- #1
zarei175
- 3
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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
$$\Big|\int f(x)\overline{g(x)}dx\Big|^2 \leq \int |f(x)|^2dx ~\cdot~\int |g(x)|^2dx $$
$$\sum\limits_{k\in\mathbb{N}}\Big(\int \big|f(x)\big|\big|g(x-k)\big|dx\Big)^2 \leq \big\|f\big\|^2 \sum\limits_{k\in\mathbb{N}}\Big (\int\big|g(x-k)\big|dx\Big)^2$$
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
$$\Big|\int f(x)\overline{g(x)}dx\Big|^2 \leq \int |f(x)|^2dx ~\cdot~\int |g(x)|^2dx $$