I have a question concerning the proof of Theorem 3.3 in Dacorogna's book "Introduction to the calculus of variations"(from 2004). This theorem deals with the standard existence and uniqueness result for minimizers of a functional. I am only interested in the uniqueness part.
Let me briefly state the (simplified) setting: Let $\Omega\subset\mathbb{R}^n$ be an open bounded domain with Lipschitz boundary and let $f\in C^0(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^n)$, $f=f(x,u,\xi)$ , be the Lagrangian of the functional $$ I(u):=\int_\Omega f(x,u(x),Du(x))dx $$ Suppose that there exists a minimizer in $W_0^{1,p}(\Omega)$ (which we can ensure by imposing some growth, convexity and coercivity assumptions on $f$). The theorem now states that if $$ (u,\xi)\mapsto f(x,u,\xi) $$ is strictly convex for every $x\in\bar{\Omega}$, then the minimizer is unique. The proof of this fact is clear to me: One takes two minimizers $u$ and $v$ and uses convexity of the above mapping to show that $(u+v)/2$ is also a minimizer. By using convexity once again, one can then deduce that $$ \frac{1}{2}f(x,u,Du) + \frac{1}{2}f(x,v,Dv) = f(x,\frac{u+v}{2},\frac{Du+Dv}{2}) \; \text{ a.e. in } \Omega. \tag{1}\label{1} $$ If $(u,\xi)\mapsto f(x,u,\xi)$ is strictly convex, then we can conclude that $u=v$ and $Du=Dv$ almost everywhere.
Now in Remark 3.4 (ii) the author claims that uniqueness holds under a slightly weaker condition, namely that $(u,\xi)\mapsto f(x,u,\xi)$ is convex and either $u\mapsto f(x,u,\xi)$ is strictly convex or $\xi\mapsto f(x,u,\xi)$ is strictly convex. The author claims that this follows directly from the proof sketched above, but I don't see how.
Convexity of $(u,\xi)\mapsto f(x,u,\xi)$ leads us again to the above equality \eqref{1}. I guess that strict convexity of $u\mapsto f(x,u,\xi)$ would give us $u=v$ a.e., and strict convexity of $\xi\mapsto f(x,u,\xi)$ would give us $Du = Dv$ a.e. which in combination with the zero boundary data implies $u=v$ almost everywhere. But I don't see how the exact argument works. Can somebody explain it to me or give me a hint?
I don't think that this is true. Let us consider the following example: $\Omega = (0,1)$ and \begin{equation*} f(x, u, \xi) = \begin{cases} \lvert u - x \, \xi \rvert^2 & \text{if } x \in (0,1/2), \\ \lvert u + (1-x) \, \xi \rvert^2 & \text{if } x \in (1/2,1). \end{cases} \end{equation*} This functional is convex and strictly convex w.r.t. $u$.
For $u \equiv 0$ and \begin{equation*} v(x) = \begin{cases} x & \text{if } x \in (0,1/2), \\ 1-x & \text{if } x \in (1/2,1) \end{cases} \end{equation*} we have $I(u) = I(v) = 0$. Hence, the minimizer is not unique.