Uniform limit of complex difference quotient Let $\mathcal{C}_0([0,1])$ be the set of continuous complex-valued functions defined on $[0,1]$ such that they vanish at the origin and let $\|\cdot\|_{MAX}$ denote the norm
$$\|x\|_{MAX} = \max_{t\in [0,1]}|x(t)|.$$
If we consider the function $f_\zeta(t) = t^{\zeta}$ for $\zeta\in \{z\in \mathbf{C}\mid \mathrm{Re}\ z > 0 \}$ then $f_{\zeta}(t)\in \mathcal{C}_0([0,1])$. Now what I want to show is that
$$\frac{f_{\zeta+\delta}-f_{\zeta}}{\delta}\rightarrow t^{\zeta}\log t$$
in the norm $\|\cdot\|_{MAX}$. And I am not quite sure how to do it. I understand that this is the point wise limit for each fix $t\in [0,1]$.
$$\left|\frac{f_{\zeta+\delta}(t)-f_{\zeta}(t)}{\delta}-t^{\zeta}\log t\right| = \left|t^{\zeta}\left(\frac{e^{\delta\log t}-1}{\delta}-\log t\right)\right|.$$
How can I show that the maximum of all $t\in [0,1]$ of the above absolute value tends to $0$? A tip on how to proceed would be very appreciated!
 A: We can assume that $\zeta$ is real (the imaginary part $t^{i \mathfrak{Im}(\zeta)} = e^{i \mathfrak{Im}(\zeta) \cdot \log(t)}$ doesn't give any contribution to the modulus). Using $$ e^{\delta \log(t)} = \sum_{n=0}^\infty \frac{\delta^n \log^n (t)}{n!} $$ we get that $$\left| t^\zeta \left( \frac{e^{\delta \log(t)} -1}{\delta} - \log(t) \right) \right| = \left| t^\zeta \sum_{n \ge 2} \frac{\delta^{n-1} \log^n (t)}{n!} \right|.$$Now consider the sequence of real-valued functions $f_n (t) := t^\zeta \log^n (t)$; clearly $f_n (1)=1$ and also all the $f_n$ can be extended by continuity at $t=0$. We will then confuse $f_n$ with their continuous extensions on $[0,1]$; with this in mind, $f_n(0)=0$ for all $n$. The functions $f_n$ have maxima belonging to $(0,1)$; this maxima satisfy $$ f_n ' (t) = \zeta t^{\zeta-1} \log^n (t) + n t^{\zeta-1} \log^{n-1} (t) = t^{\zeta -1} \log^{n-1}(t) (\zeta \log(t) + n) = 0 $$ which is true if $$t=e^{-n/\zeta} \in (0,1).$$ This means that we can bound $$|t^{\zeta} \log^n (t)| \le |t^{\zeta} | \cdot | \log^n (t)| \le e^{-n} \cdot \left( \frac{n}{\zeta} \right)^n = \left(\frac{n}{e \zeta} \right)^n. $$Hence $$ \left| t^\zeta \sum_{n \ge 2} \frac{\delta^{n-1} \log^n (t)}{n!} \right| \le \sum_{n \ge 2} \frac{\delta^{n-1}}{n!} \left(\frac{n}{e \zeta} \right)^n = \sum_{n \ge 2} \frac{\delta^{n-1} p^n }{n!} \left( \frac{ n}{p e \zeta} \right)^n  $$where $p$ is a fixed positive real (depending on $\zeta$) such that $p   > 1/\zeta$; in this way we can make use of Stirling's approximation and say that $$ \sum_{n \ge 2} \frac{\delta^{n-1} p^n }{n!} \left( \frac{ n}{p e \zeta} \right)^n \le \underbrace{\sum_{n = 2}^N \frac{\delta^{n-1} p^n }{n!} \left( \frac{ n}{p e \zeta} \right)^n}_{=I} + \underbrace{\sum_{n \ge N+1} \delta^{n-1} p^n}_{=II}  $$for some $N \in \mathbb{N}$ big enough. Notice that we need $$ p \sum_{n\ge N+1} \delta^{n-1} p^{n-1} < \infty $$ and this happens for $\delta < 1/p < \zeta$. In conclusion using the formula for the geometric series we have $$I+II = I + p \left( \frac{1}{1-\delta p} - \frac{1 - (\delta p)^{N+1}}{1- \delta p} \right) \to 0 $$ when $ \delta \to 0$ since $I$ is a finite sum. Since the RHS is independent from $t$ we have the thesis.
