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Let $f(x) =\frac{1}{x^{\alpha}}$where $x \in (0, 1)$ and $0<\alpha \le 1$. Is $f$ continuous? What about uniform continuity? Justify.

I know this function is continuous in the given domain for all such $\alpha$ but how do I show it rigourously?

For uniform continuity I need to show for any $\epsilon >0$ $ \exists \delta >0 $ for which $\vert x-y \vert < \delta\implies \vert f(x) - f(y) \vert < \epsilon $

How do I do this ?

Empty
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Rusty
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1 Answers1

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For $0<\alpha\leqslant1$, we have $$f(x) = x^{-\alpha} = \exp\left(-\alpha\log x \right), $$ so continuity follows from the composition of continuous functions.

However, given the Cauchy sequence $x_n=\frac1n$ in $(0,1)$, we have $$f(x_n) = \left(\frac1n\right)^{-\alpha} = n^\alpha\stackrel{n\to\infty}\longrightarrow\infty, $$ so $f$ is not Cauchy-continuous, and therefore not uniformly continuous. To see this more explicitly, fix $m$ and consider the behavior of $|f(x_n)-f(x_m)|$ as $n\to\infty$.

Math1000
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  • @SMath be sure to see the revised answer; my original one had an error. – Math1000 Jan 25 '16 at 05:10
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    There are many ways to show that if $S$ is a bounded non-empty subset of $R$ and $f:S\to R$ is unbounded then $f $is not uniformly continuous. – DanielWainfleet Jan 25 '16 at 11:14
  • Thanks @user254665 - although "non-empty" is not needed since any function defined on the empty set is trivially bounded :) – Math1000 Jan 27 '16 at 04:22
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    OK.I was being overly cautious. I'd rather be redundant than wrong. – DanielWainfleet Jan 27 '16 at 05:02
  • Indeed. Although the empty set is not a very interesting metric space, it is necessary to include this trivial case, cf. http://math.stackexchange.com/questions/45145/why-are-metric-spaces-non-empty – Math1000 Jan 27 '16 at 16:12