Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $S=\{f:\mathbb{R} \to \mathbb{R}\}$ that satisfies: $\forall f\in S$, $f'$ exists and $f'$ is continuous and $f(0)=f(1)=0$.

Please prove that $\exists \delta :\forall f\in S$ s.t. $\int_{0}^{1} (f(x))^2dx\leq \delta \int_{0}^{1} (f'(x))^2dx$. Thanks in advance.

share|improve this question
add comment

2 Answers

up vote 2 down vote accepted

It is enough to prove that there is an $m>0$ such that $$ J(u)\ge m \quad \forall\ u \in H_0^1(0,1),\ \|u\|_{L^2(0,1)}=1, $$ where the functional $J$ is defined by $$ J: H_0^1(0,1) \to [0,\infty),\ J(u)=\|u'\|_{L^2(0,1)}^2. $$ For every $u \in H_0^1(0,1)$ we have \begin{eqnarray} \int_0^1u^2(x)\,dx&=&\int_0^1\Big(\int_0^xu'(t)\,dt\Big)^2\,dx\le\int_0^1\Big(\int_0^x|u'(t)|\,dt\Big)^2\,dx\\ &\le&\int_0^1x\int_0^x(u'(t))^2\,dt\,dx\le\int_0^1x\int_0^1(u'(t))^2\,dt\,dx=\frac12\int_0^1(u'(x))^2\,dx. \end{eqnarray} It follows that $$ J(u) \ge 2 \quad \forall\ u \in H_0^1(0,1), \|u\|_{L^2(0,1)}=1, $$ i.e. $J$ is bounded below on $M=\{u \in H_0^1(0,1):\ \|u\|_{L^2(0,1)}^2=1\}$.

So, let $$ m=\inf_MJ. $$ If there is a $u \in M$ such that $J(u)=m$, then by the Lagrange multipliers rule there exists some scalar $\lambda$ such that $$ \int_0^1u'\phi'=\lambda\int_0^1u\phi \quad \forall\ \phi \in C_0^1(0,1), $$ i.e. $\lambda$ and $u$ solve the problem $$\tag{1} u''+\lambda u=0 \ \text{ in } (0,1), \ u(0)=0=u(1). $$ Solving (1) we get $$ u(x)=u_\lambda(x)=a\sin(\sqrt{\lambda}x),\ \lambda \in \{k^2\pi^2:\ k \in \mathbb{N}\},\ a\in \mathbb{R}. $$ Hence $m=\pi^2$, i.e. $$ \int_0^1u^2(x)\,dx\le \frac{1}{\pi^2}\int_0^1(u')^2(x)\,dx \quad \forall u \in H_0^1(0,1). $$ Another approach is to use Fourier analysis. Since $\phi_k(x)=\sqrt{2}\sin(k\pi x), k \in \mathbb{N}$ form a basis for $H_0^1(0,1)$, any $u \in H_0^1(0,1)$ can be written as $$ u=\sum_{k=1}^\infty a_k(u)\phi_k,\ a_k(u)=\int_0^1u(x)\phi_k(x)\,dx. $$ Hence, for every $u \in H_0^1(0,1)$ we have $$ \int_0^1(u')^2(x)\,dx=\sum_{k=1}^\infty k^2\pi^2a_k^2(u)\ge \pi^2\sum_{k=1}^\infty a_k^2(u)=\pi^2\int_0^1u^2(x)\,dx, $$ i.e. $$ \int_0^1u^2(x)\,dx\le \frac{1}{\pi^2}\int_0^1(u')^2(x)\,dx \quad \forall\ u \in H_0^1(0,1). $$ Remark The constant $m=\frac{1}{\pi^2}$ is optimal, and $$ \int_0^1u^2(x)\,dx=\frac{1}{\pi^2}\int_0^1(u')^2(x)\,dx \iff u(x)=a\sin(\pi x), \ a \in \mathbb{R}. $$

share|improve this answer
add comment

Since $f(0)=0$, $$|f(x)|=|\int_0^xf'(t)dt|\le \int_0^x|f'(t)|dt\le\int_0^1|f'(t)|dt .$$ Then by Cauchy-Schwarz inequality,

$$(f(x))^2\le (\int_0^1|f'(t)|dt)^2\le \int_0^1(f'(t))^2dt.$$ Integrating both sides of the inequality above over $[0,1]$, it follows that

$$\int_0^1(f(x))^2dx\le \int_0^1(f'(x))^2dx.$$

Therefore, $\delta=1$ works, but it is far from the optimal choice of $\delta$.

share|improve this answer
    
if you integrate both side of $(f(x))^2\le (\int_0^1|f'(t)|dt)^2\le \int_0^1(f'(t))^2dt$ it follow that $\int_0^1(f(x))^2dx\le \iint_0^1(f'(x))^2dx.$ and why ?($\iint_0^1(f'(x))^2dx\le\int_0^1(f'(x))^2dx$ –  Maisam Hedyelloo Jan 19 '13 at 18:00
    
@MaisamHedyelloo: The right hand side is $\int_0^1(\int_0^1(f'(t))^2dt)dx=\int_0^1(f'(t))^2dt=\int_0^1(f'(x))^2dx.$ –  23rd Jan 19 '13 at 18:03
    
@MaisamHedyelloo The right-hand side is constant. –  Matemáticos Chibchas Jan 19 '13 at 18:32
    
matematico chibchas:sorry why its constant ? –  Maisam Hedyelloo Jan 19 '13 at 18:36
    
Therefore, $δ=1$ works, but it is far from the optimal choice of $δ$. Indeed, the optimal choice $\delta=(2\pi)^{-2}$ is given by Wirtinger's inequality: en.wikipedia.org/wiki/Wirtinger's_inequality_for_functions –  Siméon Jan 22 '13 at 14:20
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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