Let $f(x)$ and $g(x)$ be two continuous functions at $[0,\infty)$.

Also, $\int _0^{\infty }f^2\left(x\right)^{ }dx$ $\;\;$and $\;\;\int _0^{\infty }g^2\left(x\right)^{ }dx\;\;$ converges.

Does $\int _0^{\infty }\left(f\left(x\right)+g\left(x\right)\right)^2dx$ converge ?

I tried to assume that it doesn't converge. and I got to :

$\:\int _0^{\infty \:}\left(f\left(x\right)+g\left(x\right)\right)^2dx\:\le\:\int _0^{\infty }f^2\left(x\right)dx\:+\int _0^{\infty \:}g^2\left(x\right)dx\:+\:2\int _0^{\infty \:}\left|f\left(x\right)\right|\left|g\left(x\right)\right|dx$

We know that the first two addents converges. What about the left addent ?

  • $\begingroup$ Just use the schwartz inequality. $\int_0^\infty f(x)g(x) dx \leq \sqrt{\int_0^\infty |f^2(x)| dx} \sqrt{\int_0^\infty |g^2(x)| dx}$ Assuming $f$ and $g$ are real. $\endgroup$ – Merkh Apr 24 '16 at 9:56

By the AM-GM inequality, $$\left|f(x)\right|\cdot\left|g(x)\right|\leq\frac{f^2(x)+g^2(x)}{2}$$ So $\int_0^\infty\left|f(x)\right|\left|g(x)\right|dx$ converges.

  • $\begingroup$ that's correct ? $\left|f\left(x\right)\right|\left|g\left(x\right)\right|=\sqrt{\left|f^2\left(x\right)\right|\left|g^2\left(x\right)\right|}\:\le \:\:\:\frac{f^2\left(x\right)+g^2\left(x\right)}{2}\:\:$ $\endgroup$ – idan di Apr 24 '16 at 10:13
  • 1
    $\begingroup$ $(a-b)^2=a^2-2ab+b^2\geq 0\implies ab\leq \dfrac{a^2+b^2}{2}$ $\endgroup$ – Chris Apostol Apr 24 '16 at 10:14

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