Fourier transform of $L^1$ function whose derivative is in $L^1$ and vanishes at infinity is in $L^1$

Question

$f \in L^1(\mathbb{R})$ is a differentiable function such that $f' \in L^1(\mathbb{R}) \cap C_0(\mathbb{R})$, prove that the fourier transform of $f$ noted $\hat{f}$ is in $L^1 (\mathbb{R})$

I know if $f,f'\in L^1(\mathbb{R})$, then $\widehat{f'}(t)=it\hat{f}(t)$ but I have no ideas how to use the condition that the derivative vanish at infinity. Any ideas will be helpful.

Answer 1

Two hints:

Use the fact that $f'$ is bounded to show that $f' \in L^2$ and the use Plancherel.

Note that $f'$ is bounded and since$|f'|^2 \le \sup |f'| |f'|$ we see that $f' \in L^2$. Then Plancherel shows that $\hat{f'} \in L^2$. Note that $\hat{f'}(\omega) = i\omega \hat{f(\omega)$.

Use Cauchy Schwartz and note that for $\omega \neq 0$ we have $\hat{f}(\omega) = \omega \hat{f}(\omega) \cdot {1 \over \omega}$.

For $\omega \neq 0$ we have $|\hat{f}(\omega)| = |\omega||\hat{f}(\omega)| \cdot {1 \over |\omega|}$ and Cauchy Schwartz gives $\int|\hat{f}| = \int |\omega||\hat{f}(\omega)| \cdot {1 \over |\omega|} d \omega = \int |\hat{f'}(\omega)| \cdot {1 \over |\omega|} d \omega \le \| \hat{f'}\| \| \omega \mapsto {1 \over |\omega|} \|$.