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 $\leq K||P||(b-a)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.