Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The Fourier Transform is in general terms a continuous mapping, as well as it's inverse when it exists. Is there any abstract result that translates convergence of the transforms of a sequence of functions into convergence of the original sequence when the inverse does not exist but convergence takes place to a function in the image of the initial space under the Fourier Transform?

Assume for instance that $f_{n}$ is a sequence in $L^{1}$ and that $\hat{f}_{n} \to \hat{f}$ uniformly, for some $f \in L^{1}$. Can something be said about the convergence of $f_{n}$ to $f$ in $L^{1}$?

That is, if for given spaces $X,Y$, the function $\mathcal{F}:X \to Y$ is continuous (I'm thinking for instance in $X = L^{p}$ and $Y=L^{p'}$, $p \in (1,2]$ ) and the image under the Fourier Transform of a sequence $f_{n}$ converges in $Y$ to a function $\hat{f} \in \mathcal{F}(X)$, does $f_n$ converge to $f$ in $X$?

share|improve this question
1  
Without further conditions, the inequivalence of the sup norm on Fourier coefficients and the L^1 norm on functions will dash hopes. For instance, take f_n to be the Dirichlet kernel divided by sqrt of log (n), then the Fourier transforms converge uniformly to 0 but the functions have L^1 norm tending to infinity –  Yemon Choi May 7 '13 at 22:51
    
Thank you! How about the case where the Fourier transforms of L^1 functions converge in L^p to the Fourier transform of another L^p function? Would that be a stronger condition? What sort of abstract, further conditions could be asked for? –  Geno Whirl May 23 '13 at 22:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.