Thread: Fréchet derivative and Gâteaux derivative

1. Fréchet derivative and Gâteaux derivative

Lets say we have the scalar field $f(x,y) = \begin{cases} \frac{xy}{\sqrt{x^2+y^2}} \ \text{if} \ (x,y) \neq (0,0) \\ 0 \ \ \ \ \ \ \ \ \ \text{if} \ (x,y) = (0,0) \end{cases}$

So by inspection, it seems that $f(x,y)$ is continuous at $(0,0)$. Now the partial derivatives of $f$ are not continuous at $(0,0)$. So $f$ is not differentiable at $(0,0)$. So this is not Fréchet differentiable? But could it be Gâteaux differentiable?

2. Originally Posted by heathrowjohnny
Lets say we have the scalar field $f(x,y) = \begin{cases} \frac{xy}{\sqrt{x^2+y^2}} \ \text{if} \ (x,y) \neq (0,0) \\ 0 \ \ \ \ \ \ \ \ \ \text{if} \ (x,y) = (0,0) \end{cases}$

So by inspection, it seems that $f(x,y)$ is continuous at $(0,0)$. Now the partial derivatives of $f$ are not continuous at $(0,0)$. So $f$ is not differentiable at $(0,0)$. So this is not Fréchet differentiable? But could it be Gâteaux differentiable?
The Frechet derivative is for functions $f:\mathbb{R}^n\mapsto \mathbb{R}^m$, and it is a matrix (or a coloumn or row matrix). If it exists it is expressed in terms of its partial derivatives. However, existence of its partial derivatives does not gaurenntee Frechet differenciablility. There is a theorem which says if the partial derivatives are continous then the function is differenciable. I am not sure about the converse. I think it is wrong. You need to justify non-Frechet differenciability in another way. I am not familar with Gateaux derivative.

Note: The function is continous because $\left| \frac{xy}{\sqrt{x^2+y^2}} \right| \leq \frac{ \frac{1}{2}(x^2+y^2) }{\sqrt{x^2+y^2}} = \frac{1}{2}\sqrt{x^2+y^2}$

3. The continuity of partial derivatives is not a necessary condition for differentiability.