4
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ sorry, I edit,This condition is $f(0)=f(2)$ $\endgroup$ – math110 Jun 5 '13 at 15:35
  • $\begingroup$ @math110 Then my argument is still valid, but not yours. $\endgroup$ – Julien Jun 5 '13 at 15:35
  • $\begingroup$ Thank you,can you use $cauchy-Schwarz$ inequality.thank you $\endgroup$ – math110 Jun 5 '13 at 15:39
  • $\begingroup$ @math110 Probably, yes, since the estimate is far from sharp. The function must vanish somewhere since it is continuous with zero integral. So using such a point, you can repeat your argument and see what bound this gives. Note however that this is a standard application of Parseval, which yields the best possible bound. $\endgroup$ – Julien Jun 5 '13 at 15:44
  • $\begingroup$ Hi, julien, thanks to your nice answer, it gives me a chance to review Wirtinger's inequality. +1. $\endgroup$ – 23rd Jun 10 '13 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.