Let $V,W$ be nonzero normed spaces over $\mathbb{K}$ such that $V$ is finite-dimensional.

Let $E$ open in $\mathbb{K}$ and $p\in E$.

Let $f:E\rightarrow W$ be Gâteaux-differentiable at $p$.

Is $f$ necessarily Fréchet-differentiable at $p$ in this case?

I think this is not true in general, but cannot find a counterexample. What would be a counterexample?


This counterexample comes from the Wikipedia page for Frechet Derivative

Consider the function $f$ that is $0$ at $(x, y) = (0, 0)$ and $$f(x, y) = \frac{x^3}{x^2+y^2}$$ otherwise.

Its Gateaux derivative $g$ as a function of the "direction" $h \in \mathbb{R}^2$ at $(0, 0)$ is $0$ if $h = (0, 0)$ and $$g(h_1, h_2) = \frac{h_1^3}{h_1^2 + h_2^2}$$ otherwise.

Since this is not a linear function of $h$, the function $f$ has no Frechet derivative.

  • $\begingroup$ I can't understand at $h=(0,0)$, $f$ is Gateaux differentiable, $g=0$ is linear , continuos , why $f$ is not Frechet differentiable at $(0,0)$. $\endgroup$ – Thu Le Jun 16 '18 at 2:57
  • $\begingroup$ @Chloe.Sannon $g(h_1, h_2)$ is not linear. Yes it is zero at $(0,0)$ but otherwise it varies non-linearly with $h_1$ and $h_2$. $\endgroup$ – muaddib Jun 20 '18 at 1:16

Define $$ f(u,v)= \begin{cases} 1 &\hbox{if $v=u^2$ and $(u,v) \neq (0,0)$} \\ 0 &\hbox{otherwise}. \end{cases} $$ It is easy to check that the Gateaux derivative at $(0,0)$ exists, but $f$ is not even continuous.

  • $\begingroup$ Sorry, you are wrong: $f(t,0)=0$. Indeed $f$ is constant on any straight line thru the origin. $\endgroup$ – Siminore Aug 19 '15 at 8:05
  • $\begingroup$ Yes, now you are right. I always forget to exclude the origin from the first case. $\endgroup$ – Siminore Aug 19 '15 at 14:26

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