• Taken from Thm 7.4.1 in the book “Sur les inégalités de Sobolev logarithmiques”.

We first assume that is a smooth and bounded function such that . Consider the Laplace transform of :

Apply the logarithmic Sobolev inequality to the function . Then: Since , we get: Because , it follows that:

Also, we have the identity: which gives:

Dividing by (which is valid since ), we obtain:

Define:

Then we have:

The function is continuous on and differentiable on . Taking the limit as gives:

Therefore, for all : that is: and hence:

Now apply Markov’s inequality:

Using the previous bound on , we get:

Optimizing this upper bound by choosing gives:

This proves the first inequality. The second follows by applying the same bound to and combining both tails:

Finally, the general case where is only Lipschitz (not smooth and bounded) is obtained by standard mollification (i.e., convolution with a smooth kernel) and a limiting argument using the semicontinuity of the Lipschitz norm and uniform upper bounds.