Is the total variation function Lipschitz

arc lengthbounded-variationlipschitz-functionsreal-analysisrough-path-theory

Let $f:\mathbb{R} \to \mathbb{R}$ be a function. Define the total variation function $V_f:[0,T] \times [0,T] \to \mathbb{R}$ over an interval $[s,s’]$ as usual: $$V_f(s,s’) = \sup_{D \subset [s,s’]} \sum_{j \in D}\lvert f(t_j)-f(t_{j-1}) \rvert$$

for all $s,s’$ in $[0,T]$. Here, $D = \{t_0, t_1, \ldots, t_n\}$, with $t_0=0$, $t_n=T$ and $t_j > t_{j-1}$ is referred to as a partition of the interval $[0,T] $.

We refer to $f$ as being of bounded variation over $[0,T]$ if $V_f(0,T)<\infty$. Moreover, we say that $V_f$ is Lipschitz if there exists a constant $C$ such that $V(s,s’) \leq C \lvert s-s’\rvert $ for all $s,s’$ in $[0,T]$.

Question: for $f$ continuous of bounded variation, is the total variation function $V_f$ Lipschitz?

I have two clues. First, a paper I saw shows that any uniformly continuous function $g$ is Lipschitz plus an epsilon, i.e. for all $\epsilon$ there exists a $C_\epsilon$ such that:

$g(x,y) \leq C_\epsilon \lvert x-y\rvert + \epsilon$

Since the total variation function of a continuous function is itself continuous (I do not know where I know this fact from, but I am quite sure it is true) on a compact, it is uniformly continuous. This is not quite what I need for my research, but it might be nice-enough that I can make things work.

The second clue is trying to make things work directly. Note that the total variation behaves, in some sense, like the arc length. This particular sense is that, if the arc length of the function is finite, then so is the total variation (an easy proof). Moreover, for any given partition $D$, the arc length of a function over any subinterval is finite. Hence, the bounded variation is finite. Hence, for that fixed partition it is easy to find a constant such that $V(t_j, t_{j-1}) \leq C_D \lvert t_j – t_{j-1} \rvert$ holds for all $j$. In fact, The following $C_D$ will do the job:

$$C_D= \sup_{j \in D} \frac{V(t_j, t_{j-1})}{ \lvert t_j – t_{j-1} \rvert} $$

The question I am asking would be thus reduced to: does the $C_D$ constant explode when we take supremum over all $D$? That is, whether the following statement is true:

$$\sup_{D \subset [0,T]} C_D < \infty$$

PS: if there was any value to the arc length detour, so to speak, is that, personally, it helps me visualise what is going on. This is because the arc length is the length of the path described by the function, and the arc length is closely related to the total variation.

EDIT: in the original question, I had forgotten to add that the function $f$ should be continuous.

Best Answer

$f(x)=\sqrt x$ for $0<x<1$, $f(x)=0$ for $x \leq 0$ and $f(x)=1$ for $x \geq 1$ defines a function of bounded variation whose total variation is not Lipschitz. Since $f$ is increasing on $(0,1)$ it is its own total variation on this interval and $f$ is not Lipschitz.

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Let $TF(x) = T_F(a,x)$. For a point $x_0\in [a,b]$ where both $F$ and $TF$ are differentiable, we have $TF'(x_0)\ge |F'(x_0)|$. Hence $$T_F(a,b)\ge \int^{b}_{a} (TF)'{\rm d}x\ge \int^{b}_{a} |F'|{\rm d}x.$$ Note that this works when the range is any finite-dimensional real normed space, in which case every BV function is almost everywhere differentiable.

Real Analysis – Total Variation Measure vs. Total Variation Function

Yes, it is the case that $\mu_f' = \mu_v$.

We will use the notation introduced in section "Attempt #2" of the original post. Additionally, we make the following two definitions that will be used throughout the proof: $$ \begin{align} Z &:= (a, b] \\ \mathcal{J} &:= \left\{(c, d] :\mid c, d\in [a, b] \right\} \end{align} $$

The proof breaks down into four steps:

1) We show that it suffices to show that $\mu_f'(E) = \mu_v(E)$ for all $E \in \mathcal{B}(Z)$ in order to conclude that $\mu_f' = \mu_v$.

