Showing that $\int_{-\pi}^{\pi} |f(x) e^{-ikx}| dx < \infty$ for $f\in L^2(-\pi,\pi)$. I have to show that $\int_{-\pi}^{\pi} |f(x) e^{-ikx}| dx < \infty$ for $f\in L^2(-\pi,\pi)$.
I have done the following so far;
$$\int_{-\pi}^{\pi} |f(x) e^{-ikx}| dx \leq \left(\int_{-\pi}^\pi|f(x)|^2 dx \right)^{1/2} \left(\int_{-\pi}^\pi|e^{-ikx}|^2 dx \right)^{1/2}, \text{ by Hölder's inequality.}$$
Because, $|e^{-ikx}| = e^{Im(kx)} = e^0$ (since $k,x\in \mathbb{R}$), we get
$$\left(\int_{-\pi}^\pi|f(x)|^2 dx \right)^{1/2} \left(\int_{-\pi}^\pi|e^{-ikx}|^2 dx \right)^{1/2} = ||f||_2 \cdot \left(\int_{-\pi}^\pi 1\; dx \right)^{1/2} = \sqrt{2\pi}||f||_2 < \infty.$$
Is this correct?
 A: What you've done is right, and the comments also suggest other ways of viewing the problem. Here's yet another generalization:

Let $(X,\mathcal{M},\mu)$ be a finite-measure space, and $0<p<q\leq \infty$. Then, $L^q(\mu)\subset L^p(\mu)$, and in fact the inclusion map is continuous because for every $f\in L^q(\mu)$, we have $\|f\|_p\leq \mu(X)^{\frac{1}{p}-\frac{1}{q}}\|f\|_q$.

In words: if we're dealing with a finite measure space (such as $[-\pi,\pi]$ with Lebesgue measure) then being in a higher Lebesgue space implies automatically that we're inside all the lower Lebesgue spaces (in particular, $L^2([-\pi,\pi])\subset L^1([-\pi,\pi])$).
The proof is straightforward: if $q=\infty$, then
\begin{align}
\|f\|_p&:=\left(\int_X|f|^p\,d\mu\right)^{1/p}\leq \bigg(\|f\|_{\infty}^p\cdot \mu(X)\bigg)^{1/p}=\|f\|_q\cdot \mu(X)^{\frac{1}{p}-\frac{1}{q}},
\end{align}
where the last equal sign is because $q=\infty$.
If $q<\infty$, then we apply Holder's inequality with the conjugate exponents $\frac{q}{p}$ and $\frac{q}{q-p}$:
\begin{align}
\|f\|_p&:=\left(\int_X|f|^p\cdot 1\,d\mu\right)^{1/p}\leq \bigg(\||f|^p\|_{q/p}\bigg)^{1/p}\cdot \bigg(\|1\|_{q/(q-p)}\bigg)^{1/p}=\|f\|_q\cdot \mu(X)^{\frac{1}{p}-\frac{1}{q}}
\end{align}
If you look over your work, then what you've done is use the fact that $|e^{ikx}|=1$ for all real $k$ and $x$, and you've actually proven the special case with $p=1$ and $q=2$ using (where Holder's inequality is just Cauchy-Schwarz).
A: For real $a,b \ge 0$, one has $2ab \le a^2 + b^2$, which follows from
$$
            0 \le (a-b)^2 = a^2 -2ab +b^2 \\
             \implies 2ab \le a^2 + b^2
$$
Therefore,
$$
        |f(x)e^{ikx}| = |f(x)||e^{ikx}| \le \frac{1}{2}(|f(x)|^2+|e^{ikx}|^2).
$$
Because $|f(x)|^2$ is assumed to be integrable on $[-\pi,\pi]$, and because $|e^{ikx}|^2$ is clearly absolutely integrable on $[-\pi,\pi]$, then it follows that $f(x)e^{ikx}$ is absolutely integrable on $[-\pi,\pi]$ by the above inequality, which is what you wanted to prove.
