A basic question on Riemann sum

Question

Suppose $f$ is a non-negative Riemann integrable function in $[a,b]$. Is this true that $$ \sup_P \sum_{j=1}^{n} |f(c_j)(x_j-x_{j-1})| = \int_{a}^{b} |f(x)|dx$$

where $c_j \in [x_{j-1}, x_j]$. I don't think so. But in the definition of bounded variation it is used http://en.wikipedia.org/wiki/Total_variation#Basic_properties

Answer 1

$$ \sup_P \sum_{j=1}^{n} |f(c_j)|\cdot(x_j-x_{j-1}) = (b-a)\cdot\sup_{[a,b]}|f|$$ Proof: For every positive $\varepsilon$, consider a partition with $(x_0,x_1,x_2)=(a,u,b)$, $|f(u)|\geqslant\sup\limits_{[a,b]}|f|-\varepsilon$, and $c_1=c_2=u$.

The only case when such $u$ do not exist is when the supremum is reached at $a$ or $b$, then a subdivision into the unique interval $(a,b)$ does the job. QED.

But in the definition of bounded variation it is used

Is it? Where?

Answer 2

Under suitable smoothness hypotheses on $f$, if $P$ is a partition, then for some $c_i \in (x_{i-1}, x_i)$ you have $\sum_P |f(x_i) -f(x_{i-1})| = \sum_P |f'(c_i)(x_i -x_{i-1})| = \sum_P |f'(c_i)|(x_i -x_{i-1})$, hence $L(|f'|,P) \le \sum_P |f(x_i) -f(x_{i-1})| \le U(|f'|,P)$.

Taking $\sup $ over all partitions gives $\int |f'| \le \operatorname{var} f$.

Let $\epsilon>0$ and select partitions $P_1,P_2$ so that $\operatorname{var} f -\epsilon < \sum_P_1 |f(x_i) -f(x_{i-1})|$ and $U(|f'|,P_2) < \int |f'| + \epsilon$. Now let $P'$ refine both $P_1,P_2$, then \begin{eqnarray} \operatorname{var} f -\epsilon &<& \sum_{P_1} |f(x_i) -f(x_{i-1})| \\ &\le& \sum_{P'} |f(x_i) -f(x_{i-1})| \\ &\le& U(|f'|,P') \\ &\le& U(|f'|,P_2) \\ &<& \int |f'| + \epsilon \end{eqnarray} from which we get $\operatorname{var} f \le \int |f'|$.

Note: the first line is from the selection of $P_1$. The second line is because $P'$ refines $P_1$, and the triangle inequality. The third line follows from the mean value theorem. The fourth line follows because $P'$ is a refinement of $P_2$, and the last from the selection of $P_2$.