Generalization of Fermat stationary point theorem Let $f:E\to\mathbb{R}$ a functional (here $E$ is a normed vector space). Is it true that if $x_0\in E$ is a local minimum for $f$, then all the directional derivatives are 0?
We have the derivative of $f$ in $x_0$ with respect to the direction $v\in E$ defined by: $df(x_0;v)=\lim\limits_{\varepsilon\to 0}\dfrac{f(x_0+\varepsilon v)-f(x_0)}{\varepsilon}$
Or if not, in what conditions is this true? 
Is there a generalization of Fermat Stationary point Theorem for such functionals? I know that in $\mathbb{R}^N$ this is a classical result, but in other spaces?
 A: If you assume in addition that $f$ is Frechet-differentiable at $x_0$, then it follows that all the directional derivatives vanish. To prove this, fix a direction $v\in E$, and define the map $\lambda : \Bbb{R} \to E$ by
\begin{align}
\lambda(t) = x_0 + tv
\end{align}
Since $f$ is assumed to have a local minimum at $x_0$, it follows that $f \circ \lambda$ has a local minimum at $0$. Since $f$ is Frechet-differentiable at $x_0$ by assumption, and $\lambda$ is Frechet differentiable everywhere (it is affine), it follows by the chain rule that the composite mapping $f \circ \lambda$ is differentiable at $0$. Hence, by the single variable version of the theorem, it follows that
\begin{align}
(f \circ \lambda)'(0) = 0.
\end{align}
The LHS is precisely the directional derivative you wish to show is zero.

You need the differentiability assumption on $f$, because otherwise it's not true. Let $E=\Bbb{R}$, and define $f(x) = |x|$. Then if $v \in \Bbb{R}$ is non-zero, only the one-sided limit exists, and equals $|v|$ (which is not zero).
A: Your (two-sided) directional derivative is zero whenever it exists.
This follows from
$$f(x_0 + \varepsilon \, v) \ge f(x_0)$$
whenever $|\varepsilon|$ is small.
Indeed, $\varepsilon \searrow 0$ implies $df(x_0;v) \ge 0$, whereas $\varepsilon \nearrow 0$ implies $df(x_0;v) \le 0$.
