Ex2-4

Chapter 2 Question 4

''Question: Let $$\{f_k\}$$ be a sequence of functions of bounded variation on $$[a,b]$$. If $$V[f_k;a,b]\leq M<+\infty$$ for all $$k$$ and if $$f_k\to f$$ pointwise on $$[a,b]$$, show that $$f$$ is of bounded variation and that $$V[f;a,b]\leq M$$. Give an example of a convergent sequence of functions of bounded variation whose limit is not of bounded variation.''

Proof: Given $$\varepsilon>0$$, suppose $$\Gamma=\{x_0,x_1,\ldots,x_n\}$$ is any partition of $$[a,b]$$, then for each $$i=0,1,\ldots,n$$we can choose a natural number $$K_i$$such that

$$|f_k(x_i)-f(x_i)|<\frac\varepsilon{2n} \quad\forall k\geq K_i.$$

(We will see why there is $$2n $$ inside the fraction later)

By letting $$K=\max\{K_0,K_1,\ldots,K_n\}$$, we have

$$|f_k(x_i)-f(x_i)|<\frac\varepsilon{2n}\quad\forall i=0,1,\ldots,n, \forall k\geq K.$$

We find that

$$\begin{align} S_\Gamma[f]&=\sum_{i=1}^n |f(x_i)-f(x_{i-1})|\\ &\leq \sum_{i=1}^n \{|f_K(x_i)-f_K(x_{i-1})|+|f_K(x_i)-f(x_i)|+|f_K(x_{i-1})-f(x_{i-1})|\}\\ &< \sum_{i=1}^n \left\{|f_K(x_i)-f_K(x_{i-1})|+\frac\varepsilon n\right\}\\ &= S_\Gamma[f_K]+\varepsilon\\ &\leq M+\varepsilon.\end{align} $$ Since $$\varepsilon$$ is arbitrary, we conclude $$S_\Gamma[f]\leq M$$ for each partition $$\Gamma$$ of $$[a,b]$$, thus the proof is complete.

What if the variations do not bounded above uniformly?

Define a sequence $$\{f_k\}_{n=1}^\infty$$ of functions on $$[0,1]$$ such that $$ f_k(x)=\begin{cases}1&\text{ if }0<x\leq \frac1{2^k},\\0&\text{ otherwise}.\end{cases}$$

We also define a sequence $$\{g_n\}_{n=1}^\infty$$ on $$[0,1]$$ such that $$g_n(x)=\sum_{k=1}^n f_k(x).$$

Clearly, each function $$f_k$$ is of bounded variation with variation $$V[f_k;0,1]=2$$, so each $$g_n$$ being a finite sum of functions of bounded variation, is also of bounded variation. One can check that $$V[g_n;0,1]=2n$$ for all $$n$$. Now $$\{g_n\}_{n=1}^\infty$$ is a sequence of functions of bounded variation. We prove this sequence has a pointwise convergence to a function $$g$$. It is clear that $$g(0)=0$$ since $$g_n(0)=0$$ for all $$n$$. For $$x\in(0,1]$$ we can choose nonnegative integer $$N$$ such that $$\frac1{2^{N+1}}N$$.

Note that for each positive integer$$N$$, there is always a positive number $$x$$ such that $$x\leq 1/2^N$$, therefore $$g$$ is unbounded and therefore it is not a function of bounded variation by Theorem 2.1(i).