Let $f$ be an integrable function on $[-\pi,\pi]$ satisfying that $f^{\prime}\in L^{2}[-\pi,\pi]$.
How can I show that
$$\sum_{n\geq 1}|\hat{f}(n)|\leq C(\|f\|_{1}+ \|f^{\prime}\|_{2})$$
where
$$\hat{f}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx.$$
Thank you in advance.
Maybe my answer is too complicated. But I use the stuff that I know to prove the inequality. First, if we knew that $f\in L^2$, then it has an absolutely continuous representative with which we can perform partial integration. So, let us prove first that $f\in L^2(-\pi,\pi)$. For this, on $D := \{\varphi' : \varphi\in C_0^\infty(-\pi,\pi)\}$ define the functional $F(\phi) := \int_{-\pi}^\pi f\phi\,dx$. Since $$ \left|\int_{-\pi}^\pi f\varphi'\,dx\right| = \left|\int_{-\pi}^\pi f'\varphi\,dx\right|\,\le\,\|f'\|_2\|\varphi\|_2\,\le\,\|f'\|_2\|\varphi'\|_2, $$ the functional $F$ is continuous w.r.t. the $L^2$-norm and hence has a continuous extension $\tilde F$ on $L^2$. By the Riesz representation theorem, $\tilde F(g) = \langle g,\overline{f_0}\rangle$ with some $f_0\in L^2$. Thus, for $\varphi\in C_0^\infty(-\pi,\pi)$ we obtain $$ \int_{-\pi}^\pi f\varphi'\,dx = F(\varphi') = \tilde F(\varphi') = \langle\varphi',\overline{f_0}\rangle = \int_{-\pi}^\pi f_0\varphi'\,dx. $$ And this proves $f = f_0 + const\in L^2(-\pi,\pi)$.
Now, we can assume that $f$ is absolutely continuous on $[-\pi,\pi]$. For $n\neq 0$ we have \begin{align*} \hat f(n) &= \frac 1 {2\pi}\int_{-\pi}^\pi f(x)e^{-inx}\,dx = -\frac 1 {2\pi in}\left(\left[f(x)e^{-inx}\right]_{-\pi}^\pi - \int_{-\pi}^\pi f'(x)e^{-inx}\,dx\right)\\ &= \frac 1 {2\pi in}\big[f(\pi)-f(-\pi)\big] + \frac{1}{in}\widehat{f'}(n). \end{align*} Hence, if $f$ is periodic, then $$ \sum_{n\neq 0}|\hat f(n)| = \sum_{n\neq 0}\frac 1 n|\widehat{f'}(n)|\,\le\,\left(\sum_{n\neq 0}\frac 1 {n^2}\right)^{1/2}\left(\sum_{n\neq 0}|\widehat{f'}(n)|^2\right)^{1/2}\,\le\,\frac\pi{\sqrt 3}\|f'\|_2. $$ If $f$ is not periodic, then $|\hat f(n)|$ is not summable and the claim is false.