Is the Fourier transform of a continuous function necessarily in $L^{\infty}$? 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
$$\widehat{f}(y)=\int_{\mathbb{R}} f(x)e^{-2\pi i xy} dx$$
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 $\lim_{T\to \infty} \int_0^T$ and $\lim_{T \to \infty} \int_{-T}^0$ exist.
 A: 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
$$f(x) = \sum_{n=1}^\infty f_n(x) = \sum_{n=1}^\infty\frac{n e^{-inx}}{x}\sin\left[(2n^3)^{-1}x \right] \chi_{[-n^4,n^4]}(x)$$
The series converges uniformly since $|f_n(x)| \leqslant \frac{1}{2n^2}$ and we can switch sum and integral to get
$$\hat{f}(m) = \sum_{n=1}^\infty n\int_{-n^4}^{n^4}\frac{e^{i(m-n)x}\sin[(2n^3)^{-1}x]}{x} \, dx$$
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 $\beta_n = (2n^3)^{-1}$,
$$\tag{*}\hat{f}(m) = \sum_{n=1}^\infty n\int_0^{n^4}\frac{\sin[(\alpha_{m.n} + \beta_n)x] - \sin[(\alpha_{m,n} - \beta_n)x]}{x} \,\,\, dx \\  = 2m \int_0^{m^4} \frac{\sin(\beta_mx)}{x} \, dx + \sum_{n\neq m} n\int_0^{n^4}\frac{\sin[(\alpha_{m,n}  + \beta_n)x] - \sin[(\alpha_{m,n} - \beta_n)x]}{x} \, dx  $$
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
$$\hat{f}(m) > \pi m - 8 -  6\sum_{n=1}^\infty \frac{1}{n^3}, $$
and $\hat{f}(m) \to \infty$ as $m \to \infty$.
