Ex2-7

Chapter 2 Question 7

Question: Suppose $$f$$ is of bounded variation on $$ [a,b]$$. If $$f$$ is continuous at a point $$x_0$$, show that $$V(x), P(x), $$ and $$N(x)$$ are also continuous at $$x_0$$. In particular, if $$f$$ is continuous on $$[a,b]$$, then so are $$V(x), P(x), $$ and $$N(x)$$.

Proof: Given a partition $$\Gamma$$ with two adjacent points $$x_0<x$$, we prove the inequality below: $$V[x_0,x]\leq |f(x)-f(x_0)|+V[a,b]-S_\Gamma.$$

Indeed, by applying Theorem 2.2(ii) twice, we have $$V[a,b]-V[x_0,x]=V[a,x_0]+V[x,b]$$. Moreover, we can write $$S_\Gamma$$ as $$S_\Gamma=S_{\Gamma_1}+|f(x)-f(x_0)|+S_{\Gamma_2},$$ where $$\Gamma_1=\Gamma\cap [a,x_0]$$ and $$\Gamma_2=\Gamma\cap [x,b]$$. Basically, we have split $$\Gamma$$ into three parts: $$\Gamma_1, \{x_0,x\}, $$ and $$\Gamma_2$$. Since $$\Gamma_1$$ is a partition of $$[a,x_0]$$, we have $$V[a,x_0]\geq S_{\Gamma_1}$$. Similarly, $$V[x,b]\geq S_{\Gamma_2}$$. Finally, we have $$S_\Gamma\leq V[a,x_0]+|f(x)-f(x_0)|+V[x,b]=|f(x)-f(x_0)|+V[a,b]-V[x_0,x].$$

Now we prove the statement. Given $$\varepsilon>0, x_0\in[a,b]$$, we choose $$\delta>0$$ such that $$|f(x)-f(x_0)|<\frac\varepsilon2\quad\text{ whenever }|x-x_0|< \delta.$$ Next, we choose a partition $$\Gamma$$ such that $$V[a,b]-S_\Gamma<\varepsilon/2$$. Without loss of generality, we might assume $$x_0\in \Gamma$$ since a refinement of partition will increase the value of $$S_\Gamma$$. We can also assume two points adjacent to $$x_0$$ (if $$x_0$$ is not boundary point) or the one point adjacent to $$x_0$$ (if $$x_0$$ is boundary point) is contained in the open interval $$(x_0-\delta,x_0+\delta)$$. If this is not true, we simply add more points in the partition.

Let $$x_1$$ be the point in $$\Gamma$$ after $$x_0$$, we have $$\begin{align}|V(x_1)-V(x_0)|&=V[x_0,x_1]\\ &\leq |f(x_1)-f(x_0)|+V[a,b]-S_{\Gamma}\\ &< \dfrac\varepsilon2 +\frac\varepsilon2\\ &=\varepsilon.\end{align}$$ Similar result holds for $$x_2$$ being the point in $$\Gamma$$ before $$x_0$$.

By monotonicity of $$V(x)$$, we conclude $$|V(x)-V(x_0)|<\varepsilon$$ for all $$x\in[x_2,x_1]$$ where $$ [x_2,x_1] $$ has $$x_0$$ as an interior point. Thus $$V$$ is continuous at $$x_0$$.

By Theorem 2.6, we find that since $$V(x),f(x)$$ are both continuous at $$x_0$$, so are $$P(x)=\frac12[V(x)+f(x)-f(a)]\quad\text{ and }\quad N(x)=\frac12[V(x)-f(x)+f(a)].$$