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Let $f:[0,\infty)\rightarrow \mathbb{R}$ be twice differentiable. Such that $\int_{0}^{\infty}f(x)^{2}dx<\infty$ and $\int_{0}^{\infty}f''(x)^{2}dx<\infty $, show that

$$\int_{0}^{\infty}f'(x)^{2}dx<\infty$$.

I'm pretty stumped on this problem. I've tried integration by parts but didn't really get anywhere. A short hint would be very appreciated. Thanks.

TheNumber23
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3 Answers3

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If $f$ and $f''$ are square integrable, then, by the Cauchy-Schwarz inequality, $ff''$ is integrable on $\mathbb{R}$. So, the limit of the following exists as $x\rightarrow\infty$: $$ \int_{0}^{x}f(t)f''(t)\,dt = f(x)f'(x)-f(0)f'(0)-\int_{0}^{x}f'(t)^{2}dt. $$ If $f'$ is not square-integrable, then $\lim_{x\rightarrow\infty}f(x)f'(x)=+\infty$ because the left side of the above is bounded and $\lim_{x\rightarrow\infty}\int_{0}^{x}f'(t)^{2}dt=\infty$. That means that, for any $M > 0$, there exists $R > 0$ such that $f(x)f'(x) > M$ whenever $x > R$. But that means $(f^{2})' > M/2$ for all $x > R$, which clearly contradicts the fact that $f$ is square-integrable. So $f'$ must be square-integrable.

Phira
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Disintegrating By Parts
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    That's the Cauchy-Schwartz inequality. If $f,g\in L^{2}$, then $fg\in L^{1}$ and $|fg|{1}\le|f|{2}|g|{2}$. It's too easy to forget that isn't obvious, which it is not. It follows from $|(f,g)| \le |f|{2}|g|_{2}$, where $(f,g)$ is the inner-product. You can argue directly from $|fg|\le \frac{1}{2}(|f|^{2}+|g|^{2})$ as well. – Disintegrating By Parts Jan 10 '14 at 05:03
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Consider $$\int_0^N f^\prime(x)^2 d x.$$ Integrate it by parts, to get: $$f(N) f^\prime(N) - f(0) f^\prime(0) - \int_0^N f(x) f^{\prime \prime}(x) dx.$$ Now note that the last integral can be bounded by Cauchy-Schwarz, so the only thing you need to bound is $f(N) f^\prime(N).$ Can you?

Phira
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Igor Rivin
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  • My idea is hopefully this should go to zero correct? If $f(x)^{2}$ is integrable then $f(x)\rightarrow 0$ for large $x$. So I'm hoping that it goes to zero faster than $f'$ is that the correct idea on the bound or am I off? – TheNumber23 Jan 10 '14 at 04:15
  • Why does $f(x)\rightarrow 0,$ if $f(x)^2$ is integrable? – Igor Rivin Jan 10 '14 at 04:17
  • I made the mistake that $f(x)$ was monotonic which it is not. – TheNumber23 Jan 10 '14 at 04:21
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A heavy handed answer: Let the Fourier transform of $f$ be $g,$ then the Fourier transform of $f^\prime$ is $s g,$ and the Fourier transform of $f^{\prime\prime}$ is $s^2 g.$ A function is in $L^2$ if and only if its Fourier transform is. But note that the Fourier transform of $f^\prime$ is dominated at $\pm \infty$ y that of $f^{\prime \prime}$ and at $0$ by that of $f,$ so must be in $L^2$ (since the other two are).

Igor Rivin
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