I am concerned with the sharp Poincaré-Friedrichs inequality which I can find proof.

Assume $f: [-a,a]\to \Bbb R$ ($a>0$) is $C^1$ such that $f(-a)=f(a)=0$. Then I want to show that

$$\int_{-a}^a|f(t)|^2d t\leq a^2\int_{-a}^a|f'(t)|^2d t.$$

When I use the fundamental theorem of calculus, I get the constant factor $2a^2$ instead of $a^2$.

$$\begin{split} f(t)= \int_t^a f'(s)((a-t) ds \implies |f(t)|^2 \leq (a-t)\int_t^a |f'(s)|^2 ds. \end{split}$$

Integrating both sides over $(-a,a)$ yileds $$\begin{split} \int_{-a}^a|f(t)|^2 dt& \leq \int_{-a}^a(a-t)\int_t^a |f'(s)|^2 dsd t\\ &\leq \int_{-a}^a(a-t)\int_{-a}^a |f'(s)|^2 dsd t\\ &= 2a^2 \int_{-a}^a |f'(s)|^2 ds. \end{split}$$

How can I optimize the above inequality with constant factor $a^2$?

More generally, as stated [https://en.wikipedia.org/wiki/Friedrichs%27s_inequality] assume $f\in C^k(\Bbb R)$ $k\in \Bbb N$ such that $supp \,\,f\subset (-a,a)$ then how can I prove that

$$\int_{-a}^a|f(t)|^2d t\leq a^{2k}\int_{-a}^a|f^{(k)}(t)|^2 d t.$$


1 Answer 1


For the second part of your question, if $f\in C^k$, for $k>1$, then $f'\in C^{k-1}$ and you can apply the first result to $\int_{-a}^a\lvert f'\rvert^2dt$. By induction then, $$\lVert f \rVert_{L^2}^2 \leq a^{2n}\lVert f^{(n)}\rVert_{L^2}^2 \text{, for all } n\leq k.$$

Concerning the first part, I can't find a way to tighten the control, but I think your first inequality is wrong, namely $f(t)\neq \int_t^a f'(s)(a-t)ds$. Instead, use Cauchy-Schwarz inequality on $$\lvert f(t)\rvert \leq \int_{-a}^t \vert f'(s)\rvert ds$$ and the fact that $(t+a)^{1/2} \leq (2a)^{1/2}$ to get $$\lVert f \rVert_{L^2} \leq 4a^2\lVert f'\rVert_{L^2}.$$


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