I have been stuck in this question for a while, and I would appreciate any new ideas. I have been considering the inequality
$$
\|f\|^2\_{H^{-1}(\mathbb{T})}\lesssim \int\_\mathbb{T} |\sin(x)f(x)|^2 \; dx,
$$
assuming that $\int\_\mathbb{T} f(x)=0$.


The inequality has the Fourier series analogue
$$
\sum\_{j\in\mathbb{Z}}\frac{|f\_j|^2}{j^2}\lesssim \sum\_{j\in\mathbb{Z}}|f\_{j}-f\_{j+2}|^2.
$$


A counterexample to the inequality is $f\_j=1$ $\forall j$, which is like taking a $f=\delta\_0$.


The question is if adding the constraint $\|f\|\_{L^2(\mathbb{T})}=1$, the inequality actually holds and which is the optimal constant.


My intuition to why should the inequality hold, is that the singularity of $\sin(x)$ at zero as a weight has the right scaling, as in Caffarelli, Luis, Robert Kohn, and Louis Nirenberg. "First order interpolation inequalities with weights." Compositio Mathematica 53, no. 3 (1984): 259-275.



$\|f\|^2_{H^{-1}(\mathbb{T})}\lesssim \int_\mathbb{T} |\sin(x)f(x)|^2 \; dx$?
The inequality does not hold. This becomes clear if we rewrite the discrete version as
$$
\sum |f\_n|^2/n^2 \lesssim \|Lf\|^2 ,
$$
with $(L f)\_n=f\_{n+1}-f\_{n-1}$. Notice that $iL$ is self-adjoint on $\ell^2$ and has spectrum $[-2,2]$ (because it becomes multiplication by $2\sin x$ after taking Fourier transforms, and in fact that is how we got $L$ in the first place).


Now an approximate eigenfunction $f$ of $iL$ for $\lambda=0$ with $f\_0=0$ will make the RHS small, but not the LHS.


**Addendum:** Given my discussion with the OP in the comments below, this perhaps deserves to be made more explicit: we can take $f\_n=0$ for even $n$ (we were asked to make $f\_0=0$) and $f\_n\simeq 1$ for odd $n$ on a large interval (we want $Lf\simeq 0$). More precisely, let's let $f\_n$ decay linearly on an interval of length $\simeq N$. Then $(Lf)\_n\simeq 1/N$, so $\|Lf\|^2\lesssim N\cdot (1/N)^2= 1/N$. However, $\sum |f\_n|^2/n^2$ clearly does not get small when $N\to\infty$.


