This particular question was asked in my Analysis quiz and I was unable to solve it . I have studied analysis from Tom M Apostol .
Let $f$ be a continuously differentiable real valued function on $[a,b]$ such that $|f'(x)|\leq K$ for all $x \in [a,b]$. For a partition $$P=\{a=a_{0} <a_{1}<...<a_{n}=b \} ,$$ let $U(f,P)$ and $L(f,P)$ denote the Upper and Lower Riemann Sums of $f$ with respect to $P$. Then which one of the following is correct?
A. $|L(f,P)|\leq K(b-a)\leq |U(f,P)|$.
B. $U(f,P)-L(f,P) \leq K(b-a)$.
C. $U(f, P)- L(f,P) \leq K\|P\|$, where $\|P\| =\max_{0\le i\le n-1} (a_{i+1} - a_{i} )$.
D. $U(f,P)-L(f,P) \leq K\|P\|(b-a)$.
Finding counterexamples to this would be tedious. So, I think its better to look for the correct option. But I am really confused.
Can anyone please tell how should I approach the problem.
D is the correct option.
$U(f,P)-L(f,P)$ $=\sum_i \sup |f(x(i))-f(y(i))||x_{i+1}-x_i|$ (I use $x(i)$ to show dependence of $x$ on $i$, because $x(i),y(i)\in [x_i,x_{i+1}]$
This equals $\sum_i \frac{|f(x)-f(y)|}{|x-y|}|x_{i+1}-x_i||x-y|$ (I assume the difference is maximised at $x,y$) which equals $\sum_i |f'(\xi_i)||x_{i+1}-x_i||x-y|$ for some $\xi_i \in (x_{i},x_{i+1})$ and hence this is $\leq K||P||(b-a)$