# How can I Prove the following Integral Inequality?

• A
I want to prove the following inequality:

$$\sum\limits_{k\in\mathbb{N}}\Big(\int \big|f(x)\big|\big|g(x-k)\big|dx\Big)^2 \leq \big\|f\big\|^2 \sum\limits_{k\in\mathbb{N}}\Big (\int\big|g(x-k)\big|dx\Big)^2$$
where

$$\|f\|^2=\int |f(x)|^2dx.$$

My attempt: Just prove the following inequality

$$\Big(\int \big|f(x)\big|\big|g(x-k)\big|dx\Big)^2 \leq \|f\|^2\Big(\int \big|g(x-k)\big|dx\Big)^2$$
I think that this inequality is different from the Cauchy–Schwarz inequality.
Cauchy–Schwarz inequality is
$$\Big|\int f(x)\overline{g(x)}dx\Big|^2 \leq \int |f(x)|^2dx ~\cdot~\int |g(x)|^2dx$$

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mathman
$$|\int f(x)\bar{g(x)}dx|^2 \le (\int|f(x)||g(x)|dx)^2$$, so Cauchy-Schwarz applies.

Erland
No, Cauchy-Schwarz Inequality does not apply in the OP.

$$\Big(\int \big|f(x)\big|\big|g(x-k)\big|dx\Big)^2 \leq \|f\|^2\Big(\int \big|g(x-k)\big|^2dx\Big)$$

then this is an application of Cauchy-Schwarz, but in the OP we have:

$$\Big(\int \big|f(x)\big|\big|g(x-k)\big|dx\Big)^2 \leq \|f\|^2\Big(\int \big|g(x-k)\big|dx\Big)^2$$

and here, Cauchy-Schwarz does not apply.

In fact, the latter inequality fails if e.g. ##k=0## and ##f(x)=g(x)=1## for ##0\le x\le 1/2## and ##0## otherwise: in this case, the left side of the inequality is ##1/4## and the right side is ##1/8## (assuming that the interval of integration is all of ##\mathbb R##).

Here, ##f=g## is not continuous, but it can easily be appoximated by a ##C^\infty##-function for which the inequality still fails.

On the other hand, the original inequality

$$\sum\limits_{k\in\mathbb{N}}\Big(\int \big|f(x)\big|\big|g(x-k)\big|dx\Big)^2 \leq \big\|f\big\|^2 \sum\limits_{k\in\mathbb{N}}\Big (\int\big|g(x-k)\big|dx\Big)^2$$

is almost trivially true: If ##f(x)=0## a.e or ##g(x)=0## a.e. then both sides are ##0##, otherwise, the right side is ##\infty## (all the terms in the sum are equal and positive).

So, it doesn't seem to me that this inequality could be very useful...