Given the function $ f(x) = \sin^2(\pi x) $, show that $ f \in BV[0, 1] $

Question

I'm learning about functions of bounded variation and need to verify my work to this problem since my textbook does not provide any solution :

Given the function $ f(x) = \sin^2(\pi x) $, show that $ f $ is of bounded variation on $ [0, 1] $.

Here's my attempt :

$ f'(x) = 2(\sin(\pi x)) \frac{d}{dx}[\sin(\pi x)] = 2\pi \sin(\pi x) \cos(\pi x) = \pi \sin(2 \pi x)$.

The function $ f $ is differentiable on $ [0, 1] $ and $ \forall x \in [0, 1] $ we have :

$$ \lvert f'(x) \rvert = \lvert \pi\sin(2 \pi x) \rvert \le \pi $$

Since the derivative of $ f $ is bounded on $ [0,1] $ this implies that $ f \in BV[0, 1] $.

Is my work correct?

Answer 1

Yes, a Lipschitz continuous function on a compact set is a function of bounded variation. Moreover $$\text{Var}_{[0,1]}f = \int_0^1|f'(x)|\,dx = \int_0^1 \pi|\sin(2\pi x)|\,dx = 2.$$

Answer 2

For a proof from scratch, suppose $f:[0,1]\rightarrow \mathbb R$ has a Lipschitz constant $M$.

Let $\mathcal P=\left \{ 0,x_{1},\cdots ,x_{n-2},1 \right \}$ be a partition of $[0,1]$.

Then using MVT, and the Lipschitz assumption, we have with $x_i<x_i^{*}<x_{i+1}$,

$\sum_{i=0}^{n-1}\left | f(x_{i+1})-f(x_i) \right |\leq \sum_{0}^{n-1}\vert f'(x^{*}_i)(x_{i+1}-x_i)\vert \leq \sum_{0}^{n-1}\vert M(x_{i+1}-x_i)\vert\leq M$ and the result follows as soon as we set $f(x)=\sin ^2(\pi x)$ and note that $f$ has a Lipschitz constant we may take to be $\pi $.