Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to prove the following:

Let $\Re z > 0$. Then $$\lim_{\varepsilon \to 0} \frac{t^{z + \varepsilon} - t^z}{\varepsilon} = t^z \log t$$ uniformly in $t \in [0,1]$.

I've tried to bound it in a straight forward way to no avail. Also, assuming the convergence is among real positive epsilons, I could reduce it to the purely real case ($\Im z$ does not matter) and apply Dini's theorem. However, it does not seem to work well for arbitary $\varepsilon$.

How to prove it for the general complex case?

share|cite|improve this question

I assume we use $t^w = e^{w\log t}$ with the real-valued logarithm of $t$ for $t > 0$ and $t^w \equiv 0$ for $t = 0$ and $\Re w > 0$ as well as $0^z\log 0 = 0$, as suggested by continuity.

Then we have

$$\begin{align} \left\lvert \frac{t^{z+\varepsilon}-t^z}{\varepsilon} - t^z\log t \right\rvert &= t^{\Re z} \left\lvert\frac{t^\varepsilon-1}{\varepsilon}-\log t\right\rvert\\ &= t^{\Re z}\left\lvert\sum_{k=1}^\infty \frac{\varepsilon^{k-1}(\log t)^k}{k!} - \log t\right\rvert\\ &\leqslant t^{\Re z}\lvert\log t\rvert\sum_{k=1}^\infty \frac{\lvert\varepsilon\log t\rvert^{k}}{(k+1)!},\tag{1} \end{align}$$

and equality holds when $\varepsilon < 0$.

So for every $0 < \lvert\varepsilon \rvert < \Re z$, the function

$$f_\varepsilon \colon t \mapsto \frac{t^{z+\varepsilon}-t^z}{\varepsilon} - t^z\log t$$

is continuous on $[0,1]$ (since $t^{z}\log t\to 0$ for $t\to 0$ and $\Re z > 0$), we have $\lvert f_\varepsilon(t)\rvert \leqslant \lvert f_{-\lvert\varepsilon\rvert}(t)\rvert$ for all $\varepsilon$ and $t$ under consideration, and from $(1)$ it is immediate that $0 < r_1 < r_2$ implies $\lvert f_{-r_1}(t)\rvert \leqslant \lvert f_{-r_2}(t)\rvert$ for all $t$.

Dini's theorem now asserts that

$$\lim_{r\searrow 0} \lvert f_{-r}(t)\rvert = 0$$

uniformly on $[0,1]$, and $\lvert f_\varepsilon\rvert \leqslant \lvert f_{-\lvert\varepsilon\rvert}\rvert$ shows that

$$\lim_{\varepsilon\to 0} \frac{t^{z+\varepsilon}-t^z}{\varepsilon} - t^z\log t = 0$$

uniformly on $[0,1]$ too.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.