There are many questions on the same topic like in this thread; however, the part where I am stuck is not explained anywhere. This problem is from the book Understanding Analysisby Stephen Abbott.
Question: Given a function $f$ on $[a,b]$, define the total variation of $f$ to be $$ Vf = \sup \left\{ \sum_{k=1}^{n} \left| f(x_k) - f(x_{k-1}) \right| \right\} $$ where the supremum is taken over all partitions $P$ of $[a, b]$.
(a) If $f$ is continuously differentiable ($f′$ exists as a continuous function), use the Fundamental Theorem of Calculus to show $Vf \leq \int_a^b |f'|$.
(b) Use the Mean Value Theorem to establish the reverse inequality and conclude that $Vf =\int_a^b |f'|$.
My attempt:
(a) Since $f$ is continuously differentiable, we can use Fundamental Theorem of Calculus to conclude that $$\left|\int_{x_{k-1}}^{x_k} f' \right| = \left|f(x_k) - f(x_{k-1}) \right| \leq \int_{x_{k-1}}^{x_k}\left| f' \right|$$ Summing the above from $k=1$ to $n$ will result in $$ \sum_{k=1}^{n} \left| f(x_k) - f(x_{k-1}) \right| \leq \int_a^b |f'|$$ RHS above is independent of the partition and forms an upper bound to the LHS. Hence $$Vf = \sup \left\{ \sum_{k=1}^{n} \left| f(x_k) - f(x_{k-1}) \right| \right\} \leq \int_a^b |f'|$$ (b) As $f$ is continuously differentiable, for a given partition $P$ of $[a,b]$, say $P=\{x_0=1,x_1,x_2,\cdots,x_n=b\}$, mean value theorem can be applied to each interval $[x_{k-1},x_k]$ giving us $$f(x_k)-f(x_{k-1}) = f'(c_k)\left[ x_k - x_{k-1} \right]$$ for some $c_k \in (x_{k-1},x_k)$. Taking absolute values of both sides and summing $k$ from $1$ to $n$ gives us $$\sum_{k=1}^{n} \left| f(x_k) - f(x_{k-1}) \right| = \sum_{k=1}^{n}\left|f'(c_k)\right|\left[ x_k - x_{k-1} \right]$$ This is the point where it is breaking for me. From here, we ultimately need to arrive at $$Vf = \sup \left\{ \sum_{k=1}^{n} \left| f(x_k) - f(x_{k-1}) \right| \right\} \geq \int_a^b |f'|$$ where the definition of integral is given in the end of this post. Please let me know if the following lines are correct and if not, please let me know how to complete part (b).
Assume that I define $n_k = \inf{|f'|}$ in the interval $[x_{k-1},x_k]$. Then we have $$\sum_{k=1}^{n} \left| f(x_k) - f(x_{k-1}) \right| = \sum_{k=1}^{n}\left|f'(c_k)\right|\left[ x_k - x_{k-1} \right] \geq \sum_{k=1}^{n} n_k \left[ x_k - x_{k-1} \right]$$ Taking supremum on both sides, we get $$Vf = \sup{\sum_{k=1}^{n} \left| f(x_k) - f(x_{k-1}) \right|} \geq \sup{\sum_{k=1}^{n} n_k \left[ x_k - x_{k-1} \right]} = \int_a^b |f'|$$
Definition of integral: Let $g$ be bounded and defined on $[a,b]$. For a given partition $P$ of $[a,b]$, say $P = \{x_0=1,x_1,x_2,\cdots,x_n=b \}$, let $U(g,P) = \sum_{k=1}^{n}M_k[x_k-x_{k-1}]$ and $L(g,P) = \sum_{k=1}^{n}m_k[x_k-x_{k-1}]$ where $M_k = \sup{g}$ and $m_k = \inf{g}$ in the interval $[x_{k-1},x_k]$. If $\inf{U(g,P)} = \sup{L(g,Q)}$ where $P$ and $Q$ can be any partition of $[a,b]$, then $g$ is said to be integrable and the integral is this common value of $U(g) = \inf{U(g,P)} = L(g) = \sup{L(g,Q)}$.
