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

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 $f(x)=\sin (\pi x)$.

