Answer to Question #235228 in Real Analysis for Moses mukonda

Question

 Let f:[a, b] → ℝ be a function and let c ∈ (a, b). If f is of bounded variation on [a, b],

prove that f is bounded on [a, b] and v(f,[a, b] = v(f,[a, c] + v(f,[c,d]).

Answer 1

Definitions. A finite subset $P=\{x_0, x_1, \dots,x_{n_P}\}\subset[a,b]$ is a partition of the segment $[a,b]$ if and only if $a=x_0<x_1<\dots,x_{n_P-1}<x_{n_P}=b$. The set of all partitions of the segment $[a,b]$ is denoted by $Part[a,b]$. If $f:[a,b]\to \mathbb{R}$ is a function, then the variation of $f(x)$ on the partition P is the sum $V(f;P)=\sum\limits_{i=1}^{n_P}|f(x_i)-f(x_{i-1})|$. A quantity $V_a^b(f):=\sup\{V(f;P): P\in Part[a,b]\}$ is called by the total variation of the function $f(x)$ on the segment $[a,b]$. The function $f(x)$ is a function of bounded variation on the segment $[a,b]$, if and only if $V_a^b(f)<+\infty$.

Theorem 1. If $f(x)$ is a function of bounded variation on the segment $[a,b]$, then $f(x)$ is bounded on $[a,b]$.

Proof. Suppose that $f(x)$ is unbounded on $[a,b]$. Then for all $k\in\mathbb{N}$ there exists $t_k\in [a,b]$ such that $|f(t_k)|\to+\infty$, as $k\to+\infty$ . Consider the partition $P_k=\{a,t_k,b\}\in Part[a,b]$.

$V(f;P_k)=|f(t_k)-f(a)|+|f(t_k)-f(b)|\geq 2|f(t_k)|-|f(a)|-|f(b)|\to+\infty$

Therefore, $V_a^b(f)=\sup\{V(f;P): P\in Part[a,b]\}=+\infty$ and $f(x)$ is not a function of bounded variation on $[a,b]$. The received contradiction proves the theorem.

Theorem 2. If $c\in(a,b)$ then $V_a^b(f)=V_a^c(f)+V_c^b(f)$.

Proof. Let $P_1=\{x_0, x_1, \dots,x_{n_1}\}\in Part[a,c]$, $P_2=\{y_0, y_1, \dots,y_{n_2}\}\in Part[c,b]$. Put $P=P_1\cup P_2\in Part[a,b]$. Then

$V(f;P)=\sum\limits_{i=1}^{n_1}|f(x_i)-f(x_{i-1})|+\sum\limits_{j=1}^{n_2}|f(y_j)-f(y_{j-1})|=V(f;P_1)+V(f;P_2)$

Hence

(*) $V_a^b(f)=\sup\{V(f;P): P\in Part[a,b]\}\geq$

$\sup\{V(f;P_1)+V(f;P_2) | P_1\in Part[a,c], P_2\in Part[c,b]\}=$

$=V_a^c(f)+V_c^b(f)$

By now, let $P\in Part[a,b]$ be arbitrary. Put $P'=P\cup\{c\}$.

If $c\in P$, then $P'=P$ and$V(f;P')=V(f;P)$.

If $c\notin P$ then $c\in(x_k, x_k+1)$ for some $k$ and, hence, $V(f;P')-V(f;P)=|f(c)-f(x_k)|+|f(x_{k+1})-f(c)|-|f(x_{k+1})-f(x_k)|\geq 0$ by the triangle inequality.

Put $LP=P'\cap [a,c]\in Part[a,c]$, $RP=P'\cap [c,b]\in Part[c,b]$. Then $P'=LP\cup RP$ and $LP\cap RP=\{c\}$. For such partitions it was proved above that $V(f; P')=V(f; LP)+V(f; RP)$

Therefore

$V(f; P)\leq V(f; P')=V(f; LP)+V(f; RP)\leq V_a^c(f)+V_c^b(f)$

Taking the supremum by $P$ , we obtain

$V_a^b(f)\leq V_a^c(f)+V_c^b(f)$

Combining this with (*), we get $V_a^b(f)= V_a^c(f)+V_c^b(f)$.

Q.E.D.