I have the following problem:
Let f a continuous function such that $f(\mathbb{S}^{1})\subseteq\mathbb{R}$, prove that \begin{equation*} \left|\int_{\mathbb{S}^{1}}f(z)dz\right| \leq 4\max_{x\in\mathbb{S}^{1}}|f(z)| \end{equation*}
What i try is: Let $\int_{\mathbb{S}^{1}}f(z)dz = Re^{i\theta} \in\mathbb{C}$, note that \begin{equation} R = \left|\int_{\mathbb{S}^{1}}f(z)dz\right| \end{equation}
Then \begin{eqnarray*} R & = & e^{-i\theta}\int_{\mathbb{S}^{1}}f(z)dz\\ & = & \int_{\mathbb{S}^{1}}e^{-i\theta}f(z)dz\\ & = & \int_{0}^{2\pi}e^{-i\theta}f(e^{it})ie^{it}dt\\ & = & \int_{0}^{2\pi}f(e^{it})ie^{i(t-\theta)}dt\\ & = & \int_{0}^{2\pi}f(e^{it})[i(\cos(t-\theta)+i\sin(t-\theta)]dt\\ & = & \text{Re}\int_{0}^{2\pi}f(e^{it})[(i\cos(t-\theta)-\sin(t-\theta)]dt\\ & = & \int_{0}^{2\pi}f(e^{it})\sin(\theta-t)dt\\ \end{eqnarray*} then \begin{eqnarray*} R & = & \left|\int_{0}^{2\pi}f(e^{it}\sin(\theta-t)dt\right|\\ &\leq& \int_{0}^{2\pi}|f(e^{it}|\cdot|\sin(\theta-t)|dt\\ &\leq& \max_{z\in\mathbb{S}^{1}}|f(z)|\cdot\int_{0}^{2\pi}|\sin(\theta-t)|dt\\ & = & 4\max_{z\in\mathbb{S}^{1}}|f(z)| \end{eqnarray*} concluding the desired. I wanted to know if there is an alternative way to do this demonstration, either with complex variable tools, or analysis in general.
Also as a comment, this type of properties is interesting because we have \begin{equation*} \left|\int_{\mathbb{S}^{1}} f(z)dz\right| \leq\text{L}(\mathbb{S}^{1})\cdot\max_{z\in\mathbb{S}^{1}}|f(z)|=2\pi\cdot\max_{z\in\mathbb{S}^{1}}|f(z)| \end{equation*} and $2\pi\sim6.28>4$, in summary, we obtain an improvement of the dimension that gives us the length of the curve.