The functional is given by:
$J(y) = \int_{a}^{b} \sqrt{1 + y'(x)^2} dx$
and I need to prove that it is lower semi-continuous with respect the norm:
$|| y || = \max_{a \leq x \leq b} |y(x)|$
and $y \in C^1(a,b)$.
My principal problem is: how can I relates the condition $||y - y_0|| < \delta$ with their respective derivatives?.
in advance thank you very much
p.d. sorry by my English.
Let $\{y_n\}$ be a sequence that converges to $y$ in the given norm. Assume $J(y) < \infty$ (the following argument can be modified without too much trouble if $J(y) = \infty$). Let $\epsilon > 0$ be given. There exists a partition $P = \{x_0,\ldots,x_N\}$ of $[a,b]$ with the property that $$ J(y) = \int_a^b \sqrt{1 + y'(x)^2} \, dx < \sum_{k=1}^N \sqrt{(x_k - x_{k-1})^2 + (y(x_k) - y(x_{k-1}))^2} + \epsilon $$ Since $$ \sqrt{(x_k - x_{k-1})^2 + (y(x_k) - y(x_{k-1}))^2} = \lim_{n \to \infty} \sqrt{(x_k - x_{k-1})^2 + (y_n(x_k) - y_n(x_{k-1}))^2} $$ and $$ \sum_{k=1}^N\sqrt{(x_k - x_{k-1})^2 + (y_n(x_k) - y_n(x_{k-1}))^2} \le \int_a^b \sqrt{1 + y_n'(x)^2} \, dx = J(y_n) $$ we obtain $$ J(y) \le \liminf_{n \to \infty} J(y_n) + \epsilon. $$ Now let $\epsilon \to 0^+$.