Prove that $X$ is relatively compact in the space of continuous functions on $[-1,1]$

Question

Let $X$ be the set of real-valued functions $g:\mathbb{R} \to\mathbb{R}$ of class $C^2$, with support in $[-1,1]$, and such that $\int_{-1}^1(g'')^2\le1$.

I am interested in proving that the closure of $X$ in the space of continuous functions on the interval $[-1,1]$ is compact, but I am absolutely stumped on how to proceed. So any help will be very useful.

Thanks in advance.

Answer 1

For any $g\in X$ and $x\in[-1,1]$, by Cauchy-Schwarz, $$|g'(x)|=\left|\int_{-1}^xg''(t)\,dt\right|\le\|1\|_{L^2[-1,1]}=\sqrt2$$ hence $X$ is equicontinuous and every $X(x)\subset\mathbb R$ is bounded (by $2\sqrt2$). The conclusion follows by Arzelà-Ascoli theorem.

Answer 2

Let $I=(-1,1)$ and let $g\in X$. Since $g,g',g''\in C(\overline{I})\subseteq L^2(I)$, you have that $g\in W_0^{2,2}(I)$. By Poincaré inequality(iterated two times), you get that $\lVert g\rVert_{W^{2,2}(I)}\leq c\lVert g''\rVert_{L^2(I)}\leq c$.

So you have that $X$ is a bounded subset of $W^{2,2}(I)$. Now use the compact embedding theorem in the case $pk>n$ and you get that: $$ W^{2,2}(I) \xrightarrow{compact} C^{o,\gamma}(\overline I)\hookrightarrow C(\overline{I}), \qquad \gamma<1/2$$ So $X$ is relatively compact in $C(\overline{I})$

Answer 3

Hint: Since each $g$ has support in $[-1,1]$ you can consider each $g \in X$ to be a solution of the Dirichlet boundary value problem $u''(t)=g''(t)$ $(t \in [-1,1])$, $u(-1)=u(1)=0$. Thus for each $g \in X$ $$ g(t)= \int_{-1}^1 G(t,s) g''(s) ds \quad (t \in [-1,1]) \quad (1) $$ where $G: [-1,1]^2 \to \mathbb{R}$ is Green's function for this boundary value problem. This also yields $$ g'(t)= \int_{-1}^1 G_t(t,s) g''(s) ds \quad (t \in [-1,1]) \quad (2) $$ and since $G$ and $G_t$ are bounded on $[-1,1]^2$ the Cauchy-Schwarz inequality gives uniform bounds $\beta_0,\beta_1$ with $\|g\|_\infty \le \beta_0$ and $\|g'\|_\infty \le \beta_1$ for each $g \in X$. Now you can apply Arzela-Ascoli.

Green's functions here is $$ G(t,s)=-(1-t)(s+1)/2 ~~ (-1 \leq s \leq t \leq 1), \quad G(t,s)=-(1-s)(t+1)/2 ~~ (-1 \leq t \leq s \leq 1). $$ With that you can also verify (1) and (2) directly.