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 $0< \alpha < 1$)
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?
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.