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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.

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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 $$ \int_0^1u^2(x)\,dx=\frac{1}{\pi^2}\int_0^1(u')^2(x)\,dx \iff u(x)=a\sin(\pi x), \ a \in \mathbb{R}. $$

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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 $\delta$.

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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:'s_inequality_for_functions – Siméon Jan 22 '13 at 14:20

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