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Let $f$ differentiable on $(0,+\infty)$ with derivative in $L^2(0,+\infty)$.

I want to show that for all $x\in (0,+\infty)$, $$f^2(x)\leq 2\|f\|_2 \|f'\|_2.$$

How to tackle this inequality ?

Thanks.

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    $f$ doesn't need to be in $L^2$, so are you assuming this part? – 2017-02-02
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    Are you sure it should not be $\frac{f^2(x)}{2}$ on the left hand side? This is what I get and for $f(x) = e^{-x}$ your inequality becomes $e^{-2x} \leq \frac{1}{2}$ which is not true close to $x=0$. – 2017-02-02
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    @Winther oh yes you are right ! i forgot :( It's changed. – 2017-02-02
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    @Winther it's not a problem to add others assumptions. At first, i supposed $f(0)=0$ to start something like $f(x) = f(x)-f(0) = \int_0^x f'(t)dt$... – 2017-02-02
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    You need some additional condition to avoid the obvious counterexample of $f$ a non zero constant. – 2017-02-02

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$$ f^2(y)-f^2(x) = \int_{x}^{y}2f(t)f'(t)dt. $$ Assuming $f,f'\in L^2$, the right side has a finite limit as $y\uparrow\infty$. So $\lim_{y\uparrow\infty}f^2(y)=L$ exists. But $L=0$ must hold because $f\in L^2$. Therefore, $$ f^2(x) = -\int_{x}^{\infty}2f(t)f'(t)dt \\ |f^2(x)| \le 2\|f\|\|f'\|. $$

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    thanks, i have a small question : why does $f\in L^2$ imply that $L=0$ ? – 2017-02-02
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    @anonymus : If $L \ne 0$, then, there exists $R > 0$ such that $|f(x)-L| < L/2$ whenever $x > R$, which would imply $|f(x)^2| > L/2$ for $x > R$, thereby preventing integrability of $|f|^2$ on $[0,\infty)$. – 2017-02-02
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    thanks again. Weird, because once a teacher told me it was not true that $f\in L^1$ implies $f(x)\to_{\infty} 0$. But in regard of your comment, the statement holds if we suppose that $f$ admits a limit. – 2017-02-06
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    @anonymus : That's right, if there is a limit at $+\infty$, then the limit must be $0$ in for the the function to be in $L^1$. Same of course at $-\infty$. But $L^1$ functions may not have limits at $\pm\infty$; you can construct examples with spikes at $n$ of width $1/n^3$ and height $n$ and still have an $L^1$ function because $\sum_n 1/n^2 < \infty$. – 2017-02-06
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    if I understand properly, you mean a (somehow periodic) function, with domain $(0,+\infty)$ where on each interval $(k,k+1)$, for $k\geq 0$, a triangle with height $k$ is drawn on the subinterval $(k/2-1/k^3,k/2+1/k^3)$. – 2017-02-07
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    @anonymus : For example, $f = \sum_{n=1}^{\infty} n\chi_{[n-1/n^3,n+1/n^3]}$ Just a bunch of step functions with areas that sum to a finite number. This step centered at $n$ has area $n \frac{2}{n^3}=\frac{2}{n^2}$, the infinite sum of which is finite. – 2017-02-07