# Complex differentiability and Frechet differentiability

I know complex differentiability for a complex function is a stronger condition than Frechet differentiability because Cauchy Riemann equation need to be satisfied.

Is there any example in the worst case that a function is Frechet differentiable everywhere but nowhere complex differentiable? If exist, it would be nice if a brief justification about how to come up with this function is provided. Thank you.

• I think the answer depends on whether you consider Fréchet differentiability over the field $\mathbb R$ or $\mathbb C$. – gerw Mar 2 '17 at 10:05
• Complex conjugation is nowhere complex differentiable. – Jochen Mar 2 '17 at 10:17
• @Jochen: But if I consider it as a map from $\mathbb C$ to $\mathbb C$ both understood as one-dimensional vector spaces over the field $\mathbb C$, then it is also not Fréchet differentiable, is it? – gerw Mar 2 '17 at 18:05
• Right. It is Frechet differentiable only if you consider $\mathbb C$ as a real vector space. – Jochen Mar 2 '17 at 18:19

## 1 Answer

The function $\Bbb C \to \Bbb C$ given by $z \mapsto \bar z$ is nowhere differentiable in $\Bbb C.$ When seeing as a transformation $\Bbb R^2 \to \Bbb R^2$ it takes the form $(x, y) \mapsto (x, -y),$ a linear transformation, hence, differentiable everywhere (even analytic).

To add an explanation of what is going on, I'll expand. Contrary to what some mathematicians believe and claim, Frechèt differentiability is the same in $\Bbb R$ or $\Bbb C.$ The issue lies not in the definition of derivative but in somewhere else, in the space of continuous linear functions. To be precise, let $\mathbf{E}$ be a normed space over the field $\Bbb C$ and $\mathbf{F}$ be the same space (set and norm) but over the field $\Bbb R.$ Let $\mathscr{L}(\mathbf{E})$ be the set of continuous linear functions $\mathbf{E} \to \Bbb C$ over the field $\Bbb C$ (taking complex values) and $\mathscr{L}(\mathbf{F})$ be the space of continuous linear functions $\mathbf{F} \to \Bbb R^2$ over the field $\Bbb R$ (taking values in $\Bbb R^2$). So, the main issue is that the set $\mathscr{L}(\mathbf{E})$ is contained in $\mathscr{L}(\mathbf{F})$ but they (usually) differ. Frechèt differntiability of a function $f:\mathbf{E} \to \Bbb C$ implies the existence of a function $\mathrm{L} \in \mathscr{L}(\mathbf{E})$ such that the order one approximation happens $$f(x + h) = f(x) + \mathrm{L} \cdot h + \mathrm{o}(\| h \|).$$ Now, $f$ can be viewed as a function $\mathbf{F} \to \Bbb R^2,$ so if $f$ is differentiable $\mathbf{E} \to \Bbb C,$ a fortiori it is differtiable $\mathbf{F} \to \Bbb R^2$ with same linear transformation as derivative (just restrict all the scalars of the field to lie in $\Bbb R$). As $\mathscr{L}(\mathbf{F})$ is larger than $\mathscr{L}(\mathbf{E})$ (usually), it might happen that $f$ is $\mathbf{F} \to \Bbb R^2$ differentiable but not $\mathbf{E} \to \Bbb C$ differentiable (even analytic, as my example showed). I hope I have clarified the issue; the concept of differentiability is exactly the same for $\Bbb R$ or $\Bbb C$, it rather happens that a linear transformation in $\Bbb R$ migth fail to be linear over $\Bbb C.$ This restriction, apparently tiny at first glance, has enormous implications.