than exponential in R 2 /δ for small δ.)

Proof of Theorem 5
The proof of Theorem 5 is based on two facts: first, the Gaussian measure γ 1 of unit variance satisfies a LSI with constant 2. Second, Lipshitz functions preserve LSIs. We give a precise statement of this second fact below. Proposition 6. Let µ be a measure on R that satisfies a LSI with constant c, and let T : R n → R n be Lipschitz. Then the push-forward measure T * µ also satisfies a LSI with constant c||T || 2 Lip . Proof. Let g : R n → R be locally Lipschitz. Then g • T is locally Lipschitz, so by the LSI for µ, But since T is Lipschitz, So by a change of variables, (1) simply becomes as desired.
We now prove Theorem 5.
Proof of Theorem 5. In light of Proposition 6, we will establish the theorem by showing that µ * γ δ is the pushforward of γ 1 under a Lipschitz map. By translation invariance of LSI, we can assume that supp(µ) ⊆ [−R, R].
We will also first assume that δ = 1 (the general case will be handled at the end of the proof by a scaling argument). Let F and G be the cumulative distribution functions of γ 1 and µ * γ 1 , i.e., Notice that q is smooth and strictly positive, so that G −1 • F is well-defined and smooth. It is readily seen that (G −1 • F ) * (γ 1 ) = µ * γ 1 , so to establish the theorem we simply need to bound the derivative of .
We will bound the above derivative in cases -when x ≥ 2R, when −2R ≤ x ≤ 2R, and when x ≤ −2R.
We first consider the case x ≥ 2R. Define Note Λ and K are smooth for x = 0.
Proof. By definition of q, p, Λ, and K, To get the other inequality, first note that e −Rx ≤ Λ(x) ≤ e Rx . (These are just the maximum and minimum values in the integrand defining Λ.) This implies that −R + R/x ≤ K(x) ≤ R + R/x, so for −R ≤ s ≤ R and x ≥ 2R, we have Proof. Recall that e −Rx ≤ Λ(x). (Again, e −Rx is the minimum value in the integrand defining Λ). We therefore have By elementary calculus, the above has a maximum value of R.
Proof. Since G and G −1 are increasing, the lemma is equivalent to The first inequality follows from the definition of G and the Fubini-Tonelli Theorem: To establish the other inequality, we use Lemmas 7 and 8: by Lemmas 7 and 8 We are almost ready to bound ( So for x ≥ 2R we have, by lemma 9, Combining this with Lemma 7, we get In the case where −2R ≤ x ≤ 2R, first note that for all x, the first inequality above was done in Lemma 9, and the second inequality is proven in the same way. So Since p has no local minima, the minimum value of the above integrand occurs at either s = R or s = −R. Without loss of generality, we assume the minimum is achieved at s = R (otherwise, we can replace (x, y) with (−x, −y) by symmetry of S and p). So Elementary calculus shows that the above infimum is equal to e −12R 2 (achieved at x = 2R, y = R). Therefore The case x ≤ −2R is dealt with in the same way as the case x ≥ 2R, the analagous statements being: and q is increasing for x ≤ −2R. The upper bound for (G −1 • F ) ′ (x) obtained in this case is the same as the one in the case x ≥ 2R. We therefore have So by Proposition 6, µ * γ 1 satisfies a LSI with constant c(1) satisfying This proves the theorem for the case δ = 1.
To establish the theorem for a general δ > 0, first observe that where h λ denotes the scaling map with factor λ, i.e., h λ (x) = λ x.