I am wondering the following problem: given $f_n\in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ and $f\in L^2(R)$, with $f_n\to f$ in $L^2$, how does $\hat{f_n}$ converge to $\hat{f}$? The $L^2$ convergence is straightforward. Also I am wondering the case of $f_n=\mathbb{1}_{[-n,n]}f$, which yields the following 2 problems:
- Is it true that for any $f_n\in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ and $f\in L^2(R)$ with $f_n\to f$ in $L^2$, we have $\hat{f_n}\to \hat{f}$ a.e. (as $n\to\infty$)? If not, what's the counterexample, and what conditions should $f$ and $f_n$ satisfy?
- Is it true that for any $f\in L^2(R)$, $\mathcal{F}[\mathbb{1}_{[-n,n]}f](k)\to\mathcal{F}[f](k)$ a.e. (as $n\to\infty$)? If not, what's the counterexample, and what conditions should $f$ satisfy?
The Plancherel transform is unitary, hence for both 1 & 2, the transforms converge in $L^2$ and hence pointwise ae. for some subsequence.
For 1, this is the best we can do without additional assumptions. To see this, let $\hat{f_n}$ be the sequence of indicator functions of sets of the form $[{k \over m}, {k+1 \over m})$, with $k=0,...,m=1$, and let $f_n$ be the inverse Plancherel transform of $\hat{f_n}$. It is straightforward to see that $\hat{f_n} \to 0$ in $L^2$ and so $f_n \to 0$ in $L^2$. However, $\hat{f_n}(\omega)$ converges nowhere (pointwise).
For 2, let $f_n = f \cdot 1_{[-n,n]}$ and note that $\hat{ \hat{f_n} }(x) = f_n(-x)$ and similarly for $f$. Applying Carleson's theorem (also here under article) and taking care of minus signs yields the desired result.