So, I have a function $f\in L^2(\mathbb R)$ and two intervals $I$ and $J$ in $(0,1)$. By $f_{I+k}$ denote the average of $f$ on $I+k$, i.e., $$ f_{I+k} := \frac 1{|I|}\int_{I+k}f(x)\,dx. $$ Now, let us periodize the intervals: $$ E := \bigcup_{k\in\mathbb Z}(I+k)\quad\text{ and }\quad F := \bigcup_{n\in\mathbb Z}(J+n) $$ and define the function $g := \chi_E\cdot(f-s)$, where $s = \sum_{k\in\mathbb Z}\chi_{I+k}f_{I+k}\in L^2(\mathbb R)$. What I would like to estimate now is the expression $$ \|\chi_F\cdot \hat g\|_{L^2(\mathbb R)}. $$ Clearly, a possible estimate is just $\|\hat g\|_{L^2(\mathbb R)} = \|g\|_{L^2(\mathbb R)}$, but I expect that there should be a better estimate . In fact, if the "rough" estimate was optimal, $\hat g$ would be concentrated on $F$, while $g$ is concentrated on $E$. It is known that this is not possible if $E$ and $F$ are bounded intervals.
Well, one could do $$ \|\chi_F\cdot \hat g\|_{L^2(\mathbb R)}^2 = \sum_n\|\chi_{J+n}\cdot \hat g\|_{L^2(\mathbb R)}^2 = \sum_n\|\hat\chi_{J+n} * g\|_{L^2(\mathbb R)}^2, $$ but as $\hat\chi_{J+n}$ (which is an exponential times a sinc function) is not $L^1$, I cannot apply Young's inequality. Any ideas anyone?
