# Fourier Coefficients of a Function of Bounded Variation

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.

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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 $$\max_{x\in[a_{k-1},a_k]}|f(x)-f(a_k)| \leq \max_{x\in[a_{k-1},a_k]}\left(|f(x)-f(a_k)| + |f(a_{k-1})-f(x)|\right) \leq T(a_{k-1},a_k)$$