If a function is separately differentiable is it diferentiable?

Question

Let $X $, $Y $ and $Z $ normed vector spaces $$f:X \times Y \to Z$$ Such that $f(x, \cdot)$ is differentiable for all $x \in X $ and $f (\cdot ,y)$ is differentiable for all $y \in Y $. Is $f $ differentiable? If no, assumpting that $f $ is bilinear can we conclude that $f $ is differentiable?

Answer 1

This is not even true for $X=Y=Z=\mathbb{R}$. Your condition, in other words, is that the partial derivatives exist. This is not sufficient for differentiability; a standard counterexample is $$f(x,y) = \frac{xy}{x^2+y^2}$$ with $f(0,0)=0$. Clearly $f(x,\cdot)$ is a differentiable function for every $x \ne 0$, and $f(0,\cdot)\equiv 0$ which is also differentiable. Similarly, $f(\cdot, y)$ is differentiable for every $y$. But $f$ is not even continuous at $(0,0)$, as you may see by approaching the origin along the line $y=x$.

At Discontinuous bilinear form separately continuous you can find an example of a bilinear form on a normed space which is separately continuous, hence separately differentiable, but not continuous.