Bounding Fourier transform of $L^{1}$ functions that don't vanish at infinity. It is fairly easy to show using integration by parts that if $f\in L^{1}$ is differentiable and $f^{'}\in L^{1}$ then under assumption that $f$ also vanishes at infinity one has $\sup_{\xi\in\mathbb{R}}\left|\xi\hat{f}\left(\xi\right)\right|\leq\left\Vert f^{'}\right\Vert _{1}$ , this is evident using integration by parts $$\hat{f^{'}}\left(\xi\right)=\int_{\mathbb{R}}e^{-i\xi x}f^{'}\left(x\right)dx=e^{-i\xi x}f\left(x\right)\vert_{-\infty}^{\infty}+\int_{\mathbb{R}}i\xi e^{-i\xi x}f\left(x\right)dx=i\xi\hat{f}\left(\xi\right)$$ My question is, can you forgo the assumption that $f$ vanishes at infinity and still get this bound? I couldn't even come up with an example of an integrable function that doesn't vanish at infinity and also has an integrable derivative.

To help with your "I couldn't even …" comment: There's a good reason you couldn't; it's impossible. We need the following lemma:

Lemma: Suppose $f\in C^1(\mathbb R)$ and $\int_\infty^\infty|f'| < \infty.$ Then $f$ is uniformly continuous on $\mathbb R.$

Proof: Let $\epsilon>0.$ Then there exists $\delta > 0$ such that $0
$$|f(y)-f(x)| = |\int_x^yf'| \le \int_x^y|f'| < \epsilon.$$

That proves the lemma.

It's therefore impossible to find $f\in C^1(\mathbb R)$ with $f,f'\in L^1(\mathbb R)$ and $f\
ot \in C_0(\mathbb R).$ Why? Because the lemma shows $f$ is uniformly continuous, and a uniformly continuous function that doesn't vanish at $\infty$ can't belong to $L^1.$ I'll leave that last part to you for now.