Calculate or estimate the norm in $L^2[-\pi, \pi]$ of the function $f(t)=\sum_{n\in \mathbb Z} c_n e^{i n t} g_n(t)$

Question

In $L^2[-\pi, \pi]$, with the inner product $$ (f, g)=\frac{1}{2 \pi} \int_{-\pi}^\pi f(t) \overline{g(t)} d t, $$ the complex trigonometric system $\left\{e^{i n t}\right\}_{-\pi}^\pi$ constitutes an orthonormal basis.

I would like to calculate/estimate the norm in $L^2[-\pi, \pi]$ of the following function: $$f(t)=\sum_{n\in \mathbb Z} c_n e^{i n t}\left(1-e^{-x_n t}\right)$$ i.e. $\|f\|^2$, where $x_n$ is a sequence of real numbers and $c_n$ is a sequence of complex numbers s.t., for simplicity's sake, $\sum_{n\in \mathbb Z} |c_n|^2=1$.

My approach: $$\|f\|^2=(f,f)=\frac{1}{2 \pi} \int_{-\pi}^\pi \left(\sum_{n\in \mathbb Z} c_n e^{i n t}\left(1-e^{-x_n t}\right)\right) \overline{\left(\sum_{m\in \mathbb Z} c_m e^{i m t}\left(1-e^{-x_m t}\right)\right)} d t$$ $$=\frac{1}{2 \pi} \sum_{m,n\in \mathbb Z} c_n \overline{c_m} \int_{-\pi}^\pi \left(1-e^{-x_m t}\right) \left(1-e^{-x_n t}\right) e^{i(n-m)t} dt$$ How can I continue? I was hoping to prove that the double series could be simplified into a series with a single index (n, for example) but it seems to me impossible for the double series to transform into a single one.

Update The sequence $x_n$ is bounded, i.e. there exists $M>0$ such that $|x_n|\leq M$ $\forall n$. To answer the @fedja's comment, I would happy to see such an estimation: $$\| f\|^2\leq C \left\|\sum_n^{+\infty} c_n e^{i n t}\right\|^2$$ where $C$ is a constant, maybe $C=\sup_{n,t}|1-e^{-x_n t}|$.

Answer 1

Ah, then it is easy. Note that $1-e^{x_n t}$ can be extended from $[-\pi,\pi]$ to a real-valued function $f_n(t)$ supported on $[-2\pi,2\pi]$ uniformly bounded in $C^2$, so we can write $$ \int_{-\pi}^\pi \left|\sum_n c_ne^{int}(1-e^{x_n t})\right|^2\,dt\le \int_{\mathbb R}\left|\sum_n c_ne^{int}f_n(t)\right|^2=\sum_{n,m}\widehat{[f_nf_m]}(m-n)c_n\bar c_m\\ \le C\sum_{m,n}\frac 1{(n-m)^2+1}|c_n||c_m|\le C'\sum_n|c_n|^2$$

This classical trick of extending the support to gain improved orthogonality can be used in many other contexts as well, so I recommend adding it to your toolbox.

Edit If you want your particular constant, you can use a slightly different trick that relies on the special expressions for your functions a bit more. Notice that $C=\sup_{n,t}|1-e^{-x_n t}|=\sum_{k\ge 1}\frac{1}{k!}X^k\pi^k$ where $X=\sup_n|x_n|$. Then we can just write $$ \left\|\sum_n c_ne^{int}(1-e^{-x_nt})\right\|_{L^2}= \left\|\sum_n \sum_{k\ge 1} c_ne^{int}\tfrac 1{k!}(-x_nt)^k\right\|_{L^2} \\ =\left\|\sum_{k\ge 1} t^k\sum_n [c_n\tfrac 1{k!}(-x_n)^k]e^{int}\right\|_{L^2}\le \sum_{k\ge 1}\left\| t^k\sum_n [c_n\tfrac 1{k!}(-x_n)^k]e^{int}\right\|_{L^2} \\ \le \sum_{k\ge 1}\pi^k\left\|\sum_n [c_n\tfrac 1{k!}(-x_n)^k]e^{int}\right\|_{L^2}=\sum_{k\ge 1}\pi^k\left[2\pi \sum_n |c_n\tfrac 1{k!}(-x_n)^k|^2\right]^{1/2} \\ \le \sum_{k\ge 1}\pi^k\left[2\pi \sum_n |c_n\tfrac 1{k!}X^k|^2\right]^{1/2} =\sum_{k\ge 1}\pi^k\tfrac 1{k!}X^k\left[2\pi \sum_n |c_n|^2\right]^{1/2} \\ =C\left\|\sum_n c_ne^{int}\right\|_{L^2}\,. $$