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.


1 Answer 1


$\renewcommand\bar\overline$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 $u\in X$.


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