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I would like to show that if $g:[a,b]\rightarrow R$ is continuous and has a bounded variation, then total Variation function is also continuous.

2 Answers2

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Take any $x_0 \in [a,b]$. Since $g$ is right-continuous, given $\epsilon > 0$ there is a $\delta > 0$ such that $0 < x - x_0 < \delta $ implies $|g(x) - g(x_0)| < \epsilon/2$.

Note that total variation of $g$ over $[a,b]$ decomposes as

$$TV(g; [x_0,b]) = TV(g;[a,b]) - TV(g;[a,x_0]) = V_g(b) - V_g(x_0)$$

There is a partition $x_0 < x_1 < \ldots < x_n = b$ such that

$$V_g(b) - V_g(x_0) \leqslant \sum_{k=1}^n |g(x_k) - g(x_{k-1})| + \epsilon/2$$

Let $\hat{\delta} = \min(\delta,x_1-x_0)$. If $x_0 < x < x_0 + \hat{\delta}$ we have both $x- x_0 < \delta $ and $x_0 < x < x_1$, whence,

$$V_g(b) - V_g(x_0) \leqslant |g(x_1) - g(x_0)| + \sum_{k=2}^n |g(x_k) - g(x_{k-1})| + \epsilon/2 \\ \leqslant |g(x) - g(x_0)| + |g(x_1) - g(x)| +\sum_{k=2}^n |g(x_k) - g(x_{k-1})| + \epsilon/2 \\ < \epsilon/2 + V_g(b) - V_g(x) + \epsilon/2,$$

which implies

$$V_g(x) < V_g(x_0) + \epsilon$$

Also, since $a \leqslant x_0 < x$, we have $V_g(x_0) < V_g(x)$.

Therefore, if $x_0 < x < x_0 + \hat{\delta}$ then $V_g(x_0) < V_g(x) < V_g(x_0) + \epsilon$, proving that $V_g$ is right-continuous at $x_0$.

RRL
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  • thank you very much! can you also explain why exactly it is posible to do the first equation? –  Jul 06 '18 at 08:20
  • @emma19: I introduced the notation $TV(g;[a,b])$ to denote the total variation of $g$ over the interval $[a,b]$. Here $TV(g;[a,b]) = \sup_P \sum_{k=1}^n |g(x_k) - g(x_{k-1}|$ where the supremum is taken over all partitions of $[a,b]$. A very basic property of bounded variation functions is that if $a < c < b$ then $TV(g;[a,b]) = TV(g;[a,c]) + TV(g;[c,b])$ - proved in most analysis books. – RRL Jul 06 '18 at 08:35
  • okay thanks again :)) have a nice day! –  Jul 06 '18 at 08:39
  • For starters if $P$ is a partition of $[a,c]$ and $Q$ is a partition of $[c,b]$ then $P \cup Q$ is a partition of $[a,b]$ and $\sum_{x_k \in P}|g(x_k) - g(x_{k-1})| + \sum_{y_k \in Q}|g(y_k) - g(y_{k-1})| =\sum_{z_k \in P \cup Q}|g(z_k) - g(z_{k-1})| \leqslant TV(g;[a,b])$ which leads to $TV(g;[a,c]) + TV(g;[c,b]) \leqslant TV(g;[a,b])$. With a little work you can prove the reverse inequality as well. QED – RRL Jul 06 '18 at 08:42
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$V_g(x)$ is easily proved to be increasing for $x\geq a.$ So for $ x>a$ let $U(x)=\lim_{y\to x^-}V_g(x).$ We have $U(x)=\sup_{y<x}V_g(y)\leq V_g(x).$

Suppose $V_g(x)$ is not right-continuous at $x>a.$ Then $U(x)<V_g(x).$ Let $d=(V_g(x)-U(x))/2.$ Take a strictly increasing finite sequence $(a_0,...,a_n)$ with $a_0=a$ and $a_n=x,$ such that $\sum_{j=0}^{n-1}|g(a_j)-g(a_{j+1})|>V_g(x)-d=U(x)+d.$

Now for any $y\in (a_{n-1},x)$ we have $$\left(|g(a_{n-1})-g(y)|+\sum_{j=0}^{n-2}|g(a_j)-g(a_{j+1})|\right)\leq V_g(y)\leq U(x)$$ and we have $$|g(y)-g(x)|+\left(|g(a_{n-1})-g(y)|+\sum_{j=0}^{n-2}|g(a_j)-g(a_{j+1})|\right)\geq$$ $$\geq |g(a_{n-1})-g(x)|+\left(\sum_{j=0}^{n-2}|g(a_j)-g(a_{j+1})|\right)=$$ $$=\sum_{j=0}^{n-1}|g(a_j)-g(a_{j+1})|>V_g(X)-d=U(x)+d.$$

These imply that $|g(y)-g(x)|>d$ for all $y\in (a_{n-1},x),$ contrary to the right-continuity of $g.$

DanielWainfleet
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  • In case $n=1$ we follow the usual convention (in the sums from $j=0$ to $j=n-2$ ) that $\sum_{j=0}^{j=-1} (Anything)_j=0.$ – DanielWainfleet Jul 06 '18 at 04:04
  • your definition of right continuity is my definition of left continuity :p –  Jul 06 '18 at 16:02
  • I always forget whether right-continuity means continuous to the right or from the right. But I think what I wrote can give you the idea of how to answer your Q. If you ask, I will re-write it for you. – DanielWainfleet Jul 06 '18 at 18:49