I need to prove the following statement: Let $\varphi \in C(\Bbb R)$ with compact support. Then, $$ \Big \Vert \sum_{k\in \Bbb Z} \varphi(k) e^{ikx} \Big \Vert_{L_1(0,2\pi)} \leq C \Vert \hat{\varphi}\Vert_{L_1(\Bbb R)}, $$ where $C$ is an absolute constant not depending on $\varphi$, and $\hat{\varphi}$ denotes the Fourier transform on $\Bbb R$.
Here's my approach: We can suppose that $\hat{\varphi}\in L_1(\Bbb R)$. Otherwise, the inequality is trivial by putting $+\infty$ on the right-hand side. Now consider $$ \psi (x) = \sum_{k\in \Bbb Z} \hat{\hat{\hat{\varphi}}}(x+2k\pi)=\sum_{k\in \Bbb Z}\hat{\varphi}(-x+2k\pi). $$ It is clear that $\psi$ is $2\pi$-periodic and belongs to $L_1(0,2\pi)$, since $\hat{\varphi}\in L_1(\Bbb R)$. Thus, it has a Fourier series $\sum_{k\in \Bbb Z} c_ne^{inx}$ associated, with coefficients \begin{align} c_n=& \frac{1}{2\pi}\int_0^{2\pi} \sum_{k\in \Bbb Z}\hat{\varphi}(-x+2k\pi)e^{-inx}\, dx=\frac{1}{2\pi} \sum_{k\in \Bbb Z}\int_0^{2\pi}\hat{\varphi}(-x+2k\pi)e^{-inx}\, dx \\ =&\frac{1}{2\pi} \int_{\Bbb R}\hat{\varphi}(-x) e^{-inx}\, dx = \frac{1}{(2\pi)^{3/2}}\varphi(n), \end{align}
where in the last step we have used inversion theorem along with continuity of $\varphi$. Now we have that the Fourier series of $\psi$ is $$ S(\psi)=\frac{1}{(2\pi)^{3/2}}\sum_{k\in \Bbb Z} \varphi(k)e^{ikx} =\frac{1}{(2\pi)^{3/2}}\sum_{k\in \Bbb Z\cap \textrm{Supp } \varphi}\varphi(k)e^{ikx}, $$ so that $S(\psi)$ is just a trigonometric polynomial, and so is $\psi$. Then, of course we have uniform convergence of the Fourier series, and moreover \begin{align} \Big \Vert \sum_{k\in \Bbb Z} \varphi(k) e^{ikx} \Big \Vert_{L_1(0,2\pi)} =& \int_0^{2\pi} \Big \vert\sum_{k\in \Bbb Z} \varphi(k) e^{ikx} \Big \vert \,dx= C\int_0^{2\pi}|\psi(x)| \, dx \\ = & C\int_0^{2\pi} \Big \vert\sum_{k\in \Bbb Z}\hat{\varphi}(-x+2k\pi) \Big\vert\, dx \leq C\Vert \hat{\varphi}\Vert_{L_1(\Bbb R)}. \end{align}