Background Information:
If $F:\mathbb{R}\rightarrow \mathbb{C}$ and $x\in\mathbb{R}$, we define $$T_{F}(x) = \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1})|:n\in\mathbb{N},-\infty<x_0<\ldots<x_n = x\}$$ $T_F$ is called the total variation of $F$. We observe that the sums in the definition of $T_F$ are made bigger if the additional subdivision points $x_j$ are added. Hence, if $a<b$, the definition of $T_F(b)$ is unaffected if we assume that $a$ is always one of the subdivision points. It follows that $$T_F(b) = T_F(a) + \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1})|:n\in\mathbb{N},a=x_0<\ldots<x_n = b\}$$ Thus $T_F$ is an increasing function with values $[0,\infty]$. If $T_F(\infty) = \lim_{x\rightarrow\infty}T_F(x)$ is finite, we say that $F$ is of bounded variation on $\mathbb{R}$, and we denote the space of all such $F$ by $BV$.
Problem 27:
a.) If $F:\mathbb{R}\rightarrow \mathbb{R}$ is bounded and increasing, then $F\in BV$ (in fact, $T_F(x) = F(x) - F(-\infty)$).
b.) If $F,G\in BV$ and $a,b\in\mathbb{C}$, then $aF + bG\in BV$.
c.) If $F$ is differentiable on $\mathbb{R}$ and $F'$ is bounded, then $F\in BV([a,b])$ for $-\infty < a < b < \infty$(by MVT).
d.) If $F(x) = \sin x$, then $F\in BV([a,b])$ for $-\infty < a < b < \infty$, but $F\notin BV$.
e.) If $F(x) = x\sin(x^{-1})$ for $x\neq 0$ and $F(0) = 0$, then $F\notin BV([a,b])$ for $a\leq 0 < b$ or $a < 0\leq b$.
Attempted proof a.) If $F:\mathbb{R} \rightarrow\mathbb{R}$ and $x\in\mathbb{R}$, then $$T_F(x) = \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1}|):n\in\mathbb{N}, -\infty < x_0 < \ldots < x_n = x \}$$ Since $F$ is increasing, the intervals $(F(x^-),F(x^+) (x\in\mathbb{R})$ are disjoint, and for $|x| < N$ they lie in the interval $(F(-N),F(N))$. Hence $$\sum_{|x| < N}|F(x^{+} - F(x^{-1}| \leq F(N) - F(-N) < \infty$$ so now we know $F$ is finite, therefore the total variation of $F$ is also finite, therefore $$\lim\limits_{x\rightarrow \infty}T_F(x) = T_F(\infty)\Rightarrow F\in BV$$
Attempted proof b.) Since $F,G\in BV \Rightarrow \lim_{x\rightarrow \infty}T_F(x) = T_F(\infty)$ and $\lim_{x\rightarrow \infty}T_G(x) = T_G(\infty)$ since $a,b$ are just scalars it follows that $$\lim_{x\rightarrow \infty} aT_F(x) + bT_G(x) = aT_F(\infty) + bT_G(\infty) \Rightarrow aF + bG\in BV$$
Attempted proof c.) Not sure
Attempted proof d.) Not sure
Attempted proof e.) Not sure
I just want to know if my two proofs are correct or not and what suggestions or hints that I could be provided for proving c,d,and e.
a). Your proof is correct.
b). Your proof is ok. You can also prove it by using the fact that $$ |aF(x_i)+bG(x_i)-(aF(x_{i-1})+bG(x_{i-1}))|\leqslant|a||F(x_i)-F(x_{i-1})|+|b||G(x_i)-G(x_{i-1})| $$ So \begin{align} T_{aF+bG}&=\sup(\sum_{i=1}^n|aF(x_i)+bG(x_i)-(aF(x_{i-1})+bG(x_{i-1}))|) \\ &\leqslant|a|\sup(\sum_{i=1}^n|F(x_i)-F(x_{i-1})|)+|b|\sup(\sum_{i=1}^n|G(x_i)-G(x_{i-1})|) \\ &<\infty \end{align} c). By MVT $$ T_F=\sup(\sum_{i=1}^n|F(x_i)-F(x_{i-1})|)=\sup(\sum_{i=1}^n|F'(\xi_i)(x_i-x_{i-1})|)\leqslant M(b-a) $$ where $|F'(x)|<M$ and $x_i>x_{i-1}$.
d). On $[a,b]$, since $F'(x)=\cos x$, by c) $F$ is of bounded variation. On $(-\infty, \infty)$, let $x_{2n}=2n\pi+\pi/2$ and $x_{2n-1}=2n\pi-\pi/2$. Then $$ T_F=\sup(\sum_{n=1}^{\infty}|F(x_n)-F(x_{n-1})|)=\infty $$
e). Note that on any $(0,\epsilon)$, $F(x)=x\sin {\frac1{x}}$ has infinite $0$. Let $x_{2n}=\frac1{2n\pi+\pi/2}$ and $x_{2n+1}=\frac1{2n\pi+\pi}$. Then $$ T_F=\sup(\sum_{n=1}^{\infty}\frac1{2n\pi+\pi/2})=\infty $$