# If $u$ is an extremal point of a functional $E$ and $γ$ is a smooth curve with $γ(0)=u$, then $0$ is an extremal point of $E\circ\gamma$

Let $$X$$ be a $$\mathbb R$$-Banach space, $$M\subseteq X$$ be open, $$M_0\subseteq M$$ be closed, $$E\in C^1(M)$$, $$u\in M_0$$ be an extremal point of $$E$$, $$\varepsilon_0>0$$ and $$\gamma\in C^1(-\varepsilon_0,\varepsilon_0),M_0)$$ with $$\gamma(0)=u$$ and $$\gamma'(0)=x_0$$ for some given $$x_0\in X$$.

Why can we conclude that $$(E\circ\gamma)'(0)=0$$? (And how can we show that $$\gamma$$ exists?)

Intuitively, it's clear to me that $$0$$ should be an extremal point of $$E\circ\gamma$$ and hence the claim holds true. The continuity should ensure that $$E$$ cannot jump along the curve away from the neighborhood of the local minimum/maximum $$u$$. But how can we prove this rigorously?

EDIT: You may assume that $$M=\{\Phi=0\}$$ for some $$\Phi\in C^1(M,Y)$$, $$Y$$ being another $$\mathbb R$$-Banach space. (I guess this implies that $$M$$ is some kind of Manifold)

• $\gamma$ does not exist if (e.g.) $x_0$ is not tangential to $M_0$. In general, $M_0$ might not have a smooth structure. – Arctic Char Aug 6 at 10:04
• @ArcticChar Please take note of my edit. – 0xbadf00d Aug 6 at 12:37