An example of $f$ and $a$ such that $f$ is Gâteaux but not Fréchet differentiable at $a$ I'm reading this lecture note about differentiability.

Let $(X, |\cdot|_X)$ and $(Y, |\cdot|_Y)$ be normed spaces. Let $A$ be an open subset of $X$ and $f: A \to Y$. The directional derivative $f^{\prime}(a) (v)$ of $f$ at $a \in A$ along direction $v \in X$ is the limit (if exists)
$$
f^{\prime}(a)(v) :=\lim _{t \rightarrow 0} \frac{f(a+t v)-f(a)}{t}.
$$
We shall say that $f$ is:

*

*Gâteaux differentiable at $a$ if there exists $x^{*} \in X^{*}$ such that $f^{\prime}(a)(v)=x^{*}(v)$ for each $v \in X$ (that is, $f^{\prime}(a)$ is everywhere defined, real-valued, linear and continuous); Then $x^{*}$ is called the Gâteaux differential (or derivative) of $f$ at $a$, and is denoted by $d f(a)$.

*Fréchet differentiable at $a$ if there exists $x^{*} \in X^{*}$ such that
$$
\lim _{h \to 0} \frac{f(a+h)-f(a)-x^{*}(h)}{|h|_X}=0 .
$$
Then $x^{*}$ is called the Fréchet differential (or derivative) of $f$ at $a$, and is denoted by $\partial f(a)$.


Could you provide an example of $f$ and $a$ such that $f$ is Gâteaux but not Fréchet differentiable at $a$?

I only find an example in which $f^{\prime}(a)$ exists and is continuous, but not linear.

For example, consider the real-valued function $F$ of two real variables defined by
$$
F(x, y)= \begin{cases}\frac{x^3}{x^2+y^2} & \text { if }(x, y) \neq(0,0), \\ 0 & \text { if }(x, y)=(0,0).\end{cases}
$$
Then
$$
d F(0,0) (a, b)=\left\{\begin{array}{ll}
\frac{F(\tau a, \tau b)-0}{\tau} & \text{if } (a, b) \neq(0,0), \\
0 & \text{if } (a, b)=(0,0).
\end{array}= \begin{cases}\frac{a^3}{a^2+b^2} & \text{if }  (a, b) \neq(0,0), \\
0 &\text{if } (a, b)=(0,0).\end{cases}\right.
$$

 A: I have just found two examples from this Wikipedia page.


*

*In another situation, the function $f$ given by
$$
f(x, y)= \begin{cases}\frac{x^3 y}{x^6+y^2} & (x, y) \neq(0,0) \\ 0 & (x, y)=(0,0)\end{cases}
$$
is Gateaux differentiable at $(0,0)$, with its derivative there being $g(a, b)=0$ for all $(a, b)$, which is a linear operator. However, $f$ is not continuous at $(0,0)$ (one can see by approaching the origin along the curve $\left(t, t^3\right)$ ) and therefore $f$ cannot be Fréchet differentiable at the origin.


*A more subtle example is
$$
f(x, y)= \begin{cases}\frac{x^2 y}{x^4+y^2} \sqrt{x^2+y^2} & (x, y) \neq(0,0) \\ 0 & (x, y)=(0,0)\end{cases}
$$
which is a continuous function that is Gateaux differentiable at $(0,0)$, with its derivative at this point being $g(a, b)=0$ there, which is again linear. However, $f$ is not Fréchet differentiable. If it were, its Fréchet derivative would coincide with its Gateaux derivative, and hence would be the zero operator $A=0$; hence the limit
$$
\lim _{\|h\|_2 \rightarrow 0} \frac{|f((0,0)+h)-f(0,0)-A h|}{\|h\|_2}=\lim _{h=(x, y) \rightarrow(0,0)}\left|\frac{x^2 y}{x^4+y^2}\right|
$$
would have to be zero, whereas approaching the origin along the curve $\left(t, t^2\right)$ shows that this limit does not exist.
