# Proof of uniform continuity of the function: $x^\alpha\log(x)$

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?

• 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 $a$. – 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.