Prove that the function $f(z) = \frac{1}{1-z}$ is not uniformly
continuous on $(-1,1)$.
Partial proof : Suppose $f$ is uniformly continuous.
$\implies \forall \epsilon > 0, \exists \delta > 0, \forall z, w \in (-1,1) :$ $ |\frac{1}{1-z} – \frac{1}{1-w}| < \epsilon$ as long as $|z-w|<\delta$
Let $\epsilon = 1$. Trying to find $0<m<1$ such that $w = mz$ for which the uniform continuity is not respected.
$|z-w| = |z|(1-m) < \delta (1-m) < \delta$
$|\frac{1}{1-z} – \frac{1}{1-w}| = |\frac{z(1-m)}{(1-z)(1-zm)}|$
I am blocked at this point. Is anyone is able to give me a hint to continue my proof?
Best Answer
Suppose by contradiction $f$ is uniformly continuous on $(-1,1)$. Let $\varepsilon = 1$ as you did, and let $\delta > 0$ be such that $\lvert f(x) - f(y)\rvert\leq \varepsilon$ for all $x,y \in(-1,1)$ such that $\lvert x-y \rvert \leq \delta$.
For $n \geq 1$, let $x_n = 1-\frac{1}{n}$, and $y_n = x_n - \delta^\prime$, where $\delta^\prime=\min(\frac{1}{100}, \delta)$. Then $\lvert x_n-y_n \rvert \leq \delta$, but $$ 1= \varepsilon \geq \lvert f(x_n) - f(y_n)\rvert = \left\lvert\frac{1}{1-x_n}-\frac{1}{1-y_n} \right\rvert = \frac{\lvert x_n-y_n \rvert}{(1-x_n)(1-y_n)} = \frac{\delta^\prime}{\frac{1}{n}(\delta^\prime + \frac{1}{n})} \xrightarrow[n\to\infty]{} \infty $$ leading to a contradiction.