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Context

The problem here is about the characterization of positivity for real or complex valued functions: $$\sigma(f)\geq 0\iff\sigma(f(x))\geq 0\text{ for all }x\in X\iff f(x)\geq 0\text{ for all }x\in X$$

The point is that the first characterization reflects its intrinsic properties allowing to apply general results in the study of positive operators as integrals...

Problem 0

Let $A$ be an algebra with identity and $\Omega$ a plain space.
Consider the algebra of functions: $$\mathcal{F}:=\{F:\Omega\to A\}$$ together with pointwise operations.

Is it true that the following characterizations of positivity are equivalent: $$\sigma(F)\geq 0\iff\sigma(F(\omega))\geq 0\text{ for all }\omega\in\Omega$$ where $\sigma(F)$ is the spectrum of $F$ in $\mathcal{F}$ whereas $\sigma(F(\omega))$ the spectrum of $F(\omega)$ in $A$?

Problem 1a

Now, let $B$ be a Banach algebra with identity.
Consider the (Banach) algebra of bounded functions: $$\mathcal{B}:=\{F:\Omega\to B:F\text{ bounded}\}$$ again with pointwise operations.
(Note that $F$ is not necessarily continuous.)

Is it still true that the characterizations of positivity are equivalent: $$\sigma(F)\geq 0\iff\sigma(F(\omega))\geq 0\text{ for all }\omega\in\Omega$$ where $\sigma(F)$ again is the spectrum of $F$ in $\mathcal{B}$ whereas $\sigma(F(\omega))$ the spectrum of $F(\omega)$ in $B$?
(Here, a function is invertible iff it has a bounded inverse!)

Problem 1b

Next, let $X$ be also topological space.
Consider the algebra of continuous functions: $$\mathcal{C}:=\{F:X\to B:F\text{ continuous}\}$$ again with pointwise operations.
(Note that $F$ is possibly unbounded.)

Is it still true that the characterizations of positivity are equivalent: $$\sigma(F)\geq 0\iff\sigma(F(x))\geq 0\text{ for all }x\in X$$ where $\sigma(F)$ again is the spectrum of $F$ in $\mathcal{C}$ whereas $\sigma(F(x))$ the spectrum of $F(x)$ in $B$?
(Here, a function is invertible iff it has a continuous inverse!)

Problem 2

Finally, consider the (Banach) algebra of bounded continuous functions: $$\mathcal{BC}:=\{F:X\to B:F\text{ bounded and continuous}\}$$ again with pointwise operations.

Is it still true that the characterizations of positivity are equivalent: $$\sigma(F)\geq 0\iff\sigma(F(x))\geq 0\text{ for all }x\in X$$ where $\sigma(F)$ again is the spectrum of $F$ in $\mathcal{BC}$ whereas $\sigma(F(x))$ the spectrum of $F(x)$ in $B$?
(Here, a function is invertible iff it has a bounded and continuous inverse!)

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I'm no expert on this, so I might say stupid things, but lets have a go. Let me assume that $A$ is unital (with unit $\mathbb{1}$). The answer to your first question is positive, since $$\sigma(F)=\bigcup_{\omega\in\Omega} \sigma(F(\omega)).$$ Indeed, the inverse $h$ (if it exists) of the function $\lambda I-F$ satisfies $h(\lambda I-F)=(\lambda I-F)h\equiv 1$ pointwise and is therefore necessarily given by \begin{equation} h(\omega):=(\lambda 1-F(\omega))^{-1}. \end{equation} From this one sees that if $\lambda \in \sigma(F(\omega))$ for some $\omega$, then $\lambda \in \sigma(F)$ and conversely, if $\lambda \in \sigma(F)$ then for some $\omega$ the existence of $(\lambda 1-F(\omega))^{-1}$ must fail.

As for the continuous case, the collection of all continuous functions do not form a Banach algebra because you cannot define a norm there (unless your topological space is compact). You need to impose a boundedness condition: $$\|F\|:=\sup_{x\in X}\|F(x)\|_B<\infty.$$ This quantity defines a norm which makes your space a Banach algebra.

Now while an inverse $h$ of some $(\lambda I-F)$ must still satisfy the identity written above, the identity itself does not necessarily define a continuous and bounded function. An example illustrating this can be seen by taking $X=\mathbb R=B$ and considering the Banach algebra $C_b(\mathbb R)$ of bounded continuous functions.

Still, the case where $X$ is compact forms an exception for the following reason. The inclusion $$\bigcup_{x\in X} \sigma(F(x))\subset \sigma(F)$$ always holds. Suppose then that $\lambda\notin \bigcup_{x\in X} \sigma(F(x))$. Since $Im(\lambda I-F)$ is a compact subset of the open set $G$ of all invertible elements of $B$, and the inversion $x\mapsto x^{-1}$ is continuous on $B$ it follows that the inverses of all the elements in $Im(\lambda I-F)$ also form a compact set (see this for all these properties), thus lying in some bounded ball of radius $a<\infty$. In other words, $$\|(\lambda I-F)^{-1}\|=\sup_{x\in X}\|(\lambda 1-F(x))^{-1}\|_B \le a,$$ thus giving a bounded, continuous inverse of $\lambda I-F$. This shows the opposite inclusion.

EDIT

Problem 0: As discussed above, the equivalence holds true in this case.

Problem 1a: In this case, for the same reasons as in the continuous and bounded case (see above) the equivalence is not true (even with compactness, since we do not have continuity).

Problem 1b: Here the equivalence is again true, since in the continuous & bounded case the only obstruction was boundedness.

Problem 2: In general no, in case $X$ is compact yes.

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  • $\begingroup$ As the spectrum is a purely algebraic concept it does not require a Banach space structure but ok let's consider that case too (I will add it as further problem). $\endgroup$ – C-Star-W-Star Aug 28 '14 at 17:45
  • $\begingroup$ Ah I see, I accidentally used the word Banach algebra for the algebra of continuous functions - corrected and extended! $\endgroup$ – C-Star-W-Star Aug 28 '14 at 18:04
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Due to the pointwise structure it formally holds: $$\sigma(F)=\bigcup_{x\in X}\sigma(F(x))$$

The difficulties arise as soon as additional requirements are tied upon the functions:

0) The formal inverse exists.
1a) The formal inverse is not necessarily bounded. (See below!)
1b) The formal inverse is continuous due to the Neumann series. (See Contuity of inversion!)
2) The formal inverse is again continuous but not necessarily bounded.

Especially for complex valued functions it holds: $\sigma(f(x))=f(x)$


Consider the bounded function: $$f:(0,1]\to\mathbb{R}:x\mapsto x$$ Then zero does not belong to its range but its formal inverse is unbounded: $$f^{-1}:(0,1)\to\mathbb{R}:x\mapsto \frac{1}{x}$$

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  • $\begingroup$ Just to be sure as someone downvoted: Does the argumentation miss a crucial point? $\endgroup$ – C-Star-W-Star Sep 9 '14 at 2:10

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