What is an example of Gâteaux differentiable but not Fréchet differentiable at a point in a finite-dimensional space?

Question

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?

Answer 1

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.

Answer 2

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.