Which real values of $\alpha$ is the function $g_{\alpha}(x) = x^{\alpha}\log(x)$ uniformly continuous?
This is a larger question and through the course of it, I have proven and used a few theorems to take care of the other cases. But there is one case in which I'm not sure if I reasoned things out correctly.
Attempt
Case: $0 < \alpha < 1$ on the interval $[1, \infty)$ where $x > y > y_{1} > 1$ Here $y_{1}$ is considered a constant as will be apparent in my proof.
By the Mean Value Theorem, there exists an $x_{*} \in (x,y)$ such that:
\begin{multline} |f(x) - f(y)|= f'(x_{*})|x-y| = [\alpha (x_{*})^{\alpha - 1} \log(x_{*}) + x_{*}^{\alpha - 1}]|x-y|\\ < [\alpha (y_{1})^{\alpha - 1} \log(y_{1}) + y_{1}^{\alpha - 1}]|x-y| = \biggl(\frac{1}{y_{1}^{1-\alpha}}\biggr)(\alpha \log(y_{1}) + 1)|x-y| \end{multline}
If we let $$\delta = \frac{y_{1}^{1-\alpha}}{(\alpha \log(y_{1}) + 1)}$$
Then we would have a satisfactory $\delta$ to establish uniform continuity on $[1,\infty)$.
Concern:
My concern revolves around the $y_{1}$ term I introduced. By the properties of Real numbers I know such a number does exist on this interval, but the purpose of uniform continuity is not being dependent on any of these values. Now I am taking the smallest $y_{1}$ and this is a constant, but it feels as if this may not be what I am aiming for. Thoughts ?
Edit:
Would the fact that I have shown that my derivative could be bounded by $f'(y_{1})$ then I have established uniform continuity?
Let $g(x) = \dfrac{\mathrm{d}}{\mathrm{d}x}(x^\alpha \ln x)$. You wish to bound $|g(x)|$ on $[1,\infty)$. Since $g(x) > 0$ on $[1,\infty)$, we can drop the absolute value bars. Happily, $g'(x) = 0$ has one solution on $(0,\infty)$, at $x = \exp \left(\dfrac{1-2\alpha}{\alpha(\alpha - 1)}\right)$, at which $g''$ is negative...
So you find $g(x)$ is bounded by an explicit constant depending only on $\alpha$.
(Carefully doing all the algebra, you will use the fact that $\alpha \in (0,1)$ repeatedly.)
Why this is a problem for your proof: What happens when $x_*$ lands on this maximum? Your inequality becomes invalid. The easiest way to see this is to pick $x$ and $y$ to straddle the maximum so that $f(x) = f(y)$, forcing $x_*$ to the critical point. But then the expression in $x_*$s is larger than the same expression with any other $y_1$ replacing $x_*$.