I am reading the proof of the statement that no non-zero multiplication operator on $L^2([0,1])$ is compact in this post. And I would like to address it as a seperate post as I am only curious about one of the step.
So We have $\{f_n\}$ orthonormal basis for $L^2([0,1])$ where $f_n=e^{2\pi inx}$. We would like to show that for every nonzero $g\in L^\infty$, the sequence $\{gf_n\}$ is not Cauchy.
We start by writing $g=\sum a_n f_n$ and $g f_k=\sum a_nf_{n+k}$. Now we want to show that $ \lVert gf_{k+j}-gf_k\lVert> c$ for some postive constant c.
now $gf_{k+j}-gf_k=\sum c_n f_n$ and by Bessels's inequality we have $ \lVert gf_{k+j}-gf_k\lVert^2> \sum |c_n|^2$. My question is in our case, how do we now any of the $|c_n|$ is non-zero?
Thanks for your help!
We have $$\begin{split} gf_{k+j}-gf_k &=\sum a_nf_{n+k+j}-\sum a_nf_{n+k}\\ &=\sum (a_{n-k-j}-a_{n-k})f_n \end{split}$$ Therefore, $c_n=a_{n-k-j}-a_{n-k}$. So, there must be at least one $c_n$ that's non-zero, because if there weren't any, you'd have $a_{n-k-j}=a_{n-k}$ for all $n$. That is, you'd have $a_n=a_{n+j}$ for all $n$, which would imply that either $a_n$ is zero for all $n$, or that $$\|g\|^2 = \sum |a_n|^2 = +\infty$$