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I am interested in building a function $f:]0,1[^2 \rightarrow \mathbb{R}$ such that

  1. $f$ is continuous
  2. $f$ has directional derivatives everywhere and in every directions
  3. $\forall t\in ]0,1[$, $$\frac{\partial f}{\partial x}(t,t)=0$$ $$\frac{\partial f}{\partial y}(t,t)=0$$
  4. the function $g(t)=f(t,t)$ is non constant.

It is clear that this function cannot be differentiable. Is it possible to build an $f$ like this? If yes, can we build an example? If no, is there a simple argument proving it?

Thank you.

Chevallier
  • 1,062
  • If it has directional derivatives everywhere then it has derivatives in all the directions of the axes which means all partial derivatives exist. Wouldn't that make it differentiable? – John Douma Jan 05 '22 at 19:55
  • @JohnDouma, in the differentiable assumption, we also require a linear link between all the directional derivative at a given point. – Chevallier Jan 09 '22 at 09:01

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