Why can you take the supremum of f out of the integral?

Question

I'm trying to follow a proof and the last step goes as follows:

$$\int_a^b|f(\gamma(t))\gamma'(t)| dt \le sup_\gamma|f|\int_a^b|\gamma'(t)|dt$$

I don't get why you can do that? Btw, $sup_\gamma|f| = sup\{|f(z)| : z \in \gamma\}$.

Extra details: $f:U\rightarrow\mathbb{C}$ is continuous on $U$ (which is an open and path-connected subset of $\mathbb{C}$). And $\gamma:[a,b]\rightarrow\mathbb{C}$ is a contour (a.k.a. a $C^1$ curve).

Thanks in advance.

Answer 1

It's because $\sup|f|$ is a constant. Since you have$$\bigl|f\bigl(\gamma(t)\bigr)\gamma'(t)\bigr|\leqslant\sup|f|.\bigl|\gamma'(t)\bigr|,$$you have$$\int_a^b\bigl|f\bigl(\gamma(t)\bigr)\gamma'(t)\bigr|\,\mathrm dt\leqslant\sup|f|\bigl|\int_a^b\gamma'(t)\bigr|\,\mathrm dt.$$