Solution for $F = \int^{b}_{a}{f(y'(x)) dx} \mbox{ min }$ with $\frac{d }{dp}f(p) = 0$ for $f(x,z,p) = f(p)$ and $f$ is convex.

Question

I tried to solve the following problem, but somehow my ideas seem to a dead end: For $D := \lbrace y \in C^1([a,b]; \mathbb{R}) |y(a) = y_1 , y(b) = y_2 \rbrace$ we define \begin{equation*} F(y) = \int^{b}_{a}{f(y'(x)) dx} \end{equation*} where \begin{equation*} \overline{y}(x) = \frac{y_2 - y_1}{b - a}(x - a) + y_1 \end{equation*} is a solution of the euler lagrange equation $\frac{d}{dx} \frac{d}{dp}f(p) = 0$ (where p = y') and which satisfie the boundary conditions. I want to show that if $f$ is convex, $\tilde{y}$ is a minimum of $F$ in $D$.
One of my ideas was to try to use the convex properties directly to show \begin{equation} F(\overline{y}) \leq F(h) \mbox{ for all } h \in D \end{equation} This leads me to another way of solving the problem: Since $f$ is convex and the integral is linear it is easy to see that $F$ is convex. Therefore I tried show that for all $h \in D$ there exists a $\delta > 0$ such that the inequality \begin{equation} F(\overline{y}) \leq F(\overline{y} + t(h-\overline{y})) \mbox{ for all } t \in [0, \delta] \end{equation} holds. I tried to figure out $\delta > 0$ which depends on $h \in D$ such that \begin{align*} F(\overline{y}) &= \int^{b}_{a}{f \left( \frac{y_2 - y_1}{b - a} \right) dx} = (b-a) f \left( \frac{y_2 - y_1}{b - a} \right) \\ &\stackrel{!}{\leq} F(\overline{y} + t(h-\overline{y})) = \int^{b}_{a}{f(t h'(x) + (1-t)\overline{y}'(x)) dx} \\ &= \int^{b}_{a}{f\left(t h'(x) + (1-t)\frac{y_2 - y_1}{b - a} \right) dx} \mbox{ for all } t \in [0,\delta]. \end{align*} From then on I tried to use the mean value theorem for $\frac{u_2 - u_1}{x_2 - x_1}$ which wasn't helpful. I also tried to do some equivalent transformation to simplify the Problem but since I can't tell anything about $f$ it doesn't help. I would be thankful if anyone can give me a hint, or can tell me if my way of approaching this problem is correct

Answer 1

I managed to proof it. If someone is interested in the proof:

By convexity of $f$ we know that

\begin{equation*} f(x) \geq f(y) + f'(y)(x-y) \end{equation*} Using this inequality we see that for any $u \in D$ \begin{equation} f(u'(x)) \geq f(\overline{y}'(x)) + f'(\overline{y}'(x))(u'(x) - \overline{y}'(x)) \end{equation} It is clear that $\overline{y}'(x) = \frac{y_2-y_1}{b - a}$ and therefore integrating both parts over $(a,b)$ we obtain \begin{equation} \int^{b}_{a}{f(u'(x)) dx} \geq \int^{b}_{a}{f(\overline{y}'(x))dx}. \end{equation} Therefore $\overline{y}$ is a minimizer of $F$ in $D$.