I wanted to relate the expression $$\left\|\sum_n c_n e^{int}\right\| \cdot \left\|\sum_n f_n(t)\right\|$$ to the norm of $\sum_n c_n e^{int} f_n(t)$ in $L^2(-\pi,\pi)$, so i thought of using the Cauchy-Schwarz inequality.
The Cauchy-Schwarz inequality states that for any two functions $g(t)$ and $h(t)$ in $L^2(-\pi,\pi)$, we have $$\left|\int_{-\pi}^{\pi} g(t)h(t) dt\right| \leq \left(\int_{-\pi}^{\pi} |g(t)|^2 dt\right)^{\frac{1}{2}} \left(\int_{-\pi}^{\pi} |h(t)|^2 dt\right)^{\frac{1}{2}}.$$
Let's consider the functions $g(t) = \sum_n c_n e^{int}$ and $h(t) = \sum_n f_n(t)$. Applying the Cauchy-Schwarz inequality to these functions, we have $$\left|\int_{-\pi}^{\pi} \left(\sum_n c_n e^{int}\right) \left(\sum_n f_n(t)\right) dt\right| \leq \left(\int_{-\pi}^{\pi} \left|\sum_n c_n e^{int}\right|^2 dt\right)^{\frac{1}{2}} \left(\int_{-\pi}^{\pi} \left|\sum_n f_n(t)\right|^2 dt\right)^{\frac{1}{2}}.$$
Since we are working in $L^2(-\pi,\pi)$, we know that the norm of a function is given by $\|g(t)\| = \left(\int_{-\pi}^{\pi} |g(t)|^2 dt\right)^{\frac{1}{2}}$. Using this notation, the inequality becomes $$\left|\int_{-\pi}^{\pi} \sum_n c_n e^{int} \sum_n f_n(t) dt\right| \leq \left\|\sum_n c_n e^{int}\right\| \cdot \left\|\sum_n f_n(t)\right\|.$$
Anyway, the LHS of the previous formula is not the norm $\|\sum_n c_n e^{int} f_n(t)\|$.
Does anyone have any suggestions for estimating the product $\left\|\sum_n c_n e^{int}\right\| \cdot \left\|\sum_n f_n(t)\right\|$ from below?
It is not possible to estimate $\left \|\sum_nc_ne^{int}f_n\right \|$ by a constant times the product of $\left \|\sum_nc_ne^{int}\right \|$ and $\left \|\sum_nf_n\right \|.$ Indeed, let $c_0=c_1=1$ and $c_n=0$ for $n\neq 0,1.$ Then for $f_0(t)\equiv 1,$ $f_1(t)\equiv -1$ and $f_n(t)\equiv0$ for $n\neq 0,1$ we get $$\sum_nf_n=0,\quad \left \|\sum_nc_ne^{int}f_n(t)\right \|=\|1-e^{it}\|=2\sqrt{\pi}$$ Similarly it is not possible to get the reversed estimate. Indeed let $c_0=c_1=1,$ $c_n=0$ for $n\neq 0,1$ and $f_0(t)\equiv 1,$ $f_1(t)=e^{-it},$ $f_n(t)\equiv 1$ for $n\neq 0,1.$ Then $$\sum_n c_ne^{int}f_n(t)=1-1=0$$ and $$\left \|\sum_nf_n\right \|=\|1+e^{-it}\|=2\sqrt{\pi}=\|1+e^{it}\|=\left \|\sum_n c_ne^{int}\right \|$$