Convexity and shortest curve I found a PDF in which there were the two following questions:

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*For any function $g \in C^1$, let $F(g) = \int_{0}^{1} \sqrt{1 + (g’(t))^2}dt$. Show that if f, g are $C^1$ and $\lambda \in ]0, 1[$, then $F(\lambda f + (1 - \lambda)g) \leq \lambda F(f) + (1-\lambda) F(g)$.


*If $a, b$ are given in $\mathbb{R}$, show that there is a function f with $f(0)=a, f(1)=b$ such that for each function g with $g(0)=a, g(1)=b$, we have $F(f) \leq F(g)$.
I’m good with question 1) (I used the convexity of $x \mapsto \sqrt{1+x^2}$).
I get the meaning of question 2 (basically, the goal is to show that amongst all $C^1$ functions, then it’s the linear function satisfying the conditions that is the shortest). However, i can’t really see how to use question 1 to answer question 2 ...
Thank you.

The function $\phi(x)= \sqrt{1+x^2}$ is convex and the map $f \to f'$ is linear.
Since $\phi$ is convex we have $\phi(x+h) \ge \phi(x) + \phi'(x)h$.
Hence $F(f+h)=\int_0^1\sqrt{1+((f+h)'(t))^2 } dt \ge F(f)+\int_0^1 {f'(t) \over \sqrt{1+(f'(t))^2}}h'(t) dt $.
Let $f(t) = a+t(b-a)$, note that $f'(t) = b-a$ and so
$F(f+h) \ge F(f) + {{b-a} \over \sqrt{1+(b-a)^2}} (h(1)-h(0))$.
If $h=g-f$ then we see that $h(0)=h(1)$ and so $F(g) \ge F(f)$.
