Let $X,Y$ be Banach spaces. A function $f :X \to Y$ is said to be Gâteaux differentiable at $x$ if there exists a bounded linear operator $A : X \to Y$ such that $$\lim_{r \to 0}\frac{\|f(x+rh)-f(x)-rAh\|}{r}=0\tag{1}$$ for every $h \in X.$

I'm trying to prove/disprove whether it is sufficient to take $r \in \mathbb R$ in $(1).$

I tried the following example:

Let $f : \mathbb C \to \mathbb C$ be defined as $f(z)=\overline{z}.$ Define $A: \mathbb C \to \mathbb C$ as $A(h)=\overline{h}.$ Then $A$ is a bounded linear operator.

Suppose $r \in \mathbb{R}.$ Then, $$\frac{\|f(z+rh)-f(z)-rAh\|}{r}=\frac{\|\overline{z+rh}-\overline{z}-r\overline{h}\|}{r}=0$$

But if $r \in \mathbb C,$ then $$\frac{\|f(z+rh)-f(z)-rAh\|}{r}=\frac{\|\overline{rh}-r\overline{h}\|}{r}=\frac{2|\overline{h}\Im{r}|}{r}\not\to 0$$ as $r \to 0.$ (where $\Im{r}$ is the imaginary part of $r.$)

Therefore, it is not sufficient to take $r \in \mathbb R$ in $(1).$

Edit: As pointed out by Daniel Fischer in the comments, my above example was wrong. I would like to know whether it is sufficient to take $r \in \mathbb R$ in $(1).$ I have tried several other examples too but I couldn't construct one succesfully disproved the claim.

  • $\begingroup$ Gâteaux. $ $ $ $ $\endgroup$ – Did Oct 26 '17 at 19:18
  • $\begingroup$ Conjugation is $\mathbb{R}$-linear, but not $\mathbb{C}$-linear. If $X,Y$ are complex vector spaces, $(1)$ can only hold for complex $r$ if $A$ is $\mathbb{C}$-linear. $\endgroup$ – Daniel Fischer Oct 26 '17 at 19:27
  • $\begingroup$ Reducing dimensions, it boils down to the question whether if $f \colon \mathbb{R}^2 \to \mathbb{R}$ has directional derivative $0$ in every direction at $(0,0)$, it follows that $f$ is differentiable (that is, Fréchet differentiable) at $(0,0)$. $\endgroup$ – Daniel Fischer Oct 26 '17 at 19:44
  • $\begingroup$ @DanielFischer I believe that is false. For Frétchet differentiability, the difference quotients needs to converge uniformly along all directions. However, having a directional derivative in every direction need not imply this. $\endgroup$ – Sahiba Arora Oct 26 '17 at 20:10
  • $\begingroup$ Right. So if you have a counterexample to that, it's not hard to modify it to get a counterexample to the criterion restricted to $r\in \mathbb{R}$. Just use $\mathbb{R}\subset \mathbb{C}$ and identify $\mathbb{R}^2$ with $\mathbb{C}$. $\endgroup$ – Daniel Fischer Oct 26 '17 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.