Let $\hat{f}(x)=\int_\mathbb{R}f(t)e^{-itx}dt$ be the Fourier transform of the function $f\in L^1(\mathbb{R})\cap L^\infty(\mathbb{R})$.
If $\hat{f}$ has bounded derivative on $\mathbb{R}$ and $$x\hat{f}(x)\in L^1(\mathbb{R})$$ can we say that $$x\hat{f}(x)\in L^\infty(\mathbb{R})?$$
Let $\phi \in C^\infty_c(-1,0)$ and $$\hat{f} = \sum_{n\ge 1} n \, 2^{-2n}\phi(2^{n} (x-2^{2n}))$$
$\hat{f}$ is $L^1$ so $f$ is $L^\infty$
$\hat{f}' = \sum_{n\ge 1}n 2^{-n} \phi'(2^n (x-2^{2n}))$ is $L^\infty$
Both $\hat{f},\hat{f}'$ are $L^2$ so $f,xf$ and $(1+|x|)f$ are $L^2$ whence $f$ is $L^1$ (Cauchy Schwarz)
$\|x\hat{f}\|_{L^1}\le \sum_{n\ge 1} \|n\phi(2^n (x-2^{2n}))\|_{L^1}= \sum_{n\ge 1} n 2^{-n} \|\phi\|_{L^1}< \infty$
$x\hat{f}$ is not $L^\infty$