I am currently trying to figure out the following estimate:
Let $\gamma: [0, 2 \pi] \to \mathbb C, \gamma(t) = e^{\mathrm i t}$, $\gamma^* := \gamma[0, 2 \pi]$ and $f: \gamma^* \to \mathbb R$ be continuous. Then it holds $$\left\vert \int_{\gamma} f(z) \, dz \right\vert \leq 4 \max_{z \in \gamma^*} \vert f(z) \vert.$$
Obviously the naive estimate yields $$\left\vert \int_{\gamma} f(z) \, dz \right\vert \leq 2 \pi \max_{z \in \gamma^*} \vert f(z) \vert.$$ So one needs to be a little bit more clever. So I wrote down \begin{align*} \int_{\gamma} f(z) \, dz = \int_0^{2\pi} f(e^{\mathrm i t}) \mathrm i e^{\mathrm i t} \, dt = \int_0^{2\pi} f(e^{\mathrm i t}) (\mathrm i \cos(t) + \sin(t)) \, dt \end{align*} Now I tried a few things. I tried to use $\vert z \vert^2 = \operatorname{Re}(z)^2 + \operatorname{Im}(z)^2$ and I wrote the integrals as double integrals achieving \begin{align*} \left\vert \int_{\gamma} f(z) \, dz \right\vert^2 &= \int_0^{2 \pi} \int_0^{2 \pi} \cos(t - s) f(e^{\mathrm i t}) f(e^{\mathrm i s}) \, dt \, ds \\ &\leq \left( \max_{z \in \gamma^*} \vert f(z) \vert \right)^2 \int_0^{2 \pi} \int_0^{2 \pi} \vert \cos(t - s) \vert \, dt \, ds = 8 \pi \left( \max_{z \in \gamma^*} \vert f(z) \vert \right)^2 \end{align*} by using the addition theorem, which is a better bound but not good enough. I guess one has to split the first integral in a clever way but I don't have any idea how. Maybe one can make some progress from using the connection between $\sin$ and $\cos$. It seems to be essential that $f$ has real values, but I don't see how to use it. I would appreciate some hints or solutions :)
We want to estimate $|\int_0^{2\pi}f(t)e^{it}\,dt|.$ This equals
$$e^{is}\int_0^{2\pi}f(t)e^{it}\,dt=\int_0^{2\pi}e^{i(s+t)}f(t)\,dt$$
for some real $s.$ Now the above is nonnegative, so it equals
$$\int_0^{2\pi}\text { Re}\left (e^{i(s+t)}f(t)\right)\,dt. $$
Since $f$ is real valued (!), the last integral equals
$$\int_0^{2\pi}\cos(s+t)f(t)\,dt \le M\int_0^{2\pi}|\cos(s+t)|\,dt. $$
Here $M$ is the maximum value of $|f|.$ We can take $s=0$ for simplicity in the last integral to see it equals $4.$ We're done.