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I am trying prove the following inequality:

Suppose $f(x)$ is a symmetric distribution about 0 (e.g. standard normal distribution), then:

$\int f^2(x)dx \geq \int f(x-a)f(x+a)dx$ for any real $a$.

My guess is it should hold, and it does on normal distribution, but I don't know how to prove it in a more general form using only properties of symmetric functions.

fnosdy
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1 Answers1

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I don't even think you need symmetricity. By Cauchy Schwartz,

$$\int f(x+a)f(x-a) dx \leq \bigg( \int f(x+a)^2 dx \bigg)^{1/2}\bigg( \int f(x-a)^2 dx \bigg)^{1/2} $$$$ =\bigg( \int f(x)^2 dx \bigg)^{1/2}\bigg( \int f(x)^2 dx \bigg)^{1/2} = \int f(x)^2 dx$$

shalop
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  • Thank you so much! I actually ran into Cauchy Schwartz when doing google search, but apparently didn't connect it to my problem. – fnosdy Jul 25 '16 at 17:06