Uniform continuity of $f(z)=1/z^2$

Question

Question: consider the function $f(z)= 1/z^2$,

a) prove that $f(z)$ is uniformly continuous in the region $1/2 ≤|z|≤1$

b)prove that $f(z)$ is not uniformly continuous in the region $|z|≤1$

My attempt:

(a): let $z, z_0$ be any points in the region $1/2≤|z|≤1$ and let $ε>0$ then,

$|f(z_0)-f(z)|= |1/z_0^2-1/z^2|$

$= |(1/z_0)^2 -(1/z)^2|$

$=|\frac{1}{z_0}- \frac{1}{z}||\frac{1}{z_0}+\frac{1}{z}|$

$≤\frac{|z-z_0|}{|z\text{ $z_0$}|} |\frac{1}{z_0}+\frac{1}{z}|$

$< \frac{δ}{ |z\text{ $z_0$}|} |\frac{1}{z_0}+\frac{1}{z}|$

$≤\frac{δ}{|z_0||z|}(\frac{1}{|z_0|}+\frac{1}{|z|})$

However, I didn't able to go further..(to show $δ$ just depends on $ε$ please can you show it.) Further, as we know in case real variable, as $g(x)= 1/x^2$ is uniformly continuous in any interval not containing $0$, hence my intension is that our function $f(x)$ is also uniformly continuous in the region $1/2≤|z|≤1$ is thinking in such way is valid here? Since we are in $\mathbb{C}$ not in $\mathbb{R}$ I think it is not good way to derive conclusion! What you say?

for (b): I don't know how to proceed.

please help me.. I am stuck from hours in proving those....otherwise I can use direct results and skip the proof but I don't want to do that...

Answer 1

Hint for (a):

For all $z$ in the region we have $1/|z| \leqslant 2$. Now apply this to the RHS of your estimate.

Hint for (b):

We have $|1/n|,|1/(2n)| \leqslant 1$ and, as $n \to \infty$,

$$\left|\frac{1}{n} - \frac{1}{2n} \right| \to 0, \,\,\,| (2n)^2 - n^2| \to \infty$$

Answer 2

Maybe you're not allowed to use this fact, but a continuous function on a compact domain is always uniformly continuous.

Answer 3

This is a proof of 'continuity in compacts $\implies$ uniform continuity' in a general setting.

Proposition: Let $(X,d_X)$ and $(Y,d_y)$ be metric spaces, $f:X\longrightarrow Y$ be continuous and $K\subset X$ be compact.

The restriction of $f$ to $K$ is uniformly continuous.

Proof: Let $\epsilon > 0$. By continuity, for every $x\in K$ there is some $\delta_x > 0$ such that $d_X\big(z,x\big)<\delta_x$ implies $d_Y\big(f(z),f(x)\big)<\epsilon/2$.

In particular, for every $x\in K$ there is some $\delta_x>0$ such that

$$z_0,z_1\in B(x;\delta_x) \implies d_Y\big(f(z_0),f(z_1)\big)<\epsilon$$

Now, the family $\mathcal F = \big\{B(x;\delta_x) : x\in K\big\}$ is an open cover of $K$, and hence $($because $K$ is compact$)$ there are $x_1,x_2,\dots, x_n \in K$ such that

$$K\subset \bigcup_{i=1}^nB\left(x_i;\delta_{x_i}\right).$$

Let $\delta$ be the Lebesgue number of this cover (see the lemma). Then for all $z_0,z_1 \in K$ we have

\begin{align} d_X(z_0,z_1) < \delta &\implies \exists i \in\{1,2,\dots,n\} \,\text{ such that }\, z_0,z_1 \in B\left(x_i;\delta_{x_i}\right) \\&\implies d_Y\big(f(z_0),f(z_1)\big)<\epsilon \end{align}

In other words, $f$ is uniformly continuous on $K$, as desired. $\square$

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Lemma (Lebesgue's number of an open cover): Let $(X,d)$ be a metric space, $K\subset X$ be compact and $\mathcal C = \bigcup_{i\in I}\,B_i$ be some (arbitrary) open cover of $K$.

There is some $\delta >0$ such that whenever $S\subset K$ has $\text{diam}(S) < \delta$, there is some $i \in I$ with $S\subset B_i$. A number with this property is called a Lebesgue number of $\mathcal C$.

Proof: Because $K$ is compact, there is some finite subcover $\{B_1,B_2,\dots,B_n\}\subset \mathcal C$. If some $B_i$ contains $K$, then any $\delta$ will do. Suppose no $B_i$ contains $K$.

For each $i\in\{1,2,\dots,n\}$ let $C_i = K\setminus B_i$. Notice that each $C_i$ is closed and nonempty. Define then $g:K\longrightarrow \mathbb R$ by

$$g(z) = \frac1n\sum_{i=1}^n\,d(z,C_i)$$

$g$ is continuous, and since $K$ is compact it must attain a minimum $\delta$. Notice that this minimum must be positive, by construction.

Now, suppose $S\subset K$ is nonempty and has $\text{diam}(S)<\delta$. Pick any $z\in S$ and note that $S\subset B(z;\delta)$.

Because $f(z) \geq \delta$, there must be some $i\in\{1,2,\dots,n\}$ such that $d(z,C_i)\geq \delta$. This implies $B(z;\delta) \subset B_i$, which in turn means $S\subset B_i$, as desired. $\square$

Answer 4

Last step is wrong. You cannot write delta unless you get it. Unboundedness of f helps. Try to use >=, instead of <=