Given $f\in L^2$, and $f'\in L^2$. I need to show that $\hat{f}$, the Fourier transform of $f$, is in $L^1$. Here's what I have so far:
\begin{align} \int _{-\infty}^{+\infty} |\hat{f}(\xi)|d\xi &= \int _{|\xi|\leq 1}|\hat{f}(\xi)|d\xi + \int _{|\xi| > 1}|\hat{f}(\xi)|d\xi \\ &= C + \int _{|\xi| > 1}\left|\frac{\hat{f'}(\xi)}{i\xi}\right|d\xi \\ &\leq C + \left(\int _{|\xi| > 1} |\hat{f'}(\xi)| d\xi\right)^{1/2}\left(\int _{|\xi| > 1} \frac{1}{\xi^2} d\xi\right)^{1/2} \\ &\leq C + ||\hat{f'}|| \\ &< \infty \end{align}
And I know $||\hat{f'}||<\infty$ due to Plancherel. However I'm not sure the first integral in the RHS of the first equation does converge, i.e. not sure that the first term is less than some $C<\infty$. But I have a feeling that if I choose to integrate over $[-\epsilon,\epsilon]$ instead of $[-1,1]$ then I can make the integral arbitrarily small. Am I on the right track?