Show that $S$ has a compact closure in $C([0,1])$ with sup-norm $\|\cdot\|_\infty$, where $\|f\|_\infty=\sup_{0\leq x\leq1}|f(x)|.$

Question

Let $S$ be the set of all functions $f$ that are continuous on $[0,1]$ and differentiable on $(0,1)$ with $\int_{0}^{1}|f^1|^2\,dm\leq1$. Show that $S$ has a compact closure in $C([0,1])$ with sup-norm $\|\cdot\|_\infty$,\ where $\|f\|_\infty=\sup_{0\leq x\leq1}|f(x)|$.

My work-

I'm going to use the Arzela Ascoli theorem. What I need to prove here is,

1.uniformly bounded 2. Equicontnuity

For that, First I need to show the uniform boundedness. let $f\in S$ then consider $|f(x)-f(0)|=|\int_{0}^{1}|f^1||\leq|\int_{0}^{1}|f|^2|$ by fundamental theorem of calculus and then applied Hölder's theorem.

then $|f(x)|=|f(x)-f(0)+f(0)|\leq|f(x)-f(0)|+|f(0)|\leq1+f(0)$. Here my concern is that $f(0)$ is not bounded for all $f$.

I'm sure there should be another way to solve this.

Answer 1

This is false. If $f_n(x)=n$ for all $x$ then $(f_n)$ is a sequence in $S$ which is not bounded.