Composition between Frechet and Gateaux derivative

Question

I am currently dealing with the two terms "Gateaux-derivation" and "Frechet-derivation" and would like to know if there is a Frechet differentiable function $f$ and Gateaux differentiable function $g$, so that $(g\circ f)$ is no longer Gateaux differentiable?

For the definitions, I'll link to the Wikipedia article:

Definition of Frechet derivative: https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative

Definition of Gateaux derivative: https://en.wikipedia.org/wiki/Gateaux_derivative .

Answer 1

Take $$ g(x,y) = \begin{cases} 1 & \text{ if } x\ne0 \text{ and } y=x^2\\ 0 & \text{ else } \end{cases} $$ and $$ f(x) = (x,x^2). $$ Then $g:\mathbb R^2 \to \mathbb R$ is Gateaux, $f:\mathbb R \to \mathbb R^2$ is Frechet differentiable, but $g\circ f : \mathbb R \to \mathbb R$ is discontinuous, hence it cannot be (Gateaux) differentiable.

Answer 2

The composition can even be discontinuous: Consider $f \colon \mathbb R^2 \to \mathbb R$ defined via $$ f(x,y) = \exp\left(-\frac{(y-x^2)^2}{x^6}\right)$$ for $x \ne 0$ and $f(0,y) = 0$. One can check that this function is Gateaux differentiable everywhere and discontinuous at zero. Together with $g \colon \mathbb R \to \mathbb R^2$, $g(t) = (t,t^2)$, we get a discontinuous function $f \circ g \colon \mathbb R \to \mathbb R$, which cannot be Gateaux differentiable.

Answer 3

Hello @grew you are right. Here are the calculations for the Gateaux derivative of the function

$$f(x,y)= \begin{cases} \exp\left(-\frac{(y-x^2)^2}{x^6}\right)&\text{for}~x\neq 0,\\ 0&\text{o.w.}\end{cases}$$ in the point $(x_0,y_0)=(0,0).$ Let $h$ be a valid direction and $\epsilon>0$. Then $$ \frac{f(x_0+\epsilon h,y_0+\epsilon h)-f(x_0,y_0)}{\epsilon}=\frac{f(x_0+\epsilon h,y_0+\epsilon h)}{\epsilon}=\frac{\exp\left(-\frac{(1-\epsilon h)^2}{\epsilon^4h^4}\right)}{\epsilon}.$$ And you get $$\frac{1}{\epsilon}\exp\left(-\frac{(1-\epsilon h)^2}{\epsilon^4h^4}\right)\stackrel{ l'Hospital}{\Rightarrow}\lim\limits_{\epsilon\to 0}\frac{f(x_0+\epsilon h,y_0+\epsilon h)-f(x_0,y_0)}{\epsilon}=0.$$ And this is independent of the direction $h$.