While doing the following practice question, I got stuck at the proof of uniform continuity of this function. (I know it should be uniformly continuous iff $0< \alpha < 1$)

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We can easily show that it will be uniformly continuous on $[0,1]$, and I believe that if it is also uniformly continuous on $(1,+\infty)$, then we are done. But how should we prove that? Also, how can we show that if $\alpha>1$ or $\alpha<0$, the function is not uniformly continuous?

  • $\begingroup$ Have you tried using the definition of uniform continuity? For every $\varepsilon>0$, try coming up with a $\delta>0$ such that $|x-y|<\delta \implies |g_a(x)-g_a(y)| < \varepsilon$, where your $\delta$ does not depend on the choice of $x$ or $y$. In order to show that it isn’t uniformly continuous on different values of $a$, all you have to do is show that for any $\delta$, you can find $x$ and $y$ such that $|x-y|<\delta$ to make $|g_a(x)-g_a(y)|$ arbitrarily large (say, greater than any $\varepsilon$) for this particular range of $a$. $\endgroup$ – Jack Crawford May 31 at 22:40

The derivative of the function is bounded on $[1,\infty)$ because $x^{\alpha -1}$ and $x^{\alpha -1} \log(x) \to 0$ as $x \to \infty$. Now apply MVT.

For $\alpha <0$ note that the function does not extend to a continuous function on $[0,\infty)$ and hence it cannot be uniformly continuous.

Let $\alpha >1$. If the function is uniformly continuous then there exists $\delta >0$ such that $|f(x)-f(x+\delta)| <1$ for all $x$. By MVT we get $\delta |f'(t)| <1$ for some $t >x$. But $|f'(x)| \to \infty$ as $ x \to \infty$ so we have a contradiction.

I will leave the case $\alpha =1$ to you.


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