7
$\begingroup$

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: $Df(x)h = 2\langle x,h\rangle$.

Questions:

  • 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?

$\endgroup$
3
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.