Show that the sequence is increasing and unbounded

Question

I am unable to proceed after a certain point in a small proof, which I present below. I can understand why the sequence under question is increasing and would like to further understand why it is unbounded. I additionally include necessary definitions and results used in the proof at the end of the post.

Let $g: [a, \infty) \rightarrow $ be right-continous with finite left limits and of finite variation on $[a, \infty)$.

Define $T_0 := a$ and let $T_1$ be the minimum of $a+1$ and the first time when $V_{[a,t]}(g)$ exceeds $\frac{1}{2}$, i.e.

$$ T_1 : = \min \left\{ a+1, \inf \left\{ t \in (a, \infty):V_{[a,t]}(g) > \frac{1}{2} \right\} \right\}, $$

with the covention that $\inf \emptyset = \infty$. Since $t \mapsto V_{[a,t]}(f)$ is right-continous, we have $T_1 >a.$ Further, define $T_2, T_3, \ldots $ inducively by

$$ T_{k+1} : = \min \left\{ a+k+1, \inf \left\{ t \in (T_k, \infty): V_{[T_k,t]}(g) > \frac{1}{2} \right\} \right\}. $$

Then $T_{k+1} > T_{k}$, and since each $g$ is of finite variation we have $\lim_{k \rightarrow \infty} T_k = +\infty.$

I understand how it is concluded that $T_{k+1} > T_k$. To elaborate, first observe $V_{[a,t]}(g)$ is right-continuous and is equal to $0$ for $t = a$. Thus, by right-continuity, there exists $\delta >0$ such that for all $t \in [a, a+\delta)$

$$ V_{[a,t]}(g) < \frac{1}{2}. $$

Therefore, $\inf \left\{ t \in (a, \infty): V_{[a,t]}(g) > \frac{1}{2} \right\} > a$, and

$$ T_1 : = \min \left\{ a+1, \inf \left\{ t \in (a, \infty): V_{[a,t]}(g) > \frac{1}{2} \right\} \right\} > a = T_0. $$

Thus, $a = T_0 < T_1 \leq a+1$.

With similar reasoning, we have

$$ T_2 : = \min \left\{ \underbrace{a+2}_{> T_1}, \underbrace{ \inf \left\{ t \in (T_1, \infty): V_{[T_1,t]}(g) > \frac{1}{2} \right\} }_{> T_1} \right\} > T_1. $$

Thus, $a = T_0 < T_1 < T_2 \leq a+2$.

Eventually, we can conclude that

$$ T_k < T_{k+1} \leq a + k + 1. $$

However, I don't understand why the last statement is true:

since each $g$ is of finite variation we have $\lim_{k \rightarrow > \infty} T_k = +\infty.$

Maybe someone could make this clear. Relevant definitions follow below.

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• Let $I$ be a real interval and $f : I \rightarrow \mathbb{R}$. The total variation of $f$ over $[a,b] \subset I$ is defined as

\begin{equation} V_{[a,b]}(f) := \sup_{ \substack{ a = t_0 < t_1 < \ldots < t_n = b \\ n \in \mathbb{N}} } \sum_{i=1}^{n} |f(t_i) - f(t_{i-1})| \end{equation}

(i.e. the supremum is taken taken over all possible partitions $a = t_0 < t_1 < \ldots < t_n = b$, $n \in \mathbb{N}$ ).

• The function $f$ is said to be of finite variation on $I$ if $V_{[a,b]}(f) < \infty$ for all compact subintervals [a,b] of $I$.
• If $f$ is a right-continuous function on $I$, then for $c \in I$, $t \geq c$, the function $t \mapsto V_{[c,t]}(f)$ is also right-continuous.

Answer 1

If $T_{k+1}-T_k<1$ then $T_{k+1}<a+k+1,$ which implies $V_{[T_k,T_{k+1}]}\geq \tfrac12.$ Note $V_{[T_0,T_k]}=V_{[T_0,T_1]}(g)+\dots+V_{[T_{k-1},T_k]}(g).$ So either:

• $T_{k+1}\geq T_k+1,$ or
• $V_{[a,T_{k+1}]}(g)\geq V_{[a,T_{k}]}(g)+\tfrac 1 2.$

This means $T_k+2V_{[a,T_k]}(g)$ increases by at least $1$ as $k$ increases by $1$: if we define $S_k=T_k+2V_{[a,T_k]}(g)$ then $S_{k+1}\geq S_k+1.$ So $S_k\geq a+k$ for all $k.$ For any $T\geq a,$ if $k\geq T-a+2V_{[a,T]}$ then $T_k+2V_{[a,T_k]}(g)=S_k\geq T+2V_{[a,T]}(g),$ which forces $T_k\geq T.$ Hence $T_k$ is not strictly bounded by $T.$

Answer 2

The sequence $(a + k)_{k \ge 0}$ is obviously unbounded, so we may ignore this part and assume that $(T_k)$ is inductively by $T_{k+1} = \inf \{t \in (T_k, \infty) \, | \, V_{[T_k, t]} > \frac12 \}$ - if we can show that the latter sequence is unbounded, then the minimum of the two unbounded sequences is also unbounded.

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Suppose that $(T_k)$ were bounded above. Then, since the sequence is increasing, it would have a limit, $L$, say (which it approaches from below).

Now, since $g$ is of finite variation, $V_{[a,L]} \leq N$ for some $N \in \mathbb{N}$. But then,

\begin{align} V_{[a,T_{2N+1}]} &= V_{[a, T_1]} + V_{[T_1, T_2]} + ... + V_{[T_{2N}, T_{2N+1}]}\\ &\geq \frac12 + \frac12 + ... + \frac12\\ &= N + \frac12 \end{align}

But clearly $V_{[a, t]}$ is an increasing function (from the definition), so this implies that $T_{2N+1} > L$ which is a contradiction.

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The general idea is that, $V_{[a,T_k]}$ is a sequence which tends to infinity, so if the $T_k$ had a limit, $L$, then the variation on $[a,L]$ would have to be infinite (it would be the sum of the variations on each $[T_k, T_{k+1}]$).