Let $C_1 > 0$ and $C_2 > 0$. We define the following set $$ M = \{f\in C^1[0,1]: |f(0)| \leq C_1, \ \int_{0}^1|f'(t)|^2dt \leq C_2 \} \subset C[0,1] $$ I want to know if $M$ a relatively compact set in $C[0,1]$.
My attempt. If we define replace the condition $\int_{0}^1|f'(t)|^2dt \leq C_2$ to condition $\int_{0}^1|f'(t)|dt \leq C_2$, then we can show that this set is not relatively compact by Arzelà–Ascoli theorem. Since the subset $\{C_2x^n, \ n\in \mathbb{N}\}$ is not uniformly equicontinuous. But I can not to give a similar example in case $\int_{0}^1|f'(t)|^2dt \leq C_2$. Also I can not show uniformly equicontinuous for $M$.
For $x <y$ we have $|f(x)-f(y)| =|\int_x^{y} f'(t)dt| \leq \sqrt {\int_x^{y}|f'(t)|^{2}dt} \sqrt {|x-y|} \leq \sqrt {C_2} \sqrt {|x-y|}$. Hence the family is equi-continuous. Together with the first condition this makes the family also uniformly bounded. Hence it is relatively compact in $C[0,1]$.