1
$\begingroup$

Apologies. I have to ask two questions in one and I will give you the reason below. The questions are:

If $f$ is entire and $g$ is continuous does it follow that $g\circ f$ is entire?

If $A$ is a unital complex Banach algebra are the elements in the continuous dual $A^\ast$ entire?

I am trying to understand the proof here of theorem 1.13. It is proved that the resolvent function $\lambda \mapsto (\lambda - a)^{-1}$ is analytic on $\mathbb C$ and then it is stated that if $\varphi \in A^\ast$ then $\lambda \mapsto \varphi((\lambda - a)^{-1})$ is analytic too. I am trying to understand why this is true.

$\endgroup$

1 Answer 1

4
+50
$\begingroup$

If $f$ is entire and $g$ is continuous does it follow that $g\circ f$ is entire?

No, unless $f$ is constant, $g$ needs to be holomorphic too, not only continuous for the composition to be holomorphic. Consider $g(z) = \overline{z}$ to see that even a very well-behaved smooth function generally doesn't preserve complex differentiability.

If $A$ is a unital complex Banach algebra are the elements in the continuous dual $A^\ast$ entire?

For $A = \mathbb{C}$, the answer is yes, otherwise it depends on what you mean by "entire". Usually, an entire function is defined as a function defined and holomorphic on all of $\mathbb{C}$ (or $\mathbb{C}^n$). Then the elements of $A^\ast$ have the wrong domain to be called entire (generally). You can, however, in this context also define an entire function to be a function that is defined and (complex) Fréchet-differentiable on all of $A$, where $f$ being Fréchet-differentiable in $a$ means that there exists a $\varphi \in A^\ast$ such that $f(a+h) - f(a) - \varphi(h) \in o(\lVert h\rVert)$. In that sense an $f\in A^\ast$ is entire, and at each point it is its own derivative, $f'(a) = h \mapsto f(h)$ for all $a\in A$.

It is proved that the resolvent function $\lambda \mapsto (\lambda-a)^{-1}$ is analytic on $\mathbb{C}$ and then it is stated that if $\varphi \in A^\ast$ then $\lambda \mapsto \varphi((\lambda - a)^{-1})$ is analytic too. I am trying to understand why this is true.

That is because a $\varphi \in A^\ast$ is $\mathbb{C}$-linear and continuous, which implies Fréchet-differentiability. Writing $R(a,\lambda) = (\lambda - a)^{-1}$, we have

$$\begin{align} \varphi(R(a,\lambda+h)) - \varphi(R(a,\lambda)) &= \varphi(R(a,\lambda+h) - R(a,\lambda))\\ &= \varphi\left(\frac{\partial R}{\partial \lambda}(a,\lambda)\cdot h + o(\lvert h\rvert)\right)\\ &= \varphi\left(\frac{\partial R}{\partial \lambda}(a,\lambda)\right)\cdot h + \varphi(o(\lvert h\rvert). \end{align}$$

Since $\varphi$ is a continuous linear map, $\varphi(o(\lvert h\rvert) \in o(\lvert h\rvert)$, so the complex differentiability of $\varphi \circ R(a,\,\cdot\,)$ follows.

$\endgroup$
8
  • $\begingroup$ Did you mean $f'(a) =\varphi$ for all $a \in A$? (in the last sentence before the third quotation) Or perhaps $f'(a) =\varphi(a)$? $\endgroup$ Commented Feb 22, 2014 at 10:50
  • 1
    $\begingroup$ No, $f'(a) = f$ is right. Not immaculately clear, though. A continuous linear map is its own derivative, we have $f'(a)(h) = f(h)$, or $f'(a) = h \mapsto f(h)$ for all $a$. $\endgroup$ Commented Feb 22, 2014 at 11:21
  • $\begingroup$ Thank you for your comment. I'm still a bit confused. I thought $\varphi$ was the linear operator $A_x$ here and $A_x$ is the derivative of $f$ at $x$. But then we'd have $\varphi = A_x = f'$ at $x$.Which of the equalities here I thought were true does not hold? $\endgroup$ Commented Feb 22, 2014 at 11:34
  • $\begingroup$ They all hold. But with $f \in A^\ast$, we additionally have $\varphi = f$. $\endgroup$ Commented Feb 22, 2014 at 11:46
  • 1
    $\begingroup$ @Student Usually, the derivative is required to be a continuous linear map, and then only continuous linear maps are Fréchet-differentiable. If one calls a map $f$ differentiable in $x_0$ if there is any linear operator $A$ such that $$\lim_{h\to 0}\frac{\lVert f(x_0+h) - f(x_0) - A(h)\rVert}{\lVert h\rVert} = 0,$$ then any linear operator is differentiable. But if one does that, differentiability no longer implies continuity, and compositions of differentiable functions are not necessarily differentiable, so one pays a big price for the greater generality. I don't remember why I put the ... $\endgroup$ Commented Jun 13, 2014 at 12:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .