Proving a weighted Poincaré inequality

Question

Prove that there exists a positive constant $C$ such that $$\int \limits_{0}^{\infty} e^{-y}[u(y) - \langle u \rangle]^2 \mathrm{d}y \le C \int \limits_{0}^{\infty} e^{-y}[u'(y)]^2 \mathrm{d}y$$ for all $u \in C^{\infty}_c (\mathbb{R})$, where $$\langle u \rangle = \int \limits_{0}^{\infty} e^{-y}u(y) \mathrm{d}y$$

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Attempt: Comparing the two sides, and seeing the limits of integration are the same, my goal is to show $$u(y) - \langle u \rangle \le Cu'(y)$$

I figured I should write $u(y)$ as an integral and follow the equations, but I got stuck with something that doesn't appear to make sense.

$$u(y) = u(0) + \int \limits_{0}^{\infty} u'(y) \mathrm{d}y$$

Then, $$u(y) - \langle u \rangle = u(0) + \int \limits_{0}^{\infty} u'(y) \mathrm{d}y - \int \limits_{0}^{\infty} e^{-y} u(y) \mathrm{d}y = u(0) + \int \limits_{0}^{\infty} u'(y)-e^{-y}u(y) \mathrm{d}y$$

Using the FTC we have that \begin{align*} u(y) - u(0) = \int \limits_{0}^{y} u'(s) \mathrm{d}s \implies & \lim \limits_{y \to \infty} u(y) = u(0) + \lim \limits_{y \to \infty} \int \limits_{0}^{y} u'(s) \mathrm{d}s \\ \implies & 0 = u(0) + \int \limits_{0}^{\infty} u'(s) \mathrm{d}s \\ \implies & u(0) = - \int \limits_{0}^{\infty} u'(s) \mathrm{d}s \end{align*} since $u \in C^{\infty}_{c}(\mathbb{R})$ means $u$ is compactly supported and so vanishes at infinity.

Substituting back in, we would then have that \begin{align*} u(y) - \langle u \rangle &= \int \limits_{0}^{\infty} -u'(y) + u'(y) - e^{-y} u(y) \mathrm{d}y \\ &= - \int \limits_{0}^{\infty} e^{-y} u(y) \mathrm{d}y = - \langle u \rangle \end{align*}

But this doesn't seem right, since it would imply that $u(y) = 0$.

Does anyone have any other approaches to this problem? Did I make a mistake in any of my assumptions?

Thanks in advance.

Answer 1

Set $g = (f-f(0))^2$. Then \begin{align*} \int_0^\infty g(x)e^{-x}\,dx &= \int_0^\infty 2(f(x)-f(0))f'(x)e^{-x}\,dx \\&\leq \left(\int_0^\infty 4 g(x)e^{-x}\,dx\right)^{\frac{1}{2}}\left(\int_0^\infty |f'(x)|^2e^{-x}\,dx\right)^{\frac{1}{2}}\,. \end{align*} We deduce that $$ \int_0^\infty g(x)e^{-x}\,dx \leq 4\int_0^\infty |f'(x)|^2e^{-x}\,dx \,. $$ Thus, $$ \mathrm{Var}_\mu[f] = \inf_{c\in \mathbb R}\int_0^\infty (f(x)-f(c))^2e^{-x}\,dx \leq \int_0^\infty g(x)e^{-x}\,dx \leq 4\mathbf E_\mu[|f'(\xi)|^2]\,, $$ where $\mu$ is the one sided exponential measure on $\mathbb R$.

The paper "Poincaré’s inequalities and Talagrand’s concentration phenomenon for the exponential distribution" discusses in detail such inequalities for the exponential distribution.

The above inequality is a special case of the more general family of Poincaré inequalities, which take the form $\mathrm{Var}_\mu f \leq \mathcal E(f,f)$, where $\mathcal E(f,g) = -(f,\mathscr L g)_\mu$ is the Dirichlet form associated to a Markov semigroup $P_t$ with stationary measure $\mu$ and self-adjoint generator $\mathscr L$ such that $P_t f \to \mu(f)$ in $L^2(\mu)$ exponentially fast in $t$.