Functions of Bounded Variation Have Left and Right Limits

Question

I have a proof of this for a real valued function
https://www.encyclopediaofmath.org/index.php/Function_of_bounded_variation#Generalizations based on the Jordan decomposition into monotonic functions.

I am looking for a more general proof in the case of a function $f: I = [a, b] \subset \mathbb R \to X$ where $X$ is a complete normed vector space (or even more generally, a complete metric space $(X, d)$ in which case substitute $||f(s) - f(t)||$ by $d(f(s) - f(t))$ in what follows) .
My own attempt follows and I would appreciate feedback on this.

Definitions:

Let $f: I = [a, b] \subset \mathbb R \to X$ where $X$ is a normed vector space and $||.||$ its norm.

A partition of a $[a, b]$ is a finite set of points $P = \{p_0 = a < p_1 < p_2 ... < p_n = b\}$

$V(f, P):= \sum_{i=1}^n ||f(p_i) - f(p_{i-1}||$ is a real non-negative number being the variation of $f$ on the partition $P$ (of the interval $I$). Note that the variation is defined for any function and any partition.

$V(f):= sup\{V(f, P): P $ is a partition of I } is an extended real (i.e. can be $+\infty$) being the (total) variation of $f$ on the interval $I$.

If $V(f)$ is finite then $f$ is a function of bounded variation.

$f$ has a left limit at $x \in (a, b]$ if given $\epsilon > 0$ there is $w \in [a, x)$ such that for all $y, z \in (w, x)$ then $||f(y) - f(z)|| < \epsilon$. This is the Cauchy condition for the existence of this limit in a complete space. The right limit is similarly defined for $x \in [a, b)$.

Proof:

Assume that $f$ does not have a left limit at some point $x \in (a, b]$ and construct a sequence of partitions of $[a, x]$ of increasing (unbounded) variation. This shows that in such cases $f$ cannot be of bounded variation (on $[a, x]$ and therefore also on $[a, b]$). The case for not having a right limit is analogous, so it follows by contra-positive that if $f$ is of bounded variation these limits must exist.

Express the absence of a left limit by negation of the Cauchy definition.......
If $f$ has no left limit at $x \in (a, b]$ then there is some $\epsilon > 0$ where for all $w \in [a, x)$ there is some $y, z \in (w, x)$ with $||f(y) - f(z)|| \ge \epsilon$. Wlg assume $y < z$.

Firstly, this applies for $w = a$ and there is $a < y_1 < z_1 < x$ with $||f(y_1) - f(z_1)|| \ge \epsilon$. Now define the partition $P_1 = \{a, y_1, z_1, x\}$ and it follows that $V(f, P_1) \ge \epsilon$.

As a second iteration, the condition applies for $w = z_1$ so there is $z_1 < y_2 < z_2 < x$ with $||f(y_2) - f(z_2)|| \ge \epsilon$. Define the partition $P_2$ by adding $y_2, z_2$ to $P_1$ (in sequence) $ = \{a, y_1, z_1, y_2, z_2, x\}$ and it follows that $V(f, P_2) \ge 2\epsilon$.

Then one can continue and define a sequence of partitions of $[a, x]$ $P_1, P_2, .....$ and for $P_n$ we have $V(f, P_n) \ge n.\epsilon$ so $sup\{V(f, P): P $ is a partition of $[a, x]\} = \infty$, I.e. $f $ is not of bounded variation on $[a, x]$.

Answer 1

Suppose the function $f$ does not have a limit from the left at some $x\in(a,b]$.

Then by your definition of left limit, there is an $\epsilon > 0$ so that if $t<x$, then there will be $a'<b' \in (t,x)$ so that $a<a'$, and $ d(f(a'),f(b'))> \epsilon$.

Proceed by induction. Let $t_0=a$, and using the above remark choose $t_0<a_1< b_1$ so that $d(f(a_1),f(b_1))> \epsilon$ and $b_1 < x$; then using induction, set $t_k = b_k$ and again using the above remark choose $a_{k+1}< b_{k+1}$ so that $d(f(a_{k+1}),f(b_{k+1}))> \epsilon$, $t_k<a_{k+1}$ and $b_{k+1} < x$.

For convenience rename the sequence ${a_1,b_1,a_2,\ldots}$ as ${x_0,x_1,x_2,\ldots}$. The sequence ${x_0,x_1,x_2,\ldots}$ are strictly increasing in $(a,b]$ and $$\sum_{k=0}^{2N} d(f(x_{k}),f(x_{k+1}))^p >\sum_{k=0}^N d(f(x_{2k}),f(x_{2k+1}))^p > N\epsilon^p$$

Recall that, although we cannot choose $\epsilon$, it is the case that $\epsilon>0$ and that the inequality holds for arbitrary positive integers $N$. We conclude that $f$ has unbounded $p$-variation for any $p$ including $p=1$.

I have only answered your question. An exercise: If $f$ is real valued and finite $p$-variation, then it is easy to show that it only has countably many points of discontinuity. Does that theorem extend to the Banach, or complete metric case?