I'm reading Van Brunt's book on the calculus of variations and I have been asking myself many questions about the behavior of some functional, but among all of them I have highlighted the following problem that I have not yet been able to answer. I would like to know if someone can give me an answer or some counterexample about this question:
Let $V \colon = C^{1}([x_0 , x_1] ; \mathbb{R})$, $\mathcal{F} \colon = \{ y \in C^{1}([x_0 , x_1] ; \mathbb{R}) \colon y(x_0) = y_0 , y(x_1) = y_1\}$ and a functional of the form \begin{align*} F \colon & V \rightarrow \mathbb{R} \\ & y \mapsto F[y] = \int_{x_0}^{x_1}{f(x,y,y')}dx \end{align*} where $f \in C([x_0 , x_1]\times \mathbb{R}^2 ; \mathbb{R})$.
Now, my question is this: If $F$ is convex, then $f^x$ is convex, $\forall x \in [x_0 , x_1]$?
Note: We define for each $ x \in [x_0, x_1] $: \begin{align*} f^x \colon & \mathbb{R}^2 \rightarrow \mathbb{R} \\ & (y,y') \mapsto f^{x}(y,y') \colon = f(x,y,y') \end{align*}
Definition:
Let us consider $V \colon = C^{1}([x_0 , x_1]; \mathbb{R})$ and a functional of the form: \begin{align*} F \colon & V \rightarrow \mathbb{R} \\ & y \mapsto F[y] = \int_{x_0}^{x_1}{f(x,y,y')}dx \end{align*} where $f \in C([x_0 , x_1]\times \mathbb{R}^2 ; \mathbb{R})$. We say that F is convex iff $F[(1-t)y + tz] \leq (1-t)F[y] + tF[z]$, $\forall y,z \in V$ y $t \in [0,1]$