Show for a uniformly continuous $f:\mathbb{C} \mapsto \mathbb{C}$ that there exists $\alpha,\beta \in \mathbb{R}$ so that the following is true

$$\forall z\in\mathbb{C}: |f(z)| \le \alpha\cdot|z|+\beta$$

I have written down the definition of uniform continuity, which is
$$\forall \epsilon \gt 0\ \ \exists \delta\gt0 \ \forall z,w\in\mathbb{C} \ \text{with} \ |w-z|<\delta : |f(w) - f(z)|<\epsilon$$
But that is as far as I got.

Any tips are appreciated.


I will take $\epsilon=1$ in the definition of uniform continuity.

By squeezing the $\delta$ a bit, we can assume that it is the $\leq$-inequality instead of $<$ and $2\delta$ instead of $\delta$.

So let us write $|w-z|\leq 2\delta$ implying that $|f(w)-f(z)|\leq 1$.

For $|w-\delta e^{i\theta}|\leq\delta$, $\theta\in[0,2\pi]$, then $|w|\leq|\delta e^{i\theta}|+\delta=2\delta$, so $|f(w)-f(0)|\leq 1$, so $|f(w)|\leq |f(0)|+1$.

For $|w-2\delta e^{i\theta}|\leq\delta$, $\theta\in[0,2\pi]$, $|f(w)|\leq|f(2\delta e^{i\theta})|+1\leq |f(0)|+2$.

For $|w-n\delta e^{i\theta}|\leq\delta$, $\theta\in[0,2\pi]$, $|f(w)|\leq |f(n\delta e^{i\theta})|+1\leq\cdots\leq|f(0)|+n$.

But for any $z$, some $n$ and $\theta\in[0,2\pi]$ are such that $|z-n\delta e^{i\theta}|\leq\delta$, then $n\delta-|z|\leq\delta$ and hence $n\leq (|z|+\delta)/\delta$, and that $|f(z)|\leq|f(0)|+n\leq |f(0)|+(|z|+\delta)/\delta$.

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  • 2
    $\begingroup$ I think if you want to pitch this appropriately at the OP's level, you should point out that you are taking $\epsilon$ to be $1$ in the usual definition of uniform continuity. $\endgroup$ – TonyK Dec 18 '19 at 22:24
  • $\begingroup$ This seems a bit beyond me. If it's not a big hassle to you, would you please try to explain step by step. Sorry $\endgroup$ – Zntzozt Dec 18 '19 at 22:52
  • $\begingroup$ Perhaps you can read this before looking at my solution. The philosophy is almost the same, just that in the real line you consider intervals but in the complex plane, you might have to consider annuli. $\endgroup$ – user284331 Dec 18 '19 at 23:03
  • $\begingroup$ I'll take look at it and make some research myself and I'll report back. $\endgroup$ – Zntzozt Dec 18 '19 at 23:19
  • $\begingroup$ @user284331 Sorry but I can't get my head around this. Is there another simpler way of achieving the same result ? $\endgroup$ – Zntzozt Dec 19 '19 at 17:57

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