Piecewise differentiable version: Total variation expressed as an integral

Question

I have already successfully found a proof of the equality $$V_{a}^{b}(f) = \int_{a}^{b} |f'(x)| \, dx$$ where $f$ is a continuously differentiable function on $(a,b)$ and continuous on $[a,b]$ as well as being of bounded variation. I tried to prove a similar statement: $$V_{0}^{2\pi}(f) = |f(0^+) - f(0^-)| + \sum\limits_{i = 1}^{n} |f(x_j^+) - f(x_j^{-})| + \int_{0}^{2\pi} |f'(x)| \, dx$$ where $f$ is piecewise continuously differentiable with jump discontinuities $(x_j)_{j=1,\ldots,n}$, $2\pi$-periodic on $[0,2\pi]$ and we impose that $f(x) = \frac{1}{2}(f(x^{+})-f(x^{-}))$ for left ($x^-$) and right ($x^{+}$) limits in $x \in [0,2\pi)$ respectively.

This showed up in an example to show that every such periodic function is of bounded variation. I tried to use the above proof of the continuously differentiable case and force it into the subintervals where $f$ is continuously differentiable, but things like the FTC and MVT break down at the boundary points, leaving me a bit in the dark how to progress.

I maybe have found a way to prove this using a contrived system of nested finite sums, but I think they would just clutter up this post if I type them up. Thus, I would like to ask if there was a more elegant way to prove this than by sheer brute force of the naive approach.

A hint or literature reference on how to get this would be enough, but I'm not upset if you put in the effort!

Answer 1

Lemma. If $z\in[a,b]$, then $V_a^b(f)=V_a^z(f)+V_z^b(f)$.
Proof. Take a partition of $[a,z]$ and one of $[z,b]$; conjoining these two gives a partition of $[a,b]$, so that $V_a^b(f)\leq V_a^z(f)+V_z^b(f)$. Conversely, any partition of $[a,b]$ can be split into a partition of $[a,z]$ and one $[z,b]$, just by adding another point, viz. $z$. So consider the supremum over all partitions with $z$: $$U_a^b(f,z)=\sup_{z\in P\vdash[a,b]}{\sum_{[x,y]\in P}{|f(y)-f(x)|}}$$ Taking suprema as above, $$U_a^b(f,z)\geq V_a^z(f)+V_z^b(f)$$ Also, by definition $U_a^b(f,z)\leq V_a^b(f)$; but since adding points to a partition increases the variation, the reverse inequality holds as well. So $V_a^b(f)=U_a^b(f,z)$ as desired. QED.

Lemma. If $f$ has an isolated discontinuity at $a$, then $\lim_{b\to a^+}{V_a^b(f)}=|f(a^+)-f(a)|$.
Proof. Certainly, if $a<z$, then $$V_a^b(f)\geq|f(z)-f(a)|$$ Taking $z\to a^+$ gives one inequality. For the reverse, note that if $P\vdash [a,b]$ has at least three points, then we can write $P=\{[a,z]\}\sqcup Q$, where $z$ is the next point in $P$. Taking suprema over $P$, \begin{align*} V_a^b(f)&=\sup_{P\vdash[a,b]}{\sum_{[x,y]\in P}{|f(y)-f(x)|}} \\ &=\sup_{a<z,Q\vdash[z,b]}{\left(|f(z)-f(a)|+\sum_{[x,y]\in Q}{|f(y)-f(x)|}\right)} \\ &=\sup_{a<z}{\left(|f(z)-f(a)|+\sup_{Q\vdash[z,b]}{\sum_{[x,y]\in Q}{|f(y)-f(x)|}}\right)} \\ &=\sup_{z\in(a,b]}{(|f(z)-f(a)|+V_z^b(f))} \end{align*} Fix $\epsilon$; since $f$ is continuous near $a$, there exists $\delta$ such that $V_z^b(f)\leq\epsilon$ for $|b-z|\leq\delta$, which is certainly true when $|b-a|\leq\delta$. So as we send $b\to a^+$, $$\limsup_{b\to a^+}{V_a^b(f)}\leq|f(a^+)-f(a)|+\epsilon$$ Now take $\epsilon\to0^+$. QED.

By considering $x\mapsto f(-x)$, it is clear that the same result holds for changing the lower limit.

Now decompose the interval as $$[a,b]=[a,x_1-\epsilon]\cup[x_1-\epsilon,x_1]\cup[x_1,x_1+\epsilon]\cup[x_1+\epsilon,x_2-\epsilon]\cup[x_2-\epsilon,x_2+\epsilon]\cup\dots\cup[x_n+\epsilon,b]$$ We can even develop a uniform notation by setting $x_0=a$ and $x_{n+1}=b$. Thus $$V_a^b(f)=\sum_{j=0}^n{V_{x_j}^{x_j+\epsilon(f)}+V_{x_j+\epsilon}^{x_{j+1}-\epsilon}(f)+V_{x_{j+1}-\epsilon}^{x_{j+1}}(f)}$$ Now send $\epsilon\to0^+$; we have \begin{gather*} V_{x_j}^{x_j+\epsilon(f)}\to|f(x_j^+)-f(x_j)| \\ V_{x_j+\epsilon}^{x_{j+1}-\epsilon}(f)\to\int_{x_j+\epsilon}^{x_{j+1}-\epsilon}{|f'(x)|\,dx} \\ V_{x_{j+1}-\epsilon}^{x_{j+1}}(f)\to|f(x_j)-f(x_j^-)| \end{gather*} where the middle term is by your cited result and the outer two are by my lemmas.

Substituting in the particular values for your $f$ at the points of discontinuity gives your claim.