prove this $\int_{0}^{2}f^2(x)dx\le\int_{0}^{2}f'^2(x)dx$

Question

let $f\in C^1[0,2]$,and such $\int_{0}^{2}f(x)dx=0,f(0)=f(2)$,

show that $$\int_{0}^{2}f^2(x)dx\le\int_{0}^{2}f'^2(x)dx$$

I think we must use $Cauchy$ inequality

my idea:I have see this let $f(x)\in C^1([a,b],R)$,and $f(a)=f(b)=0$,show that：$$\displaystyle\int_{a}^{b}f^2(x)dx\le\dfrac{(b-a)^2}{8}\displaystyle\int_{a}^{b}[f'(x)]^2dx$$ pf： $$|f(x)|=|f(x)-f(a)|\le\sqrt{x-a}\left(\displaystyle\int_{a}^{x}[f'(t)]^2dt\right)^{\frac{1}{2}}$$ then $$f^2(x)\le(x-a)\displaystyle\int_{a}^{x}[f'(t)]^2dt\le(x-a)\displaystyle\int_{a}^{b}[f'(t)]^2dt$$ so we can $a$to $b$ we have; $$\displaystyle\int_{a}^{b}f^2(x)dx\le\displaystyle\int_{a}^{b}\left[(x-a)\displaystyle\int_{a}^{b}[f'(t)]^2dt\right]dx=\dfrac{(b-a)^2}{2}\displaystyle\int_{a}^{b}[f'(x)]^2dx$$ then we use $\dfrac{a+b}{2}$to $b$,then we $$\displaystyle\int_{a}^{\frac{a+b}{2}}f^2(x)dx\le\dfrac{(b-a)^2}{8}\displaystyle\int_{a}^{\frac{a+b}{2}}[f'(x)]^2dx$$ other hand ，for any $x\in[\frac{a+b}{2},b],f(x)=-\displaystyle\int_{x}^{b}f'(t)dt$, so $$f^2(x)=\left(\displaystyle\int_{x}^{b}f'(x)dx\right)^2\le(b-x)\displaystyle\int_{x}^{b}[f'(t)]^2dt$$ we can $\dfrac{a+b}{2}$to $b$ have : \begin{align} &\displaystyle\int_{\frac{a+b}{2}}^{b}f^2(x)dx\le\displaystyle\int_{\frac{a+b}{2}}^{b}(b-x)\left(\displaystyle\int_{a}^{b}[f'(t)]^2dt\right)dx\le \displaystyle\int_{\frac{a+b}{2}}^{b}(b-x)dx\left(\displaystyle\int_{\frac{a+b}{2}}^{b}[f'(t)]^2dt\right)dx\\ &=\left(\displaystyle\int_{\frac{a+b}{2}}^{b}(b-x)dx\right)\left(\displaystyle\int_{\frac{a+b}{2}}^{b}[f'(x)]^2dx\right)\\ &=\dfrac{(b-a)^2}{8}\displaystyle\int_{\frac{a+b}{2}}^{b}[f'(x)]^2dx \end{align} then $$\displaystyle\int_{a}^{b}f^2(x)dx\le\dfrac{(b-a)^2}{8}\displaystyle\int_{a}^{b}[f'(x)]^2dx$$

let $b=2,a=0$,then we have $$2\int_{0}^{2}f^2(x)dx\le\int_{0}^{2}f'^2(x)dx$$

Answer 1

The hypothesis allow to see $f$ as a continuous $2$-periodic function, piecewise $C^1$. Working in $L^2(0,2)$, the Fourier coefficients of $f$ and $f'$ are $$ c_n(f)=\frac{1}{2}\int_0^2f(x)e^{-i\pi nx}dx\qquad c_n(f')=i\pi nc_n(f) $$ where the second formula follows from an integration by parts. In particular, $c_0(f')=0$. Note that we have $c_0(f)=0$ by assumption. Hence Parseval for $f$ and $f'$ yields $$ \frac{1}{2}\int_0^2|f(x)|^2dx=\sum_{n\geq 1}|c_n(f)|^2\leq \sum_{n\geq 1}\pi^2n^2|c_n(f)|^2=\frac{1}{2}\int_0^2|f'(x)|^2dx. $$ Note that this is strict as soon as there exists $n$ such that $c_n(f)\neq 0$. Since $f$ is piecewise $C^1$, it is equal to its Fourier series which converges normally. So we have equality if and only if $f=0$ under the given assumptions.

Finally, note that this argument yields the sharper inequality $$ \int_0^2|f(x)|^2dx\leq \frac{1}{\pi^2}\int_0^2|f'(x)|^2dx. $$ And this is optimal, considering $f(x)=\sin (\pi x)$.