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Assume that $f\colon[0,1]\to\mathbb{R}$ is a differentiable function with $f'$ square integrable (if that's too weak or unpractical, assume that $f$ is continuously differentiable) such that $f(0)=f(1)=0$. Is it the case that $$ \left(\int_0^1 |f|\right)^2 \leq \frac{1}{12}\int_0^1 {f'}^2 $$ ? It holds for the functions $f$ I have tried (e.g., the obvious $f(x)=x(1-x)$, and things like $f(x)=\sin(\pi x)$). Moreover, it is not difficult to see that $1/12$ is the best one can hope for, since it's achieved for $f(x)=x(1-x)$.

It seems to reek of Cauchy-Schwarz and/or integration by parts, but I can't see how to prove it. It feels I am one simple trick short.

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  • $\begingroup$ Now that the website is suggesting similar questions (i.e., after posting my question), it looks like this is really similar. I'm going to check if the proof allows for the absolute value. $\endgroup$ – Clement C. Sep 22 '18 at 22:24
  • $\begingroup$ Not sure. The answers to the (similar, but different) question I just linked rely on the fact that $$\int_0^1 f'(x)(x-1/2)dx = -\int_0^1 f(x)dx$$ (since $f(0)-f(1)=0$) by IPP; which is not obvious at all to me when considering $\int_0^1 |f'(x)(x-1/2)|dx$. $\endgroup$ – Clement C. Sep 22 '18 at 22:37
  • $\begingroup$ If $f$ satisfies your assumptions, then so does $|f|$, with derivative $\frac{d}{dx} |f(x)| = f'(x) 1_{f(x) >0} - f'(x) 1_{f(x) <0}$, which implies $\Big|\frac{d}{dx} |f| \Big| \leq |f'|$. $\endgroup$ – PhoemueX Sep 22 '18 at 23:12
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    $\begingroup$ @Ahmad $f(0)=f(1)=0$? $\endgroup$ – Clement C. Sep 22 '18 at 23:27
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    $\begingroup$ I was a little imprecise. If $f$ is in $W^{1,2}$, then so is $|f|$. For Details see e.g. Theorem 1.26 here: math.aalto.fi/~jkkinnun/files/sobolev_spaces.pdf $\endgroup$ – PhoemueX Sep 22 '18 at 23:44
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Fix a $c \in (0,1)$. For $0 \leq t\leq c$, since $f(0) =0$ we have $$|f(t)| = |\int_0^t f'(s) ds|\leq \int_0^t |f'(s)| ds.$$ Therefore, by Fubini theorem, we have $$ \int_0^c |f(t)| dt \leq \int_0^c \int_0^t |f'(s)| ds dt = \int_0^c |f'(s)| (c-s) ds.$$ Similarly, we get $$ \int_c^1 |f(t)| dt \leq \int_c^1 |f'(s)| (s-c) ds.$$ Hence, by summing two inequalities, we obtain $$\int_0^1 |f(t)| dt \leq \int_0^1 |f'(s)| |c -s| ds.$$ Applying Cauchy-Schwartz, we have $$(\int_0^1 |f(t)| dt)^2 \leq (\int_0^1 f'(s)^2 ds )(\int_0^1(c-s)^2 ds) =(c^2 -c + \frac13) \int_0^1 f'(s)^2 ds,$$ for any $c\in (0,1)$. Since $c^2-c + \frac13 \geq \frac1{12}$ for any $c\in (0,1)$ and attains at $c = \frac12$. By taking $c =\frac12$, we get the desired inequality.

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  • $\begingroup$ Thank you! So essentially, the key here is to replace IPP which ensures that $$\left\lvert\int_0^1 f(x)dx \right\rvert = \left\rvert \int_0^1 f(x)(x-1/2)dx\right\rvert$$ by a triangle inequality+Fubini applied to both halves to get that $$\int_0^1 \left\lvert f(x) \right\rvert dx \leq \int_0^1 \left\rvert f(x)(x-1/2)\right\rvert dx$$ $\endgroup$ – Clement C. Sep 23 '18 at 0:18

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