Supremum of (e^(i z t) - 1)/z

Question

i'm new here, so i'm not sure if this is the right place to ask this question:

I know that the following holds true: $$ \forall\, t \in \mathbb{R} \; \forall\,x\in\mathbb{R}\setminus\{0\} \qquad : \qquad \left\vert \frac{\exp(\mathbb{i} tx)-1}{x} \right\vert \leq \vert t\vert $$ but i cant remember the proof... Does anyone know a quick one?

I actually would need the stronger inequality $$ \forall\, t\in\mathbb{R} \; \forall\, z\in\mathbb{C}\setminus\{0\} \qquad : \qquad \left\vert \frac{\exp(\mathbb{i} tz)-1}{z} \right\vert \leq \vert t \vert $$ but i'm not sure, if this is correct. anyone have an idea?

Thanks for any help!

Answer 1

This is the inequality:

$$ |f(t) - f(0)|= \left|\int_0^t f'(u) du\right| \le \int_0^t \left|f'(u)\right| du \le |t| \sup |f'| $$

with here, $$ f'(t) = ix \exp itx;\\ \sup |f'| = |x|. $$