Suppose $$f(x) = \sum_{k=1}^n c_k\cos(kx)$$ satisfies $$|f(x)|\le 1,\quad \forall x\in[0,2\pi]$$ then does there exist a constant $A$ which does not rely on $n$ and the inequality $$\sum_{k=1}^n |c_k|<A$$ always holds?
I think by replacing $z = e^{ix}$, the question can be converted to a problem about a complex polynomial's sum of coefficients's norm and its infinite norm, which maybe solved by taking some kind of integration over the unit circle, but I failed in all my attempts.
The answer is no even if $c_k \in \mathbb R$ as one knows that $\sum_{k \ge 1} \frac{\sin (2k+1)x}{2k+1}$ has uniformly bounded partial sums by some constant $A$ so changing variables to $x \to \pi/2-x$ immediately gives the same for $\sum_{k \ge 1} \frac{(-1)^k\cos (2k+1)x}{2k+1}$, hence taking $f_n(x)=\frac{1}{A}\sum_{k=1}^n \frac{(-1)^k\cos (2k+1)x}{2k+1}$, one gets $|f_n(x)| \le 1$, but $\sum_{k=1}^n \frac{1}{2k+1} \to \infty$
With more care (and slightly more complicated coefficients) one can construct a counterexample where $\sum_{k=1}^n c_k \cos kx$ converges uniformly to a continous function on $\mathbb R$ and in particular its partial sums are uniformly bounded, but the convergence is not absolute so $\sum |c_k| \to \infty$