Example of function that is Gâteaux-differentiable but not Fréchet-differentiable I am looking for an example of a function that is Gateaux-differentiable but not Fréchet-differentiable. I know that there is a lot of example of function $f: \mathbb R^2 \to \mathbb R$ that satisfies this property. An example is
$$f(x, y) = 
\begin{cases}
\frac{x^3}{x^2 + y^2} & \text{if } (x, y) \neq 0,\\
0 & \text{otherwise.}
\end{cases}$$
But since the Fréchet and Gâteaux derivative are defined for Banach spaces in general, I am looking for a more fancy example such as a function from $L^p$ to $L^q$ for instance. Any idea ?
 A: The typical way of constructing Gateaux, but not Frechet differentiable functions is to consider functions which are homogeneous of degree 1, but not (bounded and) linear. See the argument in section 3.7, page 148 of Loomis and Sternberg, or this answer of mine
for a slight generalization. Note that this is precisely the reason why the $f$ you suggested has all directional derivatives, yet is not Frechet differentiable at the origin. Other examples like $f(x,y,z)=\frac{x^2yz^2}{x^4+y^4+z^4}$ if $(x,y,z)\neq (0,0,0)$ and $0$ otherwise will also have all directional derivatives, yet not be Frechet differentiable.
As a concrete example in infinite dimensions, fix a finite measure space $(X,\mathfrak{M},\mu)$, and for $p\in [1,\infty]$, let $L^p$ denote $L^p(\mu,\Bbb{R})$, so we only consider real-valued functions. Consider the function $T:L^2\times L^2\times L^{\infty}\to L^1$ defined as
\begin{align}
T(f,g,h):=
\begin{cases}
\dfrac{fgh}{\displaystyle\int_X(f^2+g^2+h^2)\,d\mu}&\text{if $(f,g,h)\neq 0$}\\\\
0 & \text{if $(f,g,h)=0$}
\end{cases}
\end{align}
Note that by Holder's (or in this case Cauchy-Schwarz) inequality, $fgh$ does indeed lies in $L^1$ and because $f,g\in L^2$ and $h\in L^{\infty}$ with $\mu$ being a finite measure, the denominator will be finite if $(f,g,h)\neq (0,0,0)$; furthermore because we're only dealing with real-valued functions, the denominator only vanishes when $(f,g,h)=0$, so everything is well-defined.
Now, $T$ is easily verified to be homogeneous of degree $1$, but is not a bounded linear transformation. Hence, all the directional derivatives exist, but $T$ is not Frechet-differentiable.
