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Fix an $0 < \alpha < 1$ and consider $f(x) = x^{\alpha}$. Show that $f$ is uniformly continuous on $[1, \infty)$.

Work so far: If $\alpha = 1/2$, then we can prove $f$ is uniformly continuous since $|\sqrt{x} - \sqrt{y}| = |x - y|/|\sqrt{x} + \sqrt{y}|$. Similarly if $\alpha = 1/n$, we can write $$|x^{1/n} - y^{1/n}| = \frac{|x - y|}{|x^{1 - 1/n} + x^{1 - 2/n}y^{1/n} + \cdots + y^{1 - 1/n}|}$$ and we are done. However, this method doesn't seem to work for let's say $\alpha = \sqrt{2}/2$. How does one show $f$ is uniformly continuous for general $\alpha$?

hjk987
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1 Answers1

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Note that $$f'(x)=\alpha x^{\alpha-1}$$ is bounded in absolute value by $\alpha$ over $[1,\infty)$. Thus $f$ is Lipschitz over $[1,\infty)$.

Pedro
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