2) We show that $\mu_f^\pm(J) \geq \mu_v^\pm(J)$, respectively, for all $J \in \mathcal{J}$.

3) We deduce that $\mu_f^\pm(E) = \mu_v^\pm(E)$, respectively, for all $E \in \mathcal{B}(Z)$.

4) We conclude that $\mu_f'(E) = \mu_v(E)$ for all $E \in \mathcal{B}(Z)$.

The proof hinges on two characterizations of the Jordan decomposition of signed measures given in proposition 6.21(i & ii) in Dshalalow's "Foundations of Abstract Algebra", 2nd edition and in the version of the Jordan decomposition theorem given in Doob's "Measure Theory".

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We show that it suffices to show that $\mu_f'(E) = \mu_v(E)$ for all $E \in \mathcal{B}(Z)$ in order to conclude that $\mu_f' = \mu_v$, as follows. (1.1) We show that $\mu_v(Z^c) = 0$. (1.2) We show that $\mu_f'(Z^c) = 0$. (1.3) We conclude that it suffices to show that $\mu_f'(E) = \mu_v(E)$ for all $E \in \mathcal{B}(Z)$ in order to conclude that $\mu_f' = \mu_v$.

1.1 We will show that $\mu_v(Z^c) = 0$. Since $Z^c = (-\infty, a] \cup (b, \infty)$, it suffices to show that $\mu_v\left((-\infty, a]\right), \mu_v\left((b, \infty)\right) = 0$.

$$ \begin{align} \mu_v\left((-\infty, a]\right) & = \mu_v\left(\bigcup_{n = 0}^\infty \left(a - (n + 1), a - n\right]\right) \\ & = \sum_{n = 0}^\infty \mu_v\left(\left(a - (n + 1), b - n\right]\right) \\ & = \sum_{n = 0}^\infty \left(v_f(a - n) - v_f(b - (n + 1))\right) \\ & = \sum_{n = 0}^\infty (0 - 0) \\ & = 0 \\ \mu_v\left((b, \infty)\right) & = \mu_v\left(\bigcup_{n = 0}^\infty \left(b + n, b + (n+1)\right]\right) \\ & = \sum_{n = 0}^\infty \mu_v\left(\left(b + n, b + (n+1)\right]\right) \\ & = \sum_{n = 0}^\infty \left(v_f(b + (n + 1)) - v_f(b + n)\right) \\ & \overset{(*)}{=} \sum_{n = 0}^\infty V_{b + n}^{b + (n + 1)}(f) \\ & \overset{(**)}{=} 0 \end{align} $$ where equality $(*)$ is a fundamental result from the theory of functions of bounded variation (see e.g. lemma 12.15(b) in Yeh's "Real Analysis", 2nd edition), and equality $(**)$ is due to the fact that, for $s, t \in [b, \infty)$ with $s \leq t$, we have $$ \begin{align} V_s^t(f) &= \sup \left\{\sum_{k = 1}^n \left|f(t_k) - f(t_{k - 1}) \right| :\mid s = t_0 \leq t_1 \leq \cdots \leq t_n = t, n \in \{1, 2, \dots\}\right\} \\ & = \sup \left\{\sum_{k = 1}^n \left|f(b) - f(b) \right| :\mid s = t_0 \leq t_1 \leq \cdots \leq t_n = t, n \in \{1, 2, \dots\}\right\} \\ & = 0 \end{align} $$

1.2 We show that $\mu_f'(Z^c) = 0$, as follows. (1.2.1) We construct finite measures $\mu^+$ and $\mu^-$ on $(\mathbb{R}, \mathcal{B})$ such that $\mu_f = \mu^+ - \mu^-$ and for which $\mu^\pm(Z^c) = 0$. (1.2.2) We apply a certain minimality condition on the Jordan decomposition of signed measures to show that $\mu_f^\pm = \mu^\pm$, respectively. (1.2.3) We conclude that $\mu_f'(Z^c) = 0$.

