I'm reading a proof of Riemann-Lebesgue lemma (about Fourier transform) on wikipedia https://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma
This uses DCT but I wonder what is the dominated function.
Let $f\in L^1$, and at the last part of the proof,
$|\widehat f(\xi)|\leqq\frac{1}{2}\int_{\mathbb R}\left|f(x)-f(x+\pi/\xi)\right|dx\to 0$ as $|\xi|\to \infty$ by the continuity of $f$ and DCT.
Of course, $|f(x)-f(x+\pi/\xi)|\to 0$ by the continuity, but I don't know why we can use DCT.
To use DCT, we have to find $F\in L^1$ (independent of $\xi$) s.t. $|f(x)-f(x+\pi/\xi)|\leqq F(x)$ for all $x$.
I have $|f(x)-f(x+\pi/\xi)|\leqq|f(x)|+|f(x+\pi/\xi)|$, and I don't think $|f(x+\pi/\xi)|$ can be dominated by a function independent of $\xi$.
How can I use DCT in this situation ?
Best Answer
The part of the proof you are looking at assumes that $f$ is continuous and has compact support, so there is an interval $[-M,M]$ outside of which $f(x)$ equals zero. And $f$ must be bounded, so put $B:=\sup_{x\in {\mathbb R}}|f(x)|<\infty$. An extremely crude, but integrable, bounding function would be
$$ F(x):=\begin{cases} B& \text{if $x\in [-M-1, M+1]$}\\ 0&\text{otherwise} \end{cases} $$ This $F$ is a suitable dominator for $|f(x+\pi/\xi)|$ whenever $|\xi|>\pi$. Remember we only care about bounding the integrand as $|\xi|\to\infty$.