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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?

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    $\begingroup$ This even fails for $E=\mathbf R$; take $f(x)=|x|$. And it is true if $df(x_0;v)$ is defined using $\lim_{\varepsilon\to 0}$. $\endgroup$
    – Dirk Werner
    Commented Jul 10, 2019 at 18:57

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As remarked by Dirk Werner in a comment, it's not true that directionals derivatives need to vanish at a local minimum point. There is a whole zoo of conditions that can be used. If you want to see the greater picture, you may consult the book "Nonsmooth Analysis" by Schirotzek. As a simple start you could look up the convex subgradient/subderivative.

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  • $\begingroup$ But if we add differentiability in the minimum point isn't it true? $\endgroup$
    – Bogdan
    Commented Jul 10, 2019 at 19:58
  • $\begingroup$ Well, the point is that there are a lot of different notions of differentiability for functions on Banach spaces (for non-compete normed spaces, I don't know anything). $\endgroup$
    – Dirk
    Commented Jul 10, 2019 at 19:59
  • $\begingroup$ The usual Frechet differentiability. In $R^N$ it is enough. $\endgroup$
    – Bogdan
    Commented Jul 10, 2019 at 20:01
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    $\begingroup$ In case of Frechet differentiability a zero derivative is necessary for a local minimum. $\endgroup$
    – Dirk
    Commented Jul 10, 2019 at 20:03
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    $\begingroup$ Gâteaux differentiability is good enough, on a normed space. (Proof: Look at $t\mapsto f(x_0+tv)$.) $\endgroup$
    – Dirk Werner
    Commented Jul 12, 2019 at 21:02

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