In Functional Analysis, we have covered Functional Calculus, that is, a way to associate, once having fixed a Banach algebra $A$ and an element $a\in A$, an element $\tilde f(a)\in A$ to every $f$ holomorphic in a neighborhood of the spectrum of $a$, via an integral formula, in such a way that $f\mapsto\tilde f(a)$ is an algebra homomorphism, $\tilde f(a)=e$ (the identity) if $f\equiv1$, $\tilde f(a)=a$ if $f(z)=z$, the subalgebra of $\tilde f(a)$'s is commutative even if $A$ is not, and these properties characterize this $f\mapsto \tilde f(a)$ map. On one occasion, the professor, being asked if this mapping was injective, replied:

Uuuh, è iniettiva, certo, fino a quando, ovviamente, tengo fisso l'elemento, quando pói cambio l'elemento, chi lo sa.


Uuuh, is it injective, of course, up till, of course, I keep the element fixed, when I change the element, who knows.

I smell rat here. After all, $\mathbb{C}$ is a B.A., and this mapping is certainly not injective, in general, since $f(z)=z$ is certainly not the only holomorphic map sending $z_0$ e.g. into itself, yet this mapping is clearly $f\mapsto f(z_0)$, in this case, if the fixed $a$ from above is $z_0$. So what did she mean? Are there conditions under which this is injective indeed? Perhaps as soon as the spectrum is not a single point for each element (i.e. as soon as we leave $\mathbb{C}$) this mapping becomes injective?

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    $\begingroup$ The domain of the map $f \mapsto \tilde{f}(a)$ is not $\mathbb{C}$, it's the algebra of functions that are holomorphic on a neighbourhood of $\sigma(a)$ (the spectrum of $a$). That is not a Banach algebra (it's not normable), but it's an inductive limit of Banach algebras (only 99.9% sure, though). The map is injective (by the identity theorem) if $\sigma(a)$ contains no isolated points. $\endgroup$ – Daniel Fischer May 17 '16 at 21:05
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    $\begingroup$ Something along those lines, but the algebra of holomorphic functions on a nonempty open sets is not normable. For a bounded nonempty open $U \subset \mathbb{C}$, let $A(U) = \mathscr{C}(\overline{U}) \cap \mathscr{O}(U)$, and endow it with the maximum norm. That's a closed subalgebra of $\mathscr{C}(\overline{U})$, so a Banach algebra. For $W \subset U$, we have the natural restriction $\rho^U_W \colon A(U) \to A(W)$. Letting $U$ range over the (open) neighbourhoods of $\sigma(a)$, we get an inductive system, and we look at its inductive limit. $\endgroup$ – Daniel Fischer May 18 '16 at 9:34
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    $\begingroup$ And I got things a bit backward at the end yesterday night. If $\sigma(a)$ has no isolated points, then $f\mapsto \tilde{f}(a)$ is injective is seen thus: By linearity, we need to see that $\tilde{f}(a) = 0 \implies f = 0$. By the spectral mapping theorem, $\{0\} = \sigma(\tilde{f}(a)) = f(\sigma(a))$, so $f\lvert_{\sigma(a)} \equiv 0$. Take a representative $f_1 \colon U_1 \to \mathbb{C}$ of $f$, and let $U_2$ be the union of components of $U_1$ intersecting $\sigma(a)$. Let $f_2 = f_1\lvert_{U_2}$. By the identity theorem $f_2 \equiv 0$, and hence $f = [f_1] = [f_2] = 0$. $\endgroup$ – Daniel Fischer May 18 '16 at 9:39
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    $\begingroup$ Urk, was not yet completely awake when I wrote that. $\endgroup$ – Daniel Fischer May 18 '16 at 12:22
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    $\begingroup$ And in fact, $\sigma(x) = \{0\} \implies x = 0$ isn't a necessary condition. Let $a = e$, then $\tilde{f}(a) = f(1)\cdot a$, so the map is not injective, whether we have the condition or not. But it's sufficient for the non-injectivity in the presence of isolated points. For then, $f\lvert_{\sigma(a)} = g\lvert_{\sigma(a)}$ implies $\sigma(\tilde{f}(a) - \tilde{g}(a)) = \sigma((\tilde{f} - \tilde{g})(a)) = \sigma(\widetilde{(f-g)}(a)) = (f-g)(\sigma(a)) = \{0\}$, and thus $\tilde{f}(a) = \tilde{g}(a)$. If there are nonzero elements with spectrum $\{0\}$, this argument doesn't work, so in $\endgroup$ – Daniel Fischer May 18 '16 at 12:45

To determine whether the map is injective, we first need to clarify the domain. If we let the domain be $\mathscr{O}(\sigma(a)) = \bigcup_{\sigma(a) \subset U} \mathscr{O}(U)$, where the $U$ are open sets, then the map $\Phi \colon f \mapsto \tilde{f}(a)$ is trivially not injective, we can let $U = D_1 \cup D_2$ consist of two disjoint open disks, with $\sigma(a) \subset D_1$, and let

$$f(z) = \begin{cases} 0 &, z \in D_1 \\ 1 &, z \in D_2. \end{cases}$$

Then clearly $\tilde{f}(a) = 0$ but $f \neq 0$.

