Find a correct form for $|u^h-1|\leq |h| |\log u| u^h$

Question

I am reading the article "A Direct Approach to the Mellin Transform". To prove Lemma 2 (see below), the authors used the following inequality \begin{equation} (I): \hspace{1cm} |u^h-1|\leq |h| |\log u| u^h, \end{equation} where $h\in\mathbb{C}$ $\left(|h|<\delta\right)$ and $u\in\mathbb{R}_+$. Then they wrote \begin{align}(II): \hspace{1cm} |f_{\mathcal{M}}^{\wedge}\left(s+h\right)-f_{\mathcal{M}}^{\wedge}\left(s\right)| &\leq \int_{0}^{\infty}|f\left(u\right)\left(u^{s+h-1}-u^{s-1}\right)|\mathrm{d}u\\ &\leq |h| \int_{0}^{\infty}|f\left(u\right)\left(\log u\right)u^{s+h-1}|\mathrm{d}u, \end{align} where $f_{\mathcal{M}}^{\wedge}\left(s\right)$ is the Mellin transform of $f$. However, I can not prove the first inequality, and I think it is wrong. I want to use the elementary inequality \begin{equation} |e^z-1|\leq |z|e^{|z|}. \end{equation} But what I can get is \begin{equation} |u^h-1|=|e^{h\log u}-1|\leq |h| |\log u| e^{|h \log u|}. \end{equation} My question: Can we find a corrected version of inequality (I) s.t. the second inequality (II) can be simply proved by using the corrected version ?

Answer 1

Ok, there seems to be a misprint in (2.5). I suppose it should be $\log u$ times the given power of $u$ (as they use below). Supposing this, if the authors have not made similar mistakes in proving (the correct version of) (2.5) then the last inequality is correct. Let $|h|<\delta$. My the MVT (using a path $th, 0\leq t\leq 1$ from 0 to $h$) we get $$ |u^h -1| = |e^{h \log u} - 1| \leq |h||\log u| \sup_{0\leq t\leq 1} |u^{th}| \leq |h||\log u| \sup_{|k|<\delta} |u^k| $$

For $0<u\leq 1$, $|k|<\delta$ you have $|u^{k}|\leq u^{-\delta}$ and for $u\geq 1$ you have $|u^k|\leq u^{\delta}$ which fits with the (correct version of) inequality (2.5) to give the stated bound. The authors (or referees) should have been more vigilant.