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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?

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  • $\begingroup$ I assume you are not allowed to use the fact that if $f$ is uniformly continuous on $(a,b)$, then it can be continuously extended on $[a,b]$? $\endgroup$
    – Clement C.
    Feb 8 '16 at 3:08
  • $\begingroup$ @ClementC. Could you be a bit more precise? Roughly, I would like to prove $f$ is not uniformly continuous with the simple general definition. Is it possible with what I did so far. $\endgroup$ Feb 8 '16 at 3:13
  • $\begingroup$ See e.g. this question (first item), or (for a weaker statement, but sufficient for your purposes) this answer. $\endgroup$
    – Clement C.
    Feb 8 '16 at 3:14
  • $\begingroup$ I can't use this theorem to show that f is not uniformly continuous $\endgroup$ Feb 8 '16 at 3:15
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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.

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  • $\begingroup$ Stating here as well for reference: this would be a direct consequence of the fact that a function uniformly continuous on a subset $(a,b)$ of $\mathbb{R}$ can be continuously extended to $[a,b]$ (follows from this more general statement, see also many questions on this website, e.g. this one). $\endgroup$
    – Clement C.
    Feb 8 '16 at 3:28
  • $\begingroup$ Why did you take $\delta' = \min(\frac{1}{100}, \delta)$? $\endgroup$ Feb 8 '16 at 3:29
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    $\begingroup$ Just to make sure that $y_n\in(-1,1)$ as well. If $\delta$ was big, say $\delta=3$ (we don't have a direct control on $\delta$), that could happen: this is just a safety check to avoid running into this (quite benign but annoying) sort of details. $\endgroup$
    – Clement C.
    Feb 8 '16 at 3:30
  • $\begingroup$ So you could take any value in $(0,1)$ instead of $\frac{1}{100}$ and it should be good, right? $\endgroup$ Feb 8 '16 at 3:33
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    $\begingroup$ Yes -- I was lazy and did not want to think about it. "$\frac{1}{100}$ always works." $\endgroup$
    – Clement C.
    Feb 8 '16 at 3:33

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