Proving compactness of a subset

Question

Let $I = [a, b]$ be an interval in $\mathbb{R}$. We consider $C(I)$ with the maximum norm and define a subset of $C(I)$

$$M_c=\left\{f \in C^1(I): \int_a^b\vert f(x) \vert ^2dx + \int_a^b\vert f'(x) \vert ^2dx \leq c \right\}.$$

The goal is to show $\bar{M_c}$ is compact in $C(I)$.

My idea is to show that $M_c$ is relatively compact using Arzela-Ascoli lemma. In order to do that I need to show $M_c$ is bounded and equicontinuous. Unfortunately, I have no idea of how to show the latter fact. I would appreciate any help to solve my problem.

Answer 1

To prove that the family is equicontinuous you can just note that all $f\in M_c$ are lipschitz of (common) constant $\frac{c}{b-a}$ (having the first derivative absolutely bounded by construction)

Answer 2

Hint. Supose that for all $n\in \mathbb{N}$ there is a $f_n\in M_c$ such that $$ \|f_n\|_{\infty}>n. $$ Then there is a sufficiently large $n_0$ such that $$ \int_{a}^{b}|f_n(x)|^2{\rm d} x + \int_{a}^{b}|f_n^{\prime}(x)|^2{\rm d} x >c $$ But this is absurd. We can only conclude that $M_c$ is limited, i.e. here is a $L>0$ such that $\|f\|_\infty<L$. Use this to show that, for all $x_0\in [a,b]$ fixed, $$ |f^\prime(x_0)|<L,\quad \forall f\in M_c $$ Now use the idea given by Daniel Fischer, i.e. $f(y) - f(x) = \int_x^y f'(t)\,dt = \int_a^b \chi_{[x,y]}(t) f'(t)\,dt $ to proof $$ |f(y)-f(x)|\leq L\cdot |y-x|, \quad \forall f\in M_c, \quad \forall x,y\in [a,b] $$