Special version of Tonelli’s theorem I am trying to prove this theorem. I have not found anything similar to it on the internet.
Special version of Tonelli’s theorem
Assume that the functions $f(x,u): [a,b] \times \mathbb{R} \to \mathbb{R}$,  $ g(x, \xi): [a,b] \times \mathbb{R} \to \mathbb{R}$ are continuous, $f$ is bounded below, $g$ is convex in $\xi$ and satisfies
$$\exists r>1,\, \exists C>0\,\, \text{such that}\,\, g(x,\xi) \ge C| \xi|^r,\,\, \forall (x, \xi) \in [a,b] \times \mathbb{R}.$$
Then there exists a minimizer of the functional
$$
J[u] = \displaystyle\int\limits_a^b \big(f(x,u(x)) + g(x,u'(x))\big) dx
$$ in the space $X= \{ u \in AC([a,b]); u(a)=\alpha, u(b)= \beta \}.$
Proof.
Since $f$ is bounded then there is a real number $m \in \mathbb{R}$ such that $m (b-a)\le f(x,u(x)), \quad \forall (x,u(x)) \in [a,b] \times \mathbb{R}$. From the properties of $g$ we get
$$m+ C \int\limits_a^b |u'(x)|^r dx \leq J[u] \Rightarrow m+ C \| u'\|_{L^r[a,b]}^r \leq J[u]\,\,\, \forall u \in X.$$
We can see that $J[u]$ is bounded below and from the definition of the infimum there is a minimizing sequence $\{u_n\}_{n\in \mathbb{N}} \subset X$ such that
$$\underset{n \to \infty}{\lim} J[u_n] = \inf \{ J[u] | u \in X \}> -\infty \,\, \text{ in } \mathbb{R}.$$
and hence, $\{ u_n'\}_{n \in \mathbb{N}}$ is uniformly bounded, i.e. there is $N>0$ such that $\forall n >N$ we have
$$\| u'_n\|_{L^r[a,b]} \leq \left(\frac{J[u_N] -m}{c} \right)^\frac{1}{r}.$$
Now, since $\{u_n\}$ is equicontinuous, and uniformly bounded in $L^r[a,b]$, then according to the Arzelà-Ascoli theorem there is a subsequence $\{ u_{n_k} \}_{k \in \mathbb{N}}$ and $\overline{u} \in AC[a,b]$ such that $u_{n_k} \to \overline{u}$ uniformly, and $u'_{n_k} \to \overline{u}'$ in the sense of $L^r[a,b]$. $\blacksquare$
I am not sure if my last argument is right. I want to make it more rigorous. Although I found the general idea of the proof on page 140 in the book of Hansjörg Kielhöfer named (Calculus of Variations
An Introduction to the One-Dimensional Theory with Examples and Exercises) I have no idea about completing the proof of the theorem. Could you please help.
 A: $\renewcommand\bar\overline$Indeed, it is not obvious why "$u'_{n_k} \to \overline{u}'$ in the sense of $L^r[a,b]$".
Look at this example: $[a,b]=[0,2\pi]$, $u_n(x)=\dfrac{\sin nx}n$, $\bar u=0$. Then $u_n\to\bar u$ uniformly, but $u_{n_k}'\not\to\bar u'$ in $L^r$ for any increasing sequence $(n_k)$ of natural numbers, because $u_n'(x)=\cos nx$ and hence $\|u_n'\|_r^r=c_r:=\int_0^{2\pi}|\cos u|^r\,du>0$ for all $n$.
Note also that your argument does not use the condition that $g$ is convex.
(Also, your post seems to have hardly anything to do with the Tonelli theorem.)

Here is how to fix this. Using, as you did, the Arzelà–Ascoli theorem and then passing to a subsequence, without loss of generality (wlog) we may assume that $u_n\to \bar u$ uniformly. Also, you showed that the sequence $(u_n')$ is bounded in (the reflexive Banach space) $L^r$.
So, by the Eberlein–Shmulyan theorem (Kôsaku Yosida, Functional Analysis, Springer 1980, Chapter V, Appendix, section 4; alternatively, see e.g. this version), passing again to a subsequence, wlog we may assume that $u_n'\to v$ for some $v\in L^r$ in the weak topology of $L^r$.
Further, by Mazur's lemma, for each natural $n$ there exist a natural $N_n\ge n$ and nonnegative real numbers $a_{n,k}$ for $k\in\{n,\dots,N_n\}$ such that $\sum_{k=n}^{N_n}a_{n,k}=1$ and
\begin{equation*}
    v_n:=\sum_{k=n}^{N_n}a_{n,k} u_k'\to v \tag{0}
\end{equation*}
in $L^r$.
For $x\in[a,b]$, let now
\begin{equation*}
    w_n(x):=u_n(a)+\int_0^x v_n(t)\,dt
    =u_n(a)-\sum_{k=n}^{N_n}a_{n,k}u_k(a)+\sum_{k=n}^{N_n}a_{n,k}u_k(x). \tag{1}
\end{equation*}
Since $u_n\to \bar u$ uniformly and the $u_n$'s are uniformly bounded, we see that $w_n\to \bar u$ uniformly and the $w_n$'s are uniformly bounded. Therefore and because $f$ is continuous,  we have
\begin{equation*}
    J_1[w_n] := \int_a^b f(x,w_n(x))\, dx\to J_1[\bar u]=\lim_n J_1[u_n]. 
\end{equation*}
Also, by the convexity of $g(x,\xi)$ in $\xi$,
\begin{equation*}
    J_2[w_n] := \int_a^b g(x,w_n'(x))\, dx
    \le\sum_{k=n}^{N_n}a_{n,k}J_2[u_k]. 
\end{equation*}
Also, $J[w_n]=J_1[w_n]+J_2[w_n]$. So,
\begin{equation*}
\begin{aligned}
    \limsup_n J[w_n]&\le \lim_n J_1[w_n]+\limsup_n J_2[w_n] \\ 
    &\le \lim_n J_1[u_n]+\sum_{k=n}^{N_n}a_{n,k}\limsup_n J_2[u_n] \\ 
    &= \lim_n J_1[u_n]+\limsup_n J_2[u_n] \\ 
    &= \limsup_n (J_1[u_n]+J_2[u_n]) \\ 
    &= \limsup_n J[u_n]= \lim_n J[u_n]=\inf_{u\in X} J[u]. 
\end{aligned}
\end{equation*}
So, passing to a subsequence, wlog we may assume that
\begin{equation*}
    J[w_n]\to\inf_{u\in X} J[u].
\end{equation*}
Recall that $w_n\to \bar u$ uniformly. So, in view of (1) and (0),
\begin{equation*}
    \bar u(x)=\bar u(a)+\int_0^x v(t)\,dt
\end{equation*}
for $x\in[a,b]$, so that $\bar u\in AC$ and $\bar u'=v$ almost everywhere (a.e.).
It also follows that $w_n'=v_n\to v=\bar u'$ in $L^r$ and hence in measure. So, by the continuity of $f$ and $g$ and the Fatou lemma,
\begin{equation}
    J[\bar u]=J[\lim_n w_n]\le\liminf_n J[w_n]=\lim_n J[w_n]=\inf_{u\in X} J[u].
\end{equation}
It is also easy to see that $\bar u\in X$. Thus, $\bar u$ is a minimizer of $J[u]$ over $u\in X$.
