Compactness of sublevels Let $(X,d)$ be a metric space, complete and separable. Suppose that there exists a function $f:X \to [0,+\infty]$ with compact sublevels. Define the function $F:C([0,1],X)\to [0,+\infty]$ in this way
$$
\gamma \mapsto\int_0^1f(\gamma_t)+|\dot{\gamma}_t|^2\, \mathrm{d}t
$$
for every $\gamma$ absolutely continuous and $F(\gamma)=+\infty$ otherwise, where $|\dot{\gamma_t}|$ is the metric speed of $\gamma$, namely
$$
|\dot{\gamma}_t|= \lim_{h\to 0} \frac{d(\gamma_{t+h},\gamma_t)}{h} 
$$
for a.e. $t\in[0,1]$. Can I conclude that sublevels of $F$ are compact?
Equicontinuity is trivial, but it is not so easy to prove (if possible) pointwise relatively compactness in order to apply Arzelà-Ascoli.
Thank you.
 A: At the least, you can conclude that the sublevel sets are pre-compact (I am not sure if they are actually closed). Let us prove this using Arzela-Ascoli.
From Cauchy-Schwarz, you can get equicontinuity (as you have already suggested). In fact you can conclude that for $\gamma \in A_K:=F^{-1}[0,K]$, one has $d(\gamma_s,\gamma_t) \leq \int_s^t |\dot \gamma(s)|ds\leq K|t-s|^{1/2}$. 
Thus, we just need to show pointwise relative compactness. Since $X$ is a complete space (hence closed subsets are complete with respect to the same metric), this amounts to showing that for each $t \in [0,1]$, the set $C_{K,t}:=\{\gamma(t):\gamma \in A_K\}$ is totally bounded.
For $R>0$ we define the compact set $L_R:=f^{-1}[0,R]$. Fix a curve $\gamma \in C([0,1],X)$ and let $M>0$. Suppose that $\int_0^1 |\dot \gamma(s)|^2 ds\leq M$, so that $d(\gamma_s,\gamma_t) \leq M|t-s|^{1/2}$ for all $s,t \in [0,1]$. Now fix $t \in [0,1]$ and suppose that dist$(\gamma_t,L_R) =:\delta$; then for all $s \in [0,1]$ we have that dist$(\gamma_s,L_R) \geq \delta - M|t-s|^{1/2}$, and thus $f(\gamma_s) \geq R$ for $s \in [t-(\delta/M)^2,t+(\delta/M)^2]\cap[0,1]$. In particular, $\int_0^1 f(\gamma_s)ds \geq (\delta/M)^2R$. Summarizing this paragraph, we showed that if $\int_0^1 |\dot \gamma(s)|^2ds \leq M$, then for all $R>0$ one has the bound $\int_0^1 f(\gamma_s)ds \geq \sup_{t \in [0,1]}(\text{dist}(\gamma_t,L_R)/M)^2R$.
In particular, if $F(\gamma)\leq K$, then $\int f(\gamma_s)ds \leq K$, therefore setting $M=K$ in the preceding paragraph gives $\sup_{t \in [0,1]} \text{dist}(\gamma_t,L_R) \leq (K/R)^{1/2}K = K^{3/2}/R^{1/2}.$ This bound holds for every $R>0$.
Now we finally prove that $C_{K,t}$ is totally bounded. Fix $K$ and $t$, and let $\epsilon>0$. Choose $R$ large enough so that $K^{3/2}/R^{1/2}<\epsilon/2$. By compactness, there exist $x_1,...,x_N \in L_R$ such that $L_R \subset \bigcup_1^N B(x_i,\epsilon/2)$. Then for any $\gamma \in A_K$, we know dist$(\gamma_t, L_R) < \epsilon/2$, thus $C_{K,t} \subset \bigcup_1^N B(x_i,\epsilon)$. This proves the claim.
