Prove that the difference of continuous and monotonically increasing functions has continuous variation

Question

Let $G:[0,\infty)\to\mathbb{R}$ be continuous and $$V^1_t(G):=\sup\bigcup_{n\in\mathbb{N}}\left\{\sum_{i=0}^{n-1}\left|G_{t_{i+1}}-G_{t_i}\right|:0=t_0\le\cdots\le t_n=t\right\}$$

be the variation until $t\ge 0$. Let $C_V$ be the set of all such $G$ with continuous variation $t\mapsto V^1_t(G)$.

I've got two questions:

1. I've read that $V^1(F+G)\le V^1(F)+V^1(G)$ and $V^1(\alpha G)=|\alpha|V^1(G)$ implies that $C_V$ is a vector space. Why? Is this a mistake?
2. Let $G=G^+-G^-$ be the difference of continuous and monotonically increasing functions $G^+,G^-$. Why do we've got $$V^1_t(G)-V^1_s(G)\le\left( G_t^+ -G_s^+\right)+\left( G_t^- -G_s^-\right)\;\;\;\text{for all }t>s$$ and why does that imply $G\in C_V$?

I know that these are basic questions which should be easily to validate. Unfortunately, I failed to solve them by myself.

Answer 1

1) a vector space ($E$) (over $\mathbb{R}$) must be such that $$f,g \in E \Rightarrow \begin{cases}f+ g \in E\\\alpha f \in E \quad \forall \alpha \in \mathbb{R} \end{cases}$$ that is exactly what you get by proving that $V(f+ g) \leq V(f) + V(g)$ and $V(\alpha f) = |\alpha| V(f)$ it implies that both $f+ g$ and $\alpha f$ belong to $C_V$ (they are both continuous and the inequalities shows they have finite variation)

2) Define $$\hat{G}^+(t) = \frac{V^1_t(G) + G(t)}{2}\\\hat{G}^+(t) = \frac{V^1_t(G) - G(t)}{2}$$

Note that $\hat{G}^+(t) - \hat{G}^+(s) = \frac{V^1_t(G) + G(t)}{2} - \frac{V^1_s(G) + G(s)}{2} \geq 0$ Since $V^1_t(G) - V^1_s(G) \geq |G(t) - G(s)|$

Therefore $\hat{G}^+$ is continuous and increasing (the same holds for $\hat{G}^-$).

Note that $$\hat{G}^+(t) + \hat{G}^-(t) = V^1_t(G) $$ therefore $$ V^1_t(G) - V^1_s(G) = \hat{G}^+(t) + \hat{G}^-(t) - \big(\hat{G}^+(s) + \hat{G}^-(s)\big)$$ Next, we claim that

claim: If $G = G^+ - G^-$ where both $G^+, G^-$ are increasing and continuous then $G^+ = \hat{G}^+ +h$ and $G^- = \hat{G}^- + h$ where $h(t)$ is increasing

Therefore $$ V^1_t(G) - V^1_s(G) = (\hat{G}^+(t) -\hat{G}^+(s))+ (\hat{G}^-(t) - \hat{G}^-(s)) = (G^+(t) - h(t) -G^+(s) + h(s))+ (G^-(t) + h(t) - G^-(s) + h(s)) \leq (G^+(t) -G^+(s))+ (G^-(t) - G^-(s))$$

So it suffices to prove the claim this follows by:

$$V^1_t(G) - V^1_s(G) = \lim_n \sum_{k = 1}^{N_n} |G(t^n_{k}) - G(t^n_{k-1})| \leq \lim_n \sum_{k = 1}^{N_n} |G^+(t^n_{k}) - G^+(t^n_{k-1})| + \sum_{k = 1}^{N_n} |G^-(t^n_{k}) - G^-(t^n_{k-1})| \leq V^1_t(G^+) + V^1_t(G^-) = G^+(t) + G^-(t) - G^+ (s) - G^-(s) $$ where the limit is taken in partitions $\mathcal{P}_n = \{s=t^n_0, t^n_1, \ldots, t^n_{N_n} = t\}$of $[s,t]$.

Now take

$h(t) - h(s) = G^+(t) -G^+(s) - (\hat{G}^+ (t) - \hat{G}^+(s)) $

$h(t) - h(s) = G^-(t) -G^-(s) - (\hat{G}^- (t) - \hat{G}^-(s)) $

Therefore $$2(h(t) - h(s)) =G^+(t) + G^-(t) -(G^+(s)+G^-(s)) - (\hat{G}^+ (t) +\hat{G}^- (t) - \hat{G}^+(s)- \hat{G}^-(s)) \\ =G^+(t) + G^-(t) -(G^+(s)+G^-(s)) - (V^1_t(G) - V^1_s(G)) \geq 0$$