Most theorems under real Banach space settings have their twin brothers for complex ones, say, the Hahn-Banach theorem. However, some theorems are not valid in complex Banach spaces, and vice versa.

I'm reading the Vol. III of "Nonlinear functional analysis and its applications" by Zeidler. Many theorems contained there assume that underling space is a real $B$-space. I have to check one by one to see whether or not it is true under complex spaces setting with some natural modifications.

Of course, it is natural to assume that the field is real in some cases. However, this setting prevents us applying the powerful tools like spectral theory, $H^\infty $-functional calculus, analytic semi-group theory and so on.

Is there any major theorem in (Linear or nonlinear) Functional analysis that works only in a real Banach space and has no similar result under a complex space setting. Of course, any theorem about "partial order, " doesn't account. Please give the ones from the theory of optimization and calculus of variations. Is there any good reference on this topic?

  • $\begingroup$ And I'd say complex waves are natural in Newtonian physics since you can see the real and imaginary part as the $\Delta$ of kinetic and potential energy (such that the conversation of energy is obvious) $\endgroup$ – reuns Nov 27 '16 at 6:49

There are some differences. For example Bishop-Phelps theorem, which holds only in real Banach spaces. In my opinion, this qualifies as a "major theorem".

Lomonosov, Victor A counterexample to the Bishop-Phelps theorem in complex spaces.
Israel J. Math. 115 (2000), 25–28.

Remark. Your statement "natural to assume that the field is real if the problem comes from physics" is completely wrong.

In fact physicists are MORE interested in the complex field than in the real field. The most fundamental theory of physics, quantum mechanics, describes the state of a system as a vector in a COMPLEX Hilbert space. From the point of view of physics, real numbers are just eigenvalues of Hermitian operators on a complex Hilbert space.

  • 2
    $\begingroup$ To throw in my two cents about the last statement - even classical electrodynamics are complex by nature. Therefore, the treatment of optical waves in the Helmholz and Schrodinger (linear and nonlinear) equations is complex by nature. $\endgroup$ – Amir Sagiv Nov 27 '16 at 6:21
  • $\begingroup$ @AmirSagiv You are right. But for the parabolic systems, like activator-Inhibitor model, semi-conduct model, Belousov–Zhabotinskii models and so on, the field should be the real field. The common way to deal with the problem is as follows: First we extend it the complex field, and then show that the solution is actually real if the initial value is real and the coefficients are real. $\endgroup$ – Ice sea Nov 27 '16 at 7:09
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    $\begingroup$ While the general Bishop–Phelps theorem is false for complex Banach spaces, it is worth noting that it holds for complex Banach space X with the Radon–Nikodym property. $\endgroup$ – anonymous Dec 3 '16 at 21:46

In optimization, you typically have a function $f \colon X \to \mathbb{R}$ which you are going to minimize. In order to apply first-order optimality conditions or first-order methods, you would like to compute the derivative of $f$. However, if $X$ is a complex Banach space, the derivative will never be $\mathbb{C}$-linear, since the range of the derivative has to be $\mathbb{R}$.

As a "major theorem", I would present:

Let $H$ be a Hilbert space. Then $x \mapsto \|x\|_H^2$ is Fréchet differentiable.

This does only hold in real Hilbert spaces, but not in complex ones (unless $H = \{0\}$).

  • $\begingroup$ Any reference for this claim? And I think you mean that Frechet derivative has to be real, right? $\endgroup$ – Henry.L May 30 '17 at 17:09
  • $\begingroup$ No, I do not have a reference, but the proof is short and easy. I do not understand your second question. $\endgroup$ – gerw May 30 '17 at 19:59
  • $\begingroup$ You said that "the range of the derivative has to be $\mathbb{R}$", you specifically mean Frechet derivative right? Can you provide a counter example in real Banach space? In Hilbert space your claim is plain. $\endgroup$ – Henry.L May 30 '17 at 20:02
  • $\begingroup$ Yes, but it should hold for all kind of derivatives. $\endgroup$ – gerw May 30 '17 at 20:03
  • $\begingroup$ I mean, what if we consider your claim in a Banach, yet not Hilbert space as the OP asked. $\endgroup$ – Henry.L May 30 '17 at 20:04

Consider a bounded operator $T$ from a Hilbert space $H$ into itself, i.e. $T \in B(H)$. You can then define the numerical radius $r(T)$ as the radius of the numerical range $W(T)$, i.e. $$ W(T) = \{ \langle Tx, x \rangle \colon \|x\| = 1 \} \quad \text{and} \quad r(T) = \sup \{ |\lambda| \colon \lambda \in W(T) \} $$ It is immediately clear that you have $r(T) \le \|T\|$ and it is natural to ask if a $T$-independent constant $c > 0$ exists with $c\|T\| \le r(T)$, so that $r$ is not just a seminorm but a proper norm on $B(H)$, equivalent to the operator norm.

If $H$ is complex (and $\dim H > 0$), the answer is yes and you can choose $c = \frac 12$.

If $H$ is real (and $\dim H \ne 1$), the answer is no: Think of a 90-degree rotation $T$ in $\mathbb R^2$; then $W(T) = \{ 0 \}$ yet $\|T\| = 1$.

Edit: According to

Martín, Miguel. A survey on the numerical index of a Banach space. III Congress on Banach Spaces (Jarandilla de la Vera, 1998). Extracta Math. 15 (2000), no. 2, 265--276. MR1823892

the proof for the complex case can be found on page 114 of

Halmos, Paul R. A Hilbert space problem book. D. Van Nostrand Co., Inc., Princeton, N.J.-Toronto, Ont.-London 1967 xvii+365 pp. MR0208368


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