I got this from Analysis II, H. Amann, J. Escher, p. 152.

Their definition of the derivative of a map $f$ between Banach spaces $E,F$ over the field $\mathbb{K}$ is a bounded linear operator $A\in\mathcal{L}(E,F)$ such that $$\lim_{x\to x_0} \frac{f(x) - f(x_0) - A(x - x_0)}{\|x - x_0\|} = 0.$$

  • Squared-Norm Example: Suppose $H$ is a Hilbert space and define $f\colon H\to\mathbb{K}$, $x\mapsto \|x\|^2$. They claim that $f$ is continuously differentiable and compute the derivative as $$Df(x) = 2\operatorname{Re}\langle x,\cdot\rangle,\ \text{for $x\in H$}.$$ For a real Hilbert space $H$, $\mathbb{K} = \mathbb{R}$, this seems ok: $Df(x)h = 2\langle x,h\rangle$.


  • For $\mathbb{K} = \mathbb{C}$, I think this is wrong. The calculation $$f(x+h) - f(x) - 2\operatorname{Re}\langle x,h\rangle = \|h\|^2 = o(\|h\|)$$ still works. For example, if we define $f\colon\mathbb{C}\to\mathbb{C}$ by $f(x) = |x|^2$, then their claim is that $Df(x)h = 2\operatorname{Re}(x\overline{h})$. This doesn't seem to be linear. So, their map $Df(x) = 2\operatorname{Re}\langle x,\cdot\rangle$ is not linear in the complex case, right?

  • Does $f(x) = \|x\|^2$ have a Frechet derivative in the complex case?


You are right. This is the same phenomenon of the derivative of $f(z)=|z|^2$ when $z\in\mathbb{C}$. If you interpret $\mathbb{C}$ as a real vector space, then $f$ is differentiable and its Jacobian matrix is $$Df(x+iy)=\begin{bmatrix} 2x & 2y\\ 0& 0\end{bmatrix}.$$ But if you interpret $\mathbb{C}$ as a complex vector space, then differentiability is the same thing as analyticity, and no, $f$ is not analytic.

The problem is exactly that the Jacobian matrix $Df(z)$ does not represent a $\mathbb{C}$-linear mapping, as you rightly point out. (If it did, $f$ would satisfy Cauchy-Riemann's equations.)

  • $\begingroup$ Thanks for your help. Thinking about this example $f(z) = |z|^2$, $z\in\mathbb{C}$, more was helpful for me. (I think the $2y$ in your Jacobian should be moved up to the first row.) $\endgroup$ – mars34 Jul 12 '15 at 2:18
  • $\begingroup$ @mars34: Of course. Edited $\endgroup$ – Giuseppe Negro Jul 12 '15 at 9:58

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