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.



Special version of Tonelli’s theorem
$\renewcommand\bar\overline$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](http://mathonline.wikidot.com/the-eberlein-smulian-theorem)), 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](https://en.wikipedia.org/wiki/Mazur%27s_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 $u\in X$.


