Is $\Bigl\vert \int_\gamma f(z) dz \Bigr\vert \leq \int_\gamma \vert f(z) \vert dz$ sometimes true?

Question

For normal integrals we know $$\Bigg\vert \int_a^b f(z) dz \Bigg\vert \leq \int_a^b \vert f(z)\vert dz$$ However, for line integrals like $\gamma:[0,2\pi]\to\mathbb{C},t\mapsto e^{it}$ we only can say $$\Bigg\vert \int_\gamma f(z) dz\Bigg\vert \leq \int_0^{2\pi}\vert f(\gamma(t))\vert\vert\gamma'(t)\vert dt$$

Now I wonder if there is any way to interchange the absolute value with a line integral directly, anything like $$\Bigg\vert \int_\gamma f(z) dz\Bigg\vert \leq \int_\gamma \vert f(z)\vert dz$$

Can anyone provide example where is is wrong? And is this maybe even true with some extra conditions? Or is this just straight up nonsense?

Answer 1

Let $f(z)=\frac1z$ and $\gamma$ be the counterclockwise unit circle centered at the origin. Then $\left\vert\oint_\gamma f(z)\mathrm dz\right\vert=2\pi$, but $\vert f(z)\vert=1$ on $\gamma$, so it's constant on the given curve. But closed line integrals of constant functions are $0$, so you'd obtain the inequality $2\pi\leq0$, which is clearly wrong.

I don't know of any "nice" conditions which would make your inequality true.

Answer 2

We have $$\int_\gamma \vert f(z)\vert dz = \int_0^{2\pi} \vert f(\gamma(t))\vert \gamma'(t) dt$$ and $\gamma'(t)$ could be not real so the inequality makes no sense.

Even if it's real, it could be negative, and then the inequality will be usually false.