Here is a problem I deal with in Calculus of Variation:
Let the functional $\;I:\mathcal A \rightarrow [0,+\infty]\;$ and assume $\;f^{*}\in \mathcal A\;$ is a minimizer of $\;I\;$. Note that $\;f^{*}:[a,b]\rightarrow \mathbb R^m\;$. Given $\;x_0,x_1\;$ such that $\;a\lt x_0 \lt x_1 \lt b\;$, for small $\;\vert h \vert \;$ define $\;f_h:[a,b+h)\rightarrow \mathbb R^m\;$ by setting:
$\;f_h(x)= \begin{cases} f^* (x),\;x\in (a,x_0]\\f^* (φ(x)), \; x\in [x_0,x_1+h]\\f^* (x-h), \; x\in [x_1+h,b+h)\\ \end{cases}\;$
where$\;φ:[x_0,x_1+h]\rightarrow[x_0,x_1]\;$is a linear function. One can show that $\;f_h\in \mathcal A\;$.
EDIT: $\;\mathcal A \subset W^{1,2}_{loc} (l^f_-,l^f_+)\;$ and here: $\;l^{f^*}_-=a,\;l^{f^*}_+=b,\;l^{f_h}_-=a,\;l^{f_h}_+=b+h\;$
My Question:
Why is $\;\frac {d}{dh} I(f_h) {\vert}_{h=0} =0\;(1)$ true?
As I know, since $\;f^*\;$ is a minimizer of $\;I\;$ , for small $\;h\in \mathbb R\;$ and for every suitable $\;g\;$, it should be $\;\frac {d}{dh} I(f^* +hg) {\vert}_{h=0} =0\;(2)$as long as $\;f^* +hg\in \mathcal A\;$. Unfortunately I can't find by myself the connection between $\;(1),(2)\;$.
Any help -even hints- would be valuable. Thanks in advance!