I am working through a proof which bounds the coefficients of the Fourier series of a function of bounded variation, and ran into some difficulty.
The proof is given here.
The only step which I am not understanding is the second to third line. How is it shown that $\int_{a_{k-1}}^{a_k} |f(x) – f(a_k)| dx \leq T(a_{k-1}, a_k) \frac{2\pi}{n}$, where T is the total variation ($\sup_{P} \Sigma |f(x_j) – f(x_{j-1})|)$?
I see that the $\frac{2\pi}{n}$ term comes from the fact that $a_k – a_{k-1} = \frac{2\pi}{n}$, but the fact that $f(a_k)$ is a constant is giving me trouble when I try to estimate the integral using Riemann sums and compare it to $T(a_{k-1}, a_k)$.
A similar question is posted here, but the answer does not give details on this calculation.
Thanks for the help.
Best Answer
\begin{align} \sum_{k=1}^{|n|}\int_{a_{k-1}}^{a_k} |f(x)-f(a_k)|\,dx &\leq \sum_{k=1}^{|n|} \max_{x\in[a_{k-1},a_k]} |f(x)-f(a_k)| \cdot\int_{a_{k-1}}^{a_k}\,dx\\ &= \sum_{k=1}^{|n|} \max_{x\in[a_{k-1},a_k]} |f(x)-f(a_k)| \cdot \frac{2\pi}{n} \end{align} Which is Holder's inequality, followed by the result they introduce before the proof.
Now, clearly $[a_{k-1},x]\cup[x,a_k]$ is a partition of $[a_{k-1},a_k]$, hence by the definition of bounded variation,
$$|f(x)-f(a_k)|+|f(a_{k-1})-f(x)| \leq T(a_{k-1},a_k)$$ and therefore $$\max_{x\in[a_{k-1},a_k]}|f(x)-f(a_k)| \leq \max_{x\in[a_{k-1},a_k]}\left(|f(x)-f(a_k)| + |f(a_{k-1})-f(x)|\right) \leq T(a_{k-1},a_k)$$