1.2.1 We will construct finite measures $\mu^+$ and $\mu^-$ on $(\mathbb{R}, \mathcal{B})$ such that $\mu_f = \mu^+ - \mu^-$ and for which $\mu^\pm(Z^c) = 0$. Define the set functions $\mu^\pm : \mathcal{B} \rightarrow [0, \infty]$ as follows: $$ \mu^\pm(E) := \mu_f^\pm(E \cap Z) $$ respectively.

The fact that $\mu^\pm$ are measures follows from the fact that $\mu_f^\pm$ are measures. The fact that $\mu^\pm$ are finite follows from the observation that $\mu^\pm \leq \mu_f^\pm$, respectively, and from the observation that $\mu_f^\pm$ are finite, since $\mu_f = \mu_g - \mu_h$ is. As for $\mu^\pm(Z^c)$, let's calculate: $\mu^\pm(Z^c) = \mu_f^\pm(Z^c \cap Z) = \mu_f^\pm(\emptyset) = 0$.

Finally, to show that $\mu_f = \mu^+ - \mu^-$ it suffices to show that $\mu_f(Z^c) = 0$, for suppose we have shown that $\mu_f(Z^c) = 0$, then for $E \in \mathcal{B}$, $$ \begin{align} \mu_f(E) &= \mu_f(E \cap Z) + \mu_f(E \cap Z^c) \\ &= \mu_f(E \cap Z) \\ &= \mu_f^+(E \cap Z) - \mu_f^-(E \cap Z) \\ &= \mu^+(E) - \mu^-(E) \end{align} $$

We will now show that $\mu_f(Z^c) = 0$. Since $Z^c = (-\infty, a] \cup (b, \infty)$, it suffices to show that $\mu_f\left((-\infty, a]\right), \mu_f\left((b, \infty)\right) = 0$.

$$ \begin{align} \mu_f\left((-\infty, a]\right) & = \mu_f\left(\bigcup_{n = 0}^\infty \left(a - (n + 1), a - n\right]\right) \\ & = \sum_{n = 0}^\infty \mu_f\left(\left(a - (n + 1), a - n\right]\right) \\ & = \sum_{n = 0}^\infty \left(\mu_g\left(\left(a - (n + 1), a - n\right]\right) - \mu_h\left(\left(a - (n + 1), a - n\right]\right) \right) \\ & = \sum_{n = 0}^\infty \left(\left(g(a - n) - g(a - (n + 1))\right) - \left(h(a - n) - h(a - (n + 1))\right)\right) \\ & = \sum_{n = 0}^\infty \left(\left(g(a) - g(a)\right) - \left(h(a) - h(a)\right)\right) \\ & = 0 \\ \mu_f\left((b, \infty)\right) & = \mu_f\left(\bigcup_{n = 0}^\infty \left(b + n, b + (n+1)\right]\right) \\ & = \sum_{n = 0}^\infty \mu_f\left(\left(b + n, b + (n+1)\right]\right) \\ & = \sum_{n = 0}^\infty \left(\mu_g\left(\left(b + n, b + (n+1)\right]\right) - \mu_h\left(\left(b + n, b + (n+1)\right]\right) \right) \\ & = \sum_{n = 0}^\infty \left(\left(g(b + (n + 1)) - g(b + n)\right) - \left(h(b + (n + 1)) - h(b + n)\right)\right) \\ & = \sum_{n = 0}^\infty \left(\left(g(b) - g(b)\right) - \left(h(b) - h(b)\right)\right) \\ & = 0 \end{align} $$

1.2.2 We will show that $\mu_f^\pm = \mu^\pm$, respectively. By definition $\mu^\pm \leq \mu_f^\pm$, respectively, so we are left to show that $\mu^\pm \geq \mu_f^\pm$, respectively. According to the version of the Jordan decomposition theorem given in Doob's "Measure Theory", if $(\Omega, \mathcal{A}, \lambda)$ is a signed measure space, if $\lambda = \lambda^+ - \lambda^-$ is the unique Jordan decomposition of $\lambda$, and if $\lambda = \lambda_1 - \lambda_2$ is any representation of $\lambda$ as the difference between two measures, then $\lambda^+ \leq \lambda_1$ and $\lambda^- \leq \lambda_2$. Therefore, since by (1.2.1) $\mu_f = \mu^+ - \mu^-$, we obtain $\mu^\pm \geq \mu_f^\pm$, respectively, as desired.

