Let $f:\mathbb C\longrightarrow \mathbb R$ be a continuous function such that $\,\lvert\, f(z)\rvert\le 1$ for all $z\in S^1\subset \mathbb C$. Prove that $$\left| \int_{\lvert z\rvert=1} f(z)\;\mathrm dz\right|\le \int_0^{2\pi}\lvert \sin(t)\rvert\,\mathrm dt.$$
I have tried to evaluate the integral on the left using the usual parametrization $\cos(t)+i\sin(t)$ and then using the triangle inequality for integrals, but with no success. Please, I'm looking for a hint.
We can view $f$ as a continuous, real-valued and $2\pi-$periodic function, absolutely bounded by 1.
Then $$ \int_{\lvert z\rvert=1}f(z)\,dz=i\int_0^{2\pi} f(t)\,\mathrm{e}^{it}\,dt=r\,\mathrm{e}^{i\vartheta}. $$ Then $$ \left|\,\int_{\lvert z\rvert=1}f(z)\,dz\,\right|=r= \int_0^{2\pi}f(t)\,\mathrm{e}^{i(t-\vartheta+\pi/2)}\,dt. $$ Hence, if $g(t)=f(t+\vartheta-\pi/2)$, then $$ \left|\,\int_{\lvert z\rvert=1}f(z)\,dz\,\right|= \int_0^{2\pi}g(t)\,\mathrm{e}^{it}\,dt=\int_0^{2\pi}g(t)\,\cos t\,dt, $$ since $\,\,\mathrm{Im}\,\int_0^{2\pi}g(t)\,\mathrm{e}^{it}\,dt=0$. But the function $g$ is also a continuous, real-valued and $2\pi-$periodic function, absolutely bounded by 1.
Hence $$ \left|\,\int_{\lvert z\rvert=1}f(z)\,dz\,\right| =\int_0^{2\pi}g(t)\,\cos t\,dt\le \int_0^{2\pi}\lvert g(t)\rvert\,\lvert\cos t\rvert\,dt \le \int_0^{2\pi}\lvert \cos t\rvert\,dt=\int_0^{2\pi}\lvert \sin t\rvert\,dt. $$ Note. In fact, the strict inequality is the one that holds, i.e, $$ \left|\,\int_{\lvert z\rvert=1}f(z)\,dz\,\right| <\int_0^{2\pi}\lvert \sin t\rvert\,dt, $$ as $$ \int_0^{2\pi}g(t)\,\cos t\,dt< \int_0^{2\pi}\lvert\cos t\rvert\,dt. $$