The following is a question from the book Fourier and Wavelet Analysis by George Bachmann, Lawrence Narici, Edward Beckenstein (Section 4.3, Question no. 8).
Question: The Fourier sine and cosine coefficients $c_n$ of $f\in BV[-\pi,\pi]\subset L^1_\Bbb R[-\pi,\pi]$ satisfy $$|c_n|\le \frac{1}{n\pi}V(f,[-\pi,\pi])$$
and these estimates cannot be improved.
But when I consider the following function $f:[-\pi,\pi]\to \Bbb R$ by $$f(x) = \left\{ \begin{array}{1 1} 0 & \mbox{if } & -\pi\le x <0 \\ 1 & \mbox{if } & 0\le x\le\pi \end{array} \right.$$
Then the Fourier cosine coefficients $a_0=1, a_n=0$ and $b_n=\frac{1}{n\pi}(1-(-1)^n)$ for $n\in\Bbb N$. Also $f$ is of bounded variation with total variation $V(f,[-\pi,\pi])=1$. Thus $f$ satisfies the above requirement of the question.
But for $n$ odd, $|b_n|=\frac{2}{n\pi}$. Thus we obtain that $|b_n|>\frac{1}{n\pi}V(f,[-\pi,\pi])$ for odd values of $n$.
Now my question is that is the above estimate in the question correct?
I think you are right. \begin{align*} \int_{0}^{\pi}\sin((2n+1)x)dx & =\frac{1}{2n+1}\int_{0}^{(2n+1)\pi}\sin t\,dt=\frac{1}{2n+1}\int_{0}^{(2n+1)\pi}\sin t\,dt\\ & =\frac{1}{2n+1}\sum_{k=0}^{n-1}\int_{2k\pi}^{(2k+2)\pi}\sin t\,dt+\frac {1}{2n+1}\int_{2n\pi}^{(2n+1)\pi}\sin t\,dt\\ & =\frac{1}{2n+1}\sum_{k=0}^{n-1}\int_{0}^{2\pi}\sin t\,dt+\frac{1}{2n+1}% \int_{0}^{\pi}\sin t\,dt\\ & =0+\frac{2}{2n+1}, \end{align*} so $b_{2n+1}=\frac{2}{(2n+1)\pi}$. I went to check the original paper Izumi and there are two mistakes. One is that $2\pi|a_n|$ should be $\pi |a_n|$ and the other is that they forgot the absolute value in $|\cos nx|$. So you get $$ \int_{-\pi/(2n)}^{\pi/(2n)}|\cos nx|dx=\frac{1}{n}\int_{-\pi/2}^{\pi /2}\left\vert \cos t\right\vert \,dt=\frac{2}{n}% $$ where $t=nx$ so $dt=ndx$. So they actually prove that $|a_n|\le 2V/\pi n$.