In my complex class we are given this problem: Take a continuous function from the complex plane to the real line: $f(z): \mathbb{C} \rightarrow \mathbb{R}$. We also know that $|f(z)| \leq 1$. Show that $$\bigr| \int_C f(z) dz\bigl| \leq 4$$ Where $C$ is the unit circle.
The way I started this was by saying: $$\bigr| \int_C f(z) dz\bigl| \leq \int_C |f(z)| dz \leq \int_C dz = \int_{0}^{2 \pi}ie^{i \theta} d{\theta} = 0$$This to me does not seem right, given what I am doing here is actually assuming the function is analytic, which was never given in the problem!
So now I am trying this: $$sup_{f}\bigr| \int_C f(z) dz\bigl| = sup_{f}\bigr| i\int_{0}^{2\pi} f(e^{i\theta}) e^{i\theta} dz\bigl|$$ While keeping in mind that $|f(z)| \leq 1$ and $f(z): \mathbb{C} \rightarrow \mathbb{R}$
I am however not too sure of how to attack this problem in a smart way, by not just plugging in functions and see what happens.
Thank you for your insight!