Let $f\in C^1([0,\pi],\mathbb R)$ such that $\displaystyle\int_0^\pi f(t) dt=0$

Prove that $\forall x\in [0,\pi],\displaystyle|f(x)|\leq \sqrt{\frac{\pi}{3}\int_0^\pi f'^2(t)dt}$

Failed natural attempt

$\int_0^\pi f(t) dt=0$ tells us that there is some $\beta\in [0,1]$ such that $f(\beta)=0$

Using the fundamental theorem of calculus and Cauchy-Schwarz inequality,

$\displaystyle |f(x)|=|f(x)-f(\beta)|\leq\int_x^\beta |f'(t)|dt\leq \int_0^\pi |f'(t)|dt \leq \sqrt{\pi} \sqrt{\int_0^\pi f'^2}$

It is not sharp enough.

This might have something to do with Fourier series.

  • $\begingroup$ An idea: Endow $C^1([0,\pi],\mathbb R)$ with the supremum norm and let $D:C^1([0,\pi],\mathbb R)\to L^2([0,\pi])$ denote the differential operator, with the image space being endowed with an $L^2$ Hilbert-space structure. Also, let $I:C^1([0,\pi],\mathbb R)\to\mathbb R$ denote the integral operator. Then, you basically need to show that if $f\in\operatorname{ker} I$, then $\|f\|^2\leq(\pi/3)\cdot\langle D(f),D(f)\rangle$. I'm not sure if this characterization is helpful, but may be worth a shot. $\endgroup$
    – triple_sec
    Jun 21, 2014 at 20:20

2 Answers 2


Without Fourier series: we have $((t-a)f(t))'=(t-a)f'(t)+f(t)$. Hence $$ \int_0^t\tau f'(\tau)d\tau=tf(t)-\int_0^tf(\tau)d\tau,\qquad \int_t^\pi(\tau-\pi)f'(\tau)d\tau=(\pi-t)f(t)-\int_t^\pi f(\tau)d\tau. $$ By adding these, we obtain $$ \pi f(t)=\int_0^\pi g(\tau)f'(\tau)d\tau,\qquad g(\tau)=\begin{cases}\tau,&\tau<t,\\\tau-\pi,&\tau>t.\end{cases} $$ Now, by Cauchy-Schwarz, $$ \pi |f(t)|\le \sqrt{\int_0^\pi |f'(\tau)|^2d\tau}\sqrt{\frac{t^3}3+\frac{(\pi-t)^3}3}\le \sqrt{\int_0^\pi |f'(\tau)|^2d\tau}\sqrt{\frac{\pi^3}{3}} $$ Dividing by $\pi$, we obtain the desired result. QED

How to guess this proof? It is easy enough: $\frac13$ can only come from $\int t^2 dt$ in this context, so (having in mind Cauchy--Schwarz) we need to estimate $f$ somehow via the integral of $tf'(t)$.

  • $\begingroup$ Note that at the interior points of the interval we obtain an estimate that is even sharper than the one requested by @G.T.R ... $\endgroup$
    – Vladimir
    Jun 21, 2014 at 21:16
  • $\begingroup$ Great Solution (+1) !! Do you think the inequality could be sharpened further? :-) $\endgroup$
    – r9m
    Jun 21, 2014 at 21:20
  • $\begingroup$ @r9m Probably not, but I am not 100% sure. $\endgroup$
    – Vladimir
    Jun 21, 2014 at 21:22

Here's an sketch of a proof using Fourier series. Let $f(x) = \sum_{n=1}^\infty a_n \cos(n x)$, then

$$ \int_0^\pi f'(t)^2dt = \frac{\pi}{2}\sum_{n=1}^\infty \left(n a_n\right)^2 $$ With this in mind, we apply Cauchy-Schwarz as follows: $$ \begin{aligned} |f(x)|^2 &=\left( \sum_{n=1}^\infty \left(n a_n\right)\left(\frac{\cos(n x)}{n}\right) \right)^2\\ &\le \left( \sum_{n=1}^\infty \left(n a_n\right)^2 \right) \left(\sum_{n=1}^\infty\left(\frac{\cos(n x)}{n}\right)^2 \right) \\ &\le \left( \sum_{n=1}^\infty \left(n a_n\right)^2 \right) \left(\sum_{n=1}^\infty\frac{1}{n^2} \right)\\ &= \left( \frac{2}{\pi}\int_0^\pi f'(t)^2dt \right) \left(\frac{\pi^2}{6} \right) \end{aligned} $$

  • $\begingroup$ It should be noted that by taking limits, the inequality also holds for $x=0,\pi$. (The fourier series need not converge to $f$ in the endpoints.) It's just a detail. $\endgroup$ Jun 21, 2014 at 20:47
  • $\begingroup$ Why can we let $f(x) = \sum_{n=1}^\infty a_n \cos(n x)$? Doesn't this mean $f(x)$ is even? $\endgroup$
    – VIVID
    Jul 30, 2021 at 6:43

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