Let $a_1, a_2, \dots, a_n$ given with $a_i = \pm 1$. Let $f(x) = \sum_{k=1}^n a_k e^{ikx}$. I need to prove that $\lVert f \rVert_\infty \geq \sqrt{n}$ where $\lVert f \rVert_\infty$ is defined as the maximum value of $|f|$ in the interval $[0,2\pi]$.
Attempt: I tried experimenting with all $a_i$ being $1$ or $-1$ but I couldn't generalize it. Any hint would be appreciated. No need for the full solution.
Hint: Define $g(x) = f(2 \pi x)$. Note that $\|g\|_\infty \geq \|g\|_2$, where $\|g\|_\infty$ is the maximum of $|g(x)|$ for $x \in [0,1]$ and $$ \|g\|_2 = \left[\int_0^1 |g(x)|^2\,dx\right]^{1/2}. $$
Second Hint: Let $f_k(x) = e^{2 \pi ikx}$. Verify that the vectors $f_1,\dots,f_n$ form an orthonormal set relative to the inner product $$ \langle f, g \rangle = \int_0^1 f(x) \overline{g(x)}\,dx, $$ where $\bar z$ denote the complex conjugate of $z$. Use this to conclude that $$ \left\| \sum _{k=1}^n a_k f_k\right\|_2 = \sqrt{\sum_{k=1}^n |a_k|^2}. $$
