Is there a function which is not absolutely integrable but which has a continuous fourier transform?
I know that if a function is absolutely integrable then the fourier transform is continuous but I want to know whether the converse is false. Any help is appreciated . Thanks
Let $$ f_n(x) = \cases{ 1 & if $|x|\le n$ \cr 0 & if $|x| \ge n+1$ \cr 1 + n - |x| & if $n \le |x| \le n + 1 $.\cr } $$ Then I think you should find that $\|\hat {f_n}\|_1 \approx \log n$. Now try something like $$ g(x) = \sum_{k=1}^\infty \frac1{k^2}f_{2^k}(x) . $$
Then $\hat g = \check g$ is not absolutely integrable, but $g$ is continuous.
Or quote something like the closed graph theorem to get a contradiction from the functions $f_n$. Sorry, I left a lot of the work for you to do.