I came across the following formulation of the problem.
Minimize the functional $L[u]$ given by
$L[u] = \int^b_a \sqrt{(1+(u'(x))^2}$ over $U = \{u\in C([a,b])\cap C^1((a,b)):u(a)=\alpha, u(b) = \beta\}$
What I am wondering is, shouldn't we require $u(x)\in C([a,b])\cap C^1([a,b])$ instead of $u(x)\in C([a,b])\cap C^1((a,b))$? That is, don't we need $u(x)$ to be continuously differentiable up to the boundary?
If a function is continuolsy differentiable only in the open interval, how do we know that $L[u]$ is well defined?
Does the continuity of u in the closed interval $[a,b]$ somehow gurantee that?
I am curious about this because I know that the function $f(x) = 1/x$ is in $C^1((0,1))$, but $\int^1_0f'(x)dx $ is not well defined.
Some people would consider $u\in C^1(a,b)$ to mean that $u$ is once continuously differentiable in $(a,b)$ and $$ \sup_{(a,b)} |u(x)| + \sup_{(a,b)} |u'(x)| <\infty. $$ In this case clearly $L$ is defined.
If no assumption is made on the derivative, you're right that $L$ will not be well defined: Take $a=0$, $b=1$ and $u(x)=x\sin(x^s)$, then for $s< -10$ say, we have $u$ is in your space with $\alpha=0$ but $u'$ is not integrable.