Rigorous proof of existence of a unique function whose arc-length is minimal.

Question

Let $a,b, \alpha, \beta \in \Bbb R$, $F = \{ f \in C^1([a,b]) \mid f(a)=\alpha, f(b)=\beta\}$ and $L : F \to \Bbb R^+, f \mapsto \int_{a}^b \sqrt{1+f'(x)^2} dx$. Prove that $L$ has a unique minimum.

This is a classical problem in calculus of variations and using the Euler-Lagrange equation one finds that the straight line passing through $(a, \alpha)$ and $(b, \beta)$ is a stationary point. But how can I show rigorously that this point is a minimum? Another question: does someone see a proof without using the Euler-Lagrange equation?

Answer 1

Consider the function $G(t) := \sqrt{1+t^2}$, so that $L(f) = \int_a^b G(f'(x))dx$. You should be able to check that

• The function $G$ is convex, so by Jensen's inequality $G(\frac{1}{b-a} \int_a^b f') \le \frac{1}{b-a}\int G(f')$. This tells you that the linear function is in fact minimizing.
• Even better, $G$ is strictly minimizing. This tells you that the minimizer is unique. Otherwise, if $f_1,f_2$ were two minimizers, $\frac{f_1+f_2}{2}$ would give an even smaller value of $L$.