How to prove equality from poincare inequality?

Question

Let $$D = \{y \in C^1(0,1) : y(0) = y(1) = 0\}$$

Suppose there exists a $C_0$ such that $$\int_{0}^{1} y^2 \ dx \leq C_0 \int_{0}^{1} (y')^2 \ dx$$ for all $y \in D$, and for all $C < C_0$ there exists a $y \in D$ such that the inequality is false.

How do I show that $\frac{1}{C_0} = \min \frac{\int_{0}^{1}y'(x)^2dx}{\int_{0}^{1}y(x)^2dx}$ in D = $\min {\int_{0}^{1}y'(x)^2dx}$ in D such that $\int_{0}^{1}y(x)^2dx$ = 1 ?

Why is $\min \frac{\int_{0}^{1}y'(x)^2dx}{\int_{0}^{1}y(x)^2dx}$ in D = $\min {\int_{0}^{1}y'(x)^2dx}$ in D such that $\int_{0}^{1}y(x)^2dx$ = 1 ?

Answer 1

The inequality with $C_0$ yields $$\frac{\int (y')^2 \, dx}{\int y^2 \, dx} \ge \frac{1}{C_0}.$$ It remains to show that there is a sequence $y_n$ with $$\frac{\int (y_n')^2 \, dx}{\int y_n^2 \, dx} \to \frac{1}{C_0}.$$ But for any $C$ with $1/C > 1/C_0$, there is a $y_C$ with $$\frac1{C_0} \le \frac{\int (y_C')^2 \, dx}{\int y_C^2 \, dx} < \frac{1}{C}.$$ This shows your desired equation.

Answer 2

Big picture: $1/C_0$ is actually the first eigenvalue for the Dirichlet boundary problem for $-\Delta y= -y'' = 0$ on $[0,1]$.

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How do I show that $\frac{1}{C_0} = \min \frac{\int_{0}^{1}y'(x)^2dx}{\int_{0}^{1}y(x)^2dx}$ in $D$?

Suppose for a $C_0$ satisfying the condition you mentioned, for all $C<C_0$ there exists a $y\in D$ such that the inequality is false. First by Poincaré inequality, $$ \frac{1}{C_0} \leq \frac{\int_{0}^{1}y'(x)^2dx}{\int_{0}^{1}y(x)^2dx}\tag{1} $$ is true. Now you need to show that the equality is attained. This is called the Minimization principle of the $n$-th eigenvalue. You can prove that the minimizer of the right hand side ratio of (1) satisfies : $$ -\Delta y = \lambda y\;\text{ in }\; [0,1], \quad y(0)=y(1) =0.\tag{$\star$} $$ in the sense of distribution ($H^2$ embedding in $C^1$). Reference can be found here Theorem 4, or any other PDE book that covers the eigenvalue problems for Laplacian in 1D. Solving $(\star)$ yields: for $k\in \mathbb{N}$ $$ \lambda = k^2\pi^2,\text{ and } y = \sin (\sqrt{\lambda }x). $$ Take the smallest $\lambda_1=\pi^2$, you can check the ratio on the right hand side of (1) for $y_1 = \sin (\pi x)$ is just $\lambda_1$. What is left to show is that $1/C_0 = \lambda_1$, i.e., one way is to show by contradition: assume there is some $C<C_0= 1/\pi^2$, so that for any $y\in D$ $$ \frac{1}{C } \leq \frac{\int_{0}^{1}y'(x)^2dx}{\int_{0}^{1}y(x)^2dx}.\tag{2} $$ Now simply plugging $y_1$ here we have: $$ \frac{1}{C } \leq \pi^2 \implies C \geq \frac{1}{\pi^2}, \quad \text{Contradiction.} $$ Therefore the converse is true, i.e., for any $C < C_0$, there exists some $y_0\in D$ such that $$ \frac{1}{C} > \frac{\int_{0}^{1}y_0'(x)^2dx}{\int_{0}^{1}y_0(x)^2dx} . $$ Lastly, since (1) holds for all $y\in D$, and equality holds for $y_1 = \sin(\pi x)$, the minimum can be attained: $$ \frac{1}{\pi^2} = \frac{1}{C_0} = \min_{y\in D} \frac{\int_{0}^{1}y'(x)^2dx}{\int_{0}^{1}y(x)^2dx}. $$

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Why is $\min \frac{\int_{0}^{1}y'(x)^2dx}{\int_{0}^{1}y(x)^2dx}$ in $D$ equal to $\min {\int_{0}^{1}y'(x)^2dx}$ in $D$ such that $\int_{0}^{1}y(x)^2dx=1$?

For any $y\in D$, let $$w(x) = y(x)/\sqrt{\int^1_0 y(\xi)^2\,d\xi},$$ i.e., normalized by its $L^2$-norm, we can check that $w\in D \iff y\in D$ provided $y\neq 0$ a.e. (we are not interested in the trivial case anyway). Keep in mind that $\int^1_0 y(x)^2\,dx$ is just a number. Then $$ \min_{y\in D} \frac{\int_{0}^{1}y'(x)^2dx}{\int_{0}^{1}y(x)^2dx} = \min_{y\in D} \int^1_0 \left(\frac{y}{\sqrt{\int^1_0 y(x)^2\,dx}}\right)'^2\,dx= \min_{w\in D} \int^1_0 w'(x)^2\,dx, $$ where $\int^1_0 w(x)^2 = 1$. For $w$ is just a dummy notation, you can say it holds for $y$ subject to the constraint $\int^1_0 w(x)^2 = 1$ as well.