For an interesting question, we need to ignore the behaviour of $f$ "far away" from $\sigma(a)$. Thus we look at $\mathscr{O}(\sigma(a))/{\sim}$, where $f \sim g$ if and only if there is an open $W \supset \sigma(a)$ with $f\lvert_W = g\lvert_W$. Note that we can also describe $\mathscr{O}(\sigma(a))/{\sim}$ as the inductive limit of the Banach algebras $A(U) = \mathscr{C}(\overline{U}) \cap \mathscr{O}(U)$ where $U$ ranges over the bounded neighbourhoods of $\sigma(a)$.

Then $\Phi$ is always injective when $\sigma(a)$ contains no isolated points: By linearity of $\Phi$, we need to prove $\tilde{f}(a) = 0$ for an $f \in \mathscr{O}(U)$ implies $f \sim 0$, i.e. $f$ vanishes on a neighbourhood of $\sigma(a)$. The spectral mapping theorem gives us $\{0\} = \sigma(0) = \sigma(\tilde{f}(a)) = f(\sigma(a))$, so $f\lvert_{\sigma(a)} \equiv 0$ for $\tilde{f}(a) = 0$. Let $W$ be a connected component of $U$ intersecting $\sigma(a)$, and $w \in W\cap \sigma(a)$. Since $\sigma(a)$ contains no isolated points, it follows that $w$ is an accumulation point of $W \cap \sigma(a)$, and by the identity theorem, $f\lvert_W \equiv 0$. Thus if $V$ is the union of the components of $U$ intersecting $\sigma(a)$, we have $f\lvert_V \equiv 0$, and since clearly $\sigma(a) \subset V$, indeed $f \sim 0$.

If $\sigma(a)$ contains isolated points, then generally $\Phi$ is not injective, for $f\lvert_{\sigma(a)} \equiv 0$ does then not imply $f \sim 0$.

If the closed subalgebra of $A$ generated by $a$ satisfies $\sigma(x) = \{0\} \implies x = 0$, then $\Phi$ is never injective if $\sigma(a)$ has isolated points, for then $\tilde{f}(a) = 0 \iff f\lvert_{\sigma(a)} \equiv 0$, and if $p$ is an isolated point of $\sigma(a)$, we can take

$$f(z) = \begin{cases} (z - p) &, \lvert z-p\rvert < \delta \\ \quad 0 &, \lvert z-p\rvert > \delta,\end{cases}$$

where $0 < \delta < \operatorname{dist}(p,\sigma(a)\setminus \{p\})$ as an $f$ with $\tilde{f}(a) = 0$ but $f \nsim 0$.

Question: Can we have $f \mapsto \tilde{f}(a)$ injective even if $\sigma(a)$ has isolated points? Answer: Yes.

Let $A = \mathscr{B}(\ell^2(\mathbb{N}))$, and consider

$$a \colon (x_0, x_1, x_2,\dotsc) \mapsto \bigl( 0, \tfrac{1}{1}x_0, \tfrac{1}{2} x_1, \tfrac{1}{3} x_2, \dotsc).$$

Then $\lVert a^k\rVert = \frac{1}{k!}$, so $\sigma(a) = \{0\}$. To see that $\Phi$ is injective for that $a$, consider a convergent power series

$$f(z) = \sum_{n = 0}^{\infty} c_n z^n.$$

Then we have

$$\tilde{f}(a) = \sum_{n = 0}^{\infty} c_n a^n,$$

since the series on the right converges absolutely in $\mathscr{B}(\ell^2(\mathbb{N}))$. If $f \neq 0$, let $m = \min \{ n \in \mathbb{N} : c_n \neq 0\}$. Then

$$\langle \tilde{f}(a) e_0, e_m\rangle = \frac{c_m}{m!} \neq 0$$

and consequently $\tilde{f}(a) \neq 0$, showing that $\Phi$ is injective.

  • $\begingroup$ Could you expand on why $\sigma(a)=\{0\}$? $\endgroup$ – MickG May 18 '16 at 13:50
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    $\begingroup$ Since $\lVert a^k\rVert = 1/k!$, the spectral radius formula $\rho(a) = \lim_{k\to\infty} \lVert a^k\rVert^{1/k}$ gives $\rho(a) = 0$. You can also directly verify that $\lambda\operatorname{id} - a$ is invertible for all $\lambda\neq 0$, but IMO, the spectral radius is quicker here. $\endgroup$ – Daniel Fischer May 18 '16 at 13:54
  • $\begingroup$ And $\sqrt[k]{\frac{1}{k!}}\to0$… why? $\endgroup$ – MickG May 18 '16 at 16:07
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    $\begingroup$ That follows for example from Stirling's formula. Less precise and more pedestrian is $k! > (k/3)^{k/2}$, which gives $\sqrt[k]{k!} > \sqrt{k/3}$. $\endgroup$ – Daniel Fischer May 18 '16 at 16:13
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    $\begingroup$ Or, for $k \geqslant 2$, we have $$k! \geqslant \prod_{m = \lfloor k/2\rfloor}^k m,$$ where each of the factors is $\geqslant \lfloor k/2\rfloor \geqslant k/3$, and there are at least $k - (\lfloor k/2\rfloor - 1) \geqslant k/2$ factors. Writing it down, I see that we could use $$\prod_{m = \lfloor k/2\rfloor + 1}^k m,$$ with each factor $> k/2$, and with $k - \lfloor k/2\rfloor = \lceil k/2\rceil \geqslant k/2$ factors, so we'd get the better bound $k! \geqslant (k/2)^{k/2}$. With a strict inequality for $k > 1$. Not that it matters. $\endgroup$ – Daniel Fischer May 18 '16 at 21:30

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