1.2.3 We will show that $\mu_f'(Z^c) = 0$. Indeed, $$ \mu_f'(Z^c) = \mu_f^+(Z^c) +\mu_f^-(Z^c) = \mu^+(Z^c) + \mu^-(Z^c) = 0 $$

1.3 We will now show that it suffices to show that $\mu_f'(E) = \mu_v(E)$ for all $E \in \mathcal{B}(Z)$ in order to conclude that $\mu_f' = \mu_v$. Indeed, suppose we have shown that, for every $E \in \mathcal{B}(Z)$, $\mu_f'(E) = \mu_v(E)$. Then, using (1.1) and (1.2) we see that for $E \in \mathcal{B} = \mathcal{B}(\mathbb{R})$ we have $$ \begin{align} \mu_f'(E) & = \mu_f'\left((E \cap Z) \cup (E \cap Z^c)\right) \\ & = \mu_f'(E \cap Z) + \mu_f'(E \cap Z^c) \\ & = \mu_f'(E \cap Z) \\ & = \mu_v(E \cap Z) \\ & = \mu_v(E \cap Z) + \mu_v(E \cap Z^c) \\ & = \mu_v\left((E \cap Z) \cup (E \cap Z^c)\right) \\ & = \mu_v(E) \end{align} $$

We show that $\mu_f^\pm(J) \geq \mu_v^\pm(J)$, respectively, for all $J \in \mathcal{J}$, as follows. (2.1) We first show that for every $J \in \mathcal{J}$, there is some collection of subsets $S_J \subseteq \mathcal{B}(J)$, such that $\mu_v^+(J) = \sup_{E \in S_J} \mu_f(E)$. (2.2) We deduce, by citing a suitable property of the Jordan decomposition of signed measures, that for every $J \in \mathcal{J}$, $\mu_f^+(J) = \sup_{E \in \mathcal{B}(J)} \mu_f(E) \geq \mu_v^+(J)$. (2.3) Using a similar argument we show that for every $J \in \mathcal{J}$, $\mu_f^-(J) \geq \mu_v^-(J)$.

2.1 We will show that, for every $J \in \mathcal{J}$, there is some collection of subsets $S_J \subseteq \mathcal{B}(J)$, such that $\mu_v^+(J) = \sup_{E \in S_J} \mu_f(E)$. Let $J \in \mathcal{J}$. If $J = \emptyset$, we can take $S_J$ to be $\{\emptyset\}$. Otherwise $J = (c, d]$ for some $c, d \in [a, b]$ such that $c < d$. Define $\mathcal{P}_J$ to be the set of strict partitions of $[c, d]$, that is to say $\mathcal{P}_J$ is the collection of all finite sets of points $\{c = t_0 < t_1 < \cdots < t_n = d\}$.

For every $Q = \{c = t_0 < t_1 < \cdots < t_n = d\} \in \mathcal{P}_J$ define $$ \begin{align} \alpha(Q) & := \bigcup_{k \in \{j \in \{1, 2, \dots, n\} \mid: f(t_j) > f(t_{j - 1})\}} (t_{k - 1}, t_k]\\ \beta(Q) & := \sum_{k = 1}^n \max\left(f(t_k) - f(t_{k - 1}), 0\right) \end{align} $$ and note that $\mu_f\left(\alpha(Q)\right) = \beta(Q)$, since for any $s, t \in \mathbb{R}$ such that $s < t$, $$ \begin{align} \mu_f\left((s, t]\right) &= \mu_g\left((s, t]\right) - \mu_h\left((s, t]\right) \\ &= \left(g(t) - g(s)\right) - \left(h(t) - h(s)\right) \\ &= \left(g(t) - h(t)\right) - \left(g(s) - h(s)\right) \\ &= f(t) - f(s) \end{align} $$

Now define $$ S_J := \{\alpha(Q) :\mid Q \in \mathcal{P}_J\} $$

Then $S_J \subseteq \mathcal{B}(J)$ and furthermore $$ \mu_v^+(J) = v_f^+(d) - v_f^+(c) \overset{(***)}{=} {V^+}_c^d(f) = \sup_{Q \in \mathcal{P}_J} \beta(Q) = \sup_{E \in S_J} \mu_f(E) $$ Equality $(***)$ can be proved similarly to equality $(*)$ above (in section 1.1).

