Vrftsjtryk

This is a followup to this earlier question. Given a (let's say continuous) function $f:mathbb{R} to mathbb{C}$ such that the Fourier Transform
$$widehat{f}(y)=int_{mathbb{R}} f(x)e^{-2pi i xy} dx$$
exists for all $y in mathbb{R}$, does it follow that $widehat{f} in L^{infty}$ i.e. there is a constant $C$ (depending on $f$) such that $vert widehat{f}(y)vert le C$ for all $y$?

This would trivially be true if $f in L^1$, but we are not assuming that. This of course also means that the integrals should better not be absolutely convergent. Instead we require convergence only in the sense that both $lim_{Tto infty} int_0^T$ and $lim_{T to infty} int_{-T}^0$ exist.

It is not necessary that $hat{f} in L^infty$.

Consider the Fourier transform defined as $hat{f}(y) = int_{-infty}^infty f(x) e^{iyx} , dx$. We can construct a counterexample where $hat{f}(m) to infty$ as $m (in mathbb{N}) to infty$, using the function

$$f(x) = sum_{n=1}^infty f_n(x) = sum_{n=1}^inftyfrac{n e^{-inx}}{x}sinleft[(2n^3)^{-1}x right] chi_{[-n^4,n^4]}(x)$$

The series converges uniformly since $|f_n(x)| leqslant frac{1}{2n^2}$ and we can switch sum and integral to get

$$hat{f}(m) = sum_{n=1}^infty nint_{-n^4}^{n^4}frac{e^{i(m-n)x}sin[(2n^3)^{-1}x]}{x} , dx$$

Note that

$$frac{e^{i(m-n)x}sin[(2n^3)^{-1}x]}{x} = frac{cos[(m-n)x] sin[(2n^3)^{-1}x]}{x}+ i frac{sin[(m-n)x]sin[(2n^3)^{-1}x]}{x}$$

The second term on the RHS is an odd function whose integral over $[-n^4,n^4]$ vanishes and the first term reduces to

$$frac{cos[(m-n)x] sin[(2n^3)^{-1}x]}{x} = frac{sin[(|m-n| + (2n^3)^{-1})x] - sin[(|m-n| - (2n^3)^{-1})x]}{2x}$$

Since this is an even function, the integral over $[-n^4,n^4]$ is twice the integral over $[0,n^4]$, and we have with $alpha_{m,n} = |m-n|$ and $beta_n = (2n^3)^{-1}$,

$$tag{*}hat{f}(m) = sum_{n=1}^infty nint_0^{n^4}frac{sin[(alpha_{m.n} + beta_n)x] - sin[(alpha_{m,n} - beta_n)x]}{x} ,,, dx \ = 2m int_0^{m^4} frac{sin(beta_mx)}{x} , dx + sum_{nneq m} nint_0^{n^4}frac{sin[(alpha_{m,n} + beta_n)x] - sin[(alpha_{m,n} - beta_n)x]}{x} , dx $$

We have the well-known results (where $c > 0$):

$$tag{**}int_0^z frac{sin (cx)}{x} , dx = frac{pi}{2} - int_z^infty frac{sin (cx)}{x} , dx , \ - frac{2}{cz} leqslant int_z^infty frac{sin(cx)}{x} , dx leqslant frac{2}{cz}$$

Since $alpha_{m,n} - beta_n = |m-n| - beta_n geqslant 1 - 1/2 > 0$ if $m neq n$, we can apply (**) to (*) and obtain

$$hat{f}(m) > pi m - 8 - sum_{n neq m} frac{2n}{n^4}left(frac{1}{alpha_{m,n}+beta_n} + frac{1}{alpha_{m,n} - beta_n} right)$$

Since

$$- frac{1}{alpha_{m,n} + beta_n} - frac{1}{alpha_{m,n} - beta_n}geqslant - frac{1}{1 + 0}- frac{1}{1 - 1/2} = -3,$$

we have

$$hat{f}(m) > pi m - 8 - 6sum_{n=1}^infty frac{1}{n^3}, $$

and $hat{f}(m) to infty$ as $m to infty$.