I would like to know if there is a relation between bounded variation of functions and bounded variation in sequences. I know one can not directly correlate the concepts but if we consider the Fourier series of the function of bounded variation, what can be said about Fourier coefficients, would they be of bounded variation (sequence definition) and what about the vice versa case? Are there any articles or books related to interlinking these concepts?
Analysis – Bounded Variation of Functions and Sequences
analysisbounded-variationfourier analysisreference-requestsequences-and-series
Best Answer
No, there is no relation between bounded variation of functions and sequences via Fourier transform.
The function $f(x) = x$ on $(-\pi, \pi)$ has bounded variation, but its Fourier coefficients do not; indeed, for every $n\in\mathbb Z$ we have $\int_{-\pi}^\pi f(x) \sin(n x) d x = 2\pi\frac{ (-1)^{n+1}}{n}$.
Conversely, bounded variation for sequences does not seem to be the way to go for Fourier coefficients. If a sequence converges to $1$, when applying the inverse Fourier transform you don’t even get a function, only a distribution. But if you assume that the sequence is $L^1$, i.e. that $S = \sum_{n\in \mathbb Z} |a_n| < \infty$, then the inverse Fourier transform $f$ is continuous and bounded by $S$.
This is already good, but unfortunately, you cannot deduce that $f$ has bounded variation. Indeed, take $a_{2^n} = n^{-2}$ for every $n\geq 1$, and $a_k = 0$ if $k$ is not a power of $2$. The sequence $a_n$ is in $L^1$ (thus has bounded variation) but $n a_n$ is not bounded. But if $f$ had bounded variation, then it should be bounded, as $$ na_n = n\left| \int_{-\pi}^\pi f(x) \sin(nx) dx \right| = \left| \int_{-\pi}^\pi (\cos (nx)-1) df(x) \right| \leq 4\pi \mathrm{Variation}(f) . $$
In conclusion, a function with bounded variation can have Fourier coefficients that do not have bounded variation, and a function with Fourier coefficients having bounded variation may not, itself, have bounded variation.