2.2 We will show that for every $J \in \mathcal{J}$, $\mu_f^+(J) \geq \mu_v^+(J)$. According to proposition 6.21(i) in Dshalalow's "Foundations of Abstract Algebra", 2nd edition, for every $D \in \mathcal{B} = \mathcal{B}(\mathbb{R})$, $\mu_f^+(D) = \sup_{E \in \mathcal{B}(D)} \mu_f(E)$. Therefore, using the result of the previous step, $$ \mu_f^+(J) = \sup_{E \in \mathcal{B}(J)} \mu_f(E) \geq \sup_{E \in S_J} \mu_f(E) = \mu_v^+(J) $$

2.3 We can show that $\mu_f^-(J) \geq \mu_v^-(J)$ for all $J \in \mathcal{J}$ using arguments analogous to those presented in (2.1) and (2.2) (this time invoking Dshalalow's proposition 6.21(ii)).
We show that $\mu_f^\pm(E) = \mu_v^\pm(E)$, respectively, for all $E \in \mathcal{B}(Z)$, as follows. (3.1) Firstly we show that $\mu_f^\pm(J) = \mu_v^\pm(J)$, respectively, for all $J \in \mathcal{J}$. We do so by applying a certain minimality property of the Jordan decomposition of signed measures to the results obtained in step (2) above. (3.2) We generalize (3.1) using Dynkin's $\pi$-$\lambda$ theorem to show that $\mu_f^\pm(E) = \mu_v^\pm(E)$, respectively, for all $E \in \mathcal{B}(Z)$.

3.1 We will show that $\mu_f^\pm(J) = \mu_v^\pm(J)$, respectively, for all $J \in \mathcal{J}$. Recall that in step (2) above we showed that $\mu_f^\pm(J) \geq \mu_v^\pm(J)$, respectively, for all $J \in \mathcal{J}$. Therefore, it is left to show that, for all $J \in \mathcal{J}$, $\mu_f^\pm(J) \leq \mu_v^\pm(J)$. Indeed, according to the version of the Jordan decomposition theorem given in Doob's "Measure Theory", if $(\Omega, \mathcal{A}, \lambda)$ is a signed measure space, if $\lambda = \lambda^+ - \lambda^-$ is the unique Jordan decomposition of $\lambda$, and if $\lambda = \lambda_1 - \lambda_2$ is any representation of $\lambda$ as the difference between two measures, then $\lambda^+ \leq \lambda_1$ and $\lambda^- \leq \lambda_2$. Therefore, recalling from section "Attempt #2" of the original post that $\mu_f = \mu_v^+ - \mu_v^-$, we obtain that $\mu_f^\pm \leq \mu_v^\pm$, respectively.

3.2 We will show that $\mu_f^\pm(E) = \mu_v^\pm(E)$, respectively, for all $E \in \mathcal{B}(Z)$. $\mu_f^+$ and $\mu_v^+$ are positive measures that coincide on $\mathcal{J}$. Since $\mathcal{J}$ is a $\pi$-system that generates $Z$ and that includes $Z$, we conclude (for instance, by using Dynkin's $\pi$-$\lambda$ theorem) that $\mu_f^+(E) = \mu_v^+(E)$ for all $E \in \mathcal{B}(Z)$. By the same reasoning, $\mu_f^-(E) = \mu_v^-(E)$ for all $E \in \mathcal{B}(Z)$.
We will show that or all $E \in \mathcal{B}(Z)$, $\mu_f'(E) = \mu_v(E)$. Recall from section "Attempt #2" of the original post that $\mu_v = \mu_v^+ + \mu_v^-$. Therefore, using (3.2), for all $E \in \mathcal{B}(Z)$, $$ \mu_f'(E) = \mu_f^+(E) + \mu_f^-(E) = \mu_v^+(E) + \mu_v^-(E) = \mu_v(E) $$ Q.E.D.

Acknowledgements

The proof idea was suggested to me by the last sentence of section X.6, "Functions of bounded variation vs. signed measures", on p. 164 of Doob's "Measure Theory", Springer 1993.

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