Theorem: Let $A$ be a unital Banach algebra. Then for $a \in A$ the spectrum $\sigma (a) \neq \varnothing$.

Consider the following proof:

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The first step that seems unnecessary to me:

Let's say we have shown that

$$f: \mathbb C \to A, \lambda \mapsto (a-\lambda)^{-1}$$

is bounded entire. My next step would be to conclude: by Liouville $f$ is constant therefore $a = a-1$. However, this proof continues by arguing that for every $\tau \in A^\ast$ the map $\tau \circ f$ is bounded entire and hence $a = a-1$.

Why is this step using $A^\ast$ done? Is it necessary? To me it seems to be enough that $f$ is differentiable and bounded on $\mathbb C$ to derive the desired contradiction.

The other step that is not clear to me are the two lines of inequalities before the ''Consequently, ...''. After $1-|\lambda^{-1}| \|a\| > {1\over 2}$ my next step would be to take the inverse on both sides to get ${1 \over 1- |\lambda^{-1}| \|a\| } < 2$. What am I missing?

Thank you for clarification.


1 Answer 1


Liouville's theorem (the one about bounded entire functions) is in complex analysis only proved for complex-valued functions. The generalisation to $\mathbb{C}^n$-valued functions is immediate, but for general Banach-space-valued (or Banach-algebra-valued) functions, it needs to be proved before it is used. The proof in that general case is by reducing it to the $\mathbb{C}$-valued case via continuous linear forms, and then by Hahn-Banach the conclusion that $f$ itself is constant is reached.

If the general form is available, that step is not necessary, but usually, it isn't, since the non-emptiness of spectra is - in books on functional analysis - typically the only place where it is used.

The other step is necessary, because we don't have $\lVert a^{-1}\rVert = \lVert a\rVert^{-1}$ in general. Consider the operator $M$ on $\ell^2$ that multiplies every even-indexed term with $2$, and every odd-indexed term with $\frac12$. Then $\lVert M\rVert = \lVert M^{-1}\rVert = 2$, so from $1-\lvert \lambda^{-1}\rvert\lVert a\rVert < \frac12$ we cannot directly conclude $\lVert (1-\lambda^{-1}a)^{-1}\rVert < 2$, the special structure of the inverse is needed for that conclusion.

  • $\begingroup$ But I'm confused because I think I can prove the taking inverses step as follows:$${}$$ We want: $$\left \| {1 \over 1 - \lambda^{-1}a} \right \| = {1\over \|1-\lambda^{-1}a\|} < \infty$$ and we have: $$\\$$ $$\begin{align} \|1-\lambda^{-1}a\| &\ge | \|1\| - \|\lambda^{-1}a\|| \\ &= |1- \|\lambda^{-1}a\||\\ &\ge 1-\|\lambda^{-1}a\| \end{align}$$ hence $$ {1\over \|1-\lambda^{-1}a\| } \le {1\over 1-\|\lambda^{-1}a\| } < 2$$ $\endgroup$
    – Student
    Feb 21, 2014 at 14:17
  • $\begingroup$ No sorry. I get it. The mistake happens in the first equality above where I use $\|a^{-1}\| = \|a\|^{-1}$. Sorry, ignore my previous comment. $\endgroup$
    – Student
    Feb 21, 2014 at 14:23
  • $\begingroup$ Okay. Just: The notation $\frac{1}{1-\lambda^{-1}a}$ for $(1-\lambda^{-1}a)^{-1}$ is a bit dangerous. It may seduce you to write $\frac{x}{1-\lambda^{-1}a}$, and then you don't know whether that is $x(1-\lambda^{-1}a)^{-1}$ or $(1-\lambda^{-1}a)^{-1}x$; in general, the two are different. The clumsier $(1-\lambda^{-1}a)^{-1}$ avoids that ambiguity. $\endgroup$ Feb 21, 2014 at 14:26
  • $\begingroup$ You are right, I will write $()^{-1}$ from now on. Also, I found a slightly shorter proof: By assumption, we have $$(1 - \| \lambda^{-1}a\|)^{-1} < 2$$ hence $$ \left \| \sum_{n=0}^\infty \lambda^{-n}a^n \right \| \le \sum_{n=0}^\infty |\lambda|^{-n}\|a\|^n = (1- \|\lambda^{-1} a\|)^{-1} < 2$$ and therefore $ \left \| \sum_{n=0}^\infty \lambda^{-n}a^n \right \| = \| (1 - \lambda^{-1}a)^{-1} \| < 2$ which concludes the proof that $f$ is bounded. $\endgroup$
    – Student
    Feb 21, 2014 at 16:02
  • 1
    $\begingroup$ Yes, that works to show $\lVert(1-\lambda^{-1}a)^{-1}\rVert < 2$, and is slightly shorter than the book version. I guess the book version comes directly from the way one proves that $x\mapsto x^{-1}$ is continuous (and then, that it is differentiable), where the distance to $1$ matters of course, while here it doesn't. $\endgroup$ Feb 21, 2014 at 16:41

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