Integral inequality between function and derivative

Question

Let $f \in \mathcal{C}^1([0,1])$ such that $f(0)=f(1)=0$. I want to proof the following inequality: $$\int_0^1 f'(x)^2 \,\mathrm{d}x \geq \pi^2 \int_0^1 f(x)^2 \,\mathrm{d}x $$ I proceeded by integration by parts to get $$\int_0^1 f''(x)f(x) \,\mathrm{d}x \geq \pi^2 \int_0^1 f(x)^2 \,\mathrm{d}x ,$$ but I have no idea what to do now. Any help please ?

Answer 1

For each $n\in\mathbb{Z}_{> 0}$, define $$a_n:=2\,\int_0^1\,\cos(2n\pi\,x)\,f(x)\,\text{d}x$$ and $$b_n:=2\,\int_0^1\,\sin(2n\pi\,x)\,f(x)\,\text{d}x\,.$$ Furthermore, let $$a_0:=\int_0^1\,f(x)\,\text{d}x\,.$$ Therefore, $$f(x)=a_0+\sum_{n=1}^\infty\,a_n\,\cos(2n\pi\,x)+\sum_{n=1}^\infty\,b_n\,\sin(2n\pi\,x)\text{ for }x\in[0,1]\,.$$ Because $f(0)=0$ and $f(1)=0$, we must have $$\sum_{n=1}^\infty\,a_{n}=-a_0\,.$$

We have $$f'(x)=-\sum_{n=1}^\infty\,2n\pi\,a_n\,\sin(2n\pi\,x)+\sum_{n=1}^\infty\,2n\pi\,b_n\,\cos(2n\pi\,x)\text{ for }x\in[0,1]\,.$$ That is, $$\int_0^1\,\big|f'(x)\big|^2\,\text{d}x=2\pi^2\, \sum_{n=1}^\infty\,n^2\,\big(|a_n|^2+|b_n|^2\big)$$ and $$\int_0^1\,\big|f(x)\big|^2\,\text{d}x=|a_0|^2+\frac{1}{2}\,\sum_{n=1}^\infty\,\big(|a_n|^2+|b_n|^2\big)\,.$$ Thus, $$\int_0^1\,\big|f'(x)\big|^2\,\text{d}x- \pi^2\,\int_0^1\,\big|f(x)\big|^2\,\text{d}x\geq \frac{\pi^2}{2}\,\left(\sum_{n=1}^\infty\,(4n^2-1)\,|a_{n}|^2-2\,|a_0|^2\right)\,.$$ Let $$w_n:=\frac{2}{4n^2-1}\text{ for }n=1,2,3,\ldots\,.$$ Note that $\sum\limits_{n=1}^\infty\,w_n=1$. Then, $$\int_0^1\,\big|f'(x)\big|^2\,\text{d}x- \pi^2\,\int_0^1\,\big|f(x)\big|^2\,\text{d}x\geq \frac{\pi^2}{2}\,\Biggl(\frac12\,\sum_{n=1}^\infty\,w_n\,\big((4n^2-1)|a_{n}|\big)^2-2\,|a_0|^2\Biggr)\,.$$ Since $\sum\limits_{n=1}^\infty\,w_n=1$, the Power-Mean Inequality implies that $$\int_0^1\,\big|f'(x)\big|^2\,\text{d}x- \pi^2\,\int_0^1\,\big|f(x)\big|^2\,\text{d}x\geq \frac{\pi^2}{2}\,\Biggl(\frac12\,\left(\sum_{n=1}^\infty\,w_n\,(4n^2-1)|a_{n}|\right)^2-2\,|a_0|^2\Biggr)\,.$$ It follows immediately that $$\int_0^1\,\big|f'(x)\big|^2\,\text{d}x- \pi^2\,\int_0^1\,\big|f(x)\big|^2\,\text{d}x\geq \frac{\pi^2}{2}\,\left(2\,\left(\sum_{n=1}^\infty\,|a_{n}|\right)^2-2\,|a_0|^2\right)\,.$$ Because $$\sum_{n=1}^\infty\,|a_{n}|\geq \left|\sum_{n=1}^\infty\,a_{n}\right|=|-a_0|=|a_0|\,,$$ we obtain $$\int_0^1\,\big|f'(x)\big|^2\,\text{d}x- \pi^2\,\int_0^1\,\big|f(x)\big|^2\,\text{d}x\geq \frac{\pi^2}{2}\,\left(2-2\right)\,|a_0|^2=0\,.$$

Ergo, $$\int_0^1\,\big|f'(x)\big|^2\,\text{d}x\geq \pi^2\,\int_0^1\,\big|f(x)\big|^2\,\text{d}x\,.$$ The equality becomes an equality if and only if there exist constant $A$ such that $$f(x)=A\,\left(\frac12-\sum_{n=1}^\infty\,\frac{\cos(2n\pi\,x)}{4n^2-1}\right)=a\,\sin(\pi\,x)$$ for $x\in[0,1]$, where $a:=\dfrac{4}{\pi}\,A$.