Can a bounded analytic function have unbounded derivative?

Question

Inspired by a physics problem, can a bounded analytic function have unbounded derivative?

I have seen, for instance, this question, but that one and the others all seem to take advantage of the fact that the function is not differentiable around the point where it has an infinite derivative, but in physics we usually deal with analytic function that ¨behave nicely¨ on ¨nice looking domains¨.

Addendum:

Please consider functions that are analytic on either $\mathbb{R}$ or $[0,1]$.

There are obvious examples (e.g., $(x-1)^{-1}$) on non-compact sets. I can't think of an example on a compact set immediately. — MathematicsStudent1122, Sep 26 '21 at 07:26
@AlohaSine the functions that are considered in physics are usually defined on compact sets or on the whole real axis. — Our, Sep 26 '21 at 07:29
Having regard to the addendum, the question should not have been closed as the linked question is completely different. — Adayah, Sep 26 '21 at 09:59

score 0 · Answer 1 · answered Sep 26 '21 at 07:51

When a function $f$ is analytic on an open interval I, then the derivative of $f$ is bounded on every closed sub-interval of $I$.

But notice the following examples, where an analytic function is bounded but its derivative is not bounded like:

$\sqrt x$ at $x=0$.
$\arcsin x$ at $x=\pm1$.

In these specific cases, the $x$'s are called branch points, and the derivatives are $+\infty$ in either case.

Notice however, that the functions are not analytic at these branch points, even though they are continuous there. The functions are only analytic on $\Bbb R^+$ resp. $(-1,1)$ Theses examples carry over to $\Bbb C$ by the way; and even though the square-root extends to negative numbers, the square-root is still not analytic in $0$.

Can a bounded analytic function have unbounded derivative?

1 Answers1