I've met with a multiple select question:
Q. Let $f:[a, b] \rightarrow \mathbb{R}$ be a differentiable map and let $m \leq|f(x)| \leq M$ and $k \leq\left|f^{\prime}(x)\right| \leq K, \forall x \in[a, b]$, then, which of the following is/are true?
[note: $U(P,f)$ and $L(P,f)$ denote the Riemann upper sum and lower sum respectively relative to the partition $P=\{a=x_0, x_1,...,x_n=b\}$ on $[a,b]$. The notation $||P||$ stands for the maximum interval size in the partition $P$.]
- $\left|U(P, f)-\int_a^b f(x) d x\right| \leq M(b-a)$
- $\left|L(P, f)-\int_a^b f(x) d x\right| \geq m(b-a)$
- $\left|L(P, f)-\int_a^b f(x) d x\right| \geq k(b-a)^2$
- $|U(P, f)-{L}(P, f)| \leq K|| P \|(b-a)$
I have eliminated option 1 using the function $f(x)=x^2-\frac{1}{2}$ and I am sure about the option 4, as $$\sup_{[x_{i-1},x_i]}f(t)-\inf_{[x_{i-1},x_i]}f(t)=f(t')-f(t'')=f'(t^*)(t'-t'') \leq K ||P||,$$ for some $t',t'' \in [x_{i-1},x_i]$ and $t^* \in (t',t'')$ (w.l.g).
But I couldn't prove or disprove options 2 and 3.
Thanks in advance for the help
A counterexample for option 2 is $f(x) = x+1$, $[a,b] = [0,1]$ and $P = \{0,1\}$.
In this case, $\int_a^b f(x) \, dx = \int_0^1 (1+x) \, dx = \frac{3}{2}$ and $L(P,f) = 1 \cdot (1-0) = 1$. We then have for the lower bound $m = \inf \{|f(x)| : a\leqslant x \leqslant b\}= 1$, $$\left| L(P,f) - \int_a^b f(x) \, dx\right| = \frac{1}{2} \leqslant 1 = m(b-a)$$
This also serves as a counterexample for option 3, since $f'(x) \equiv 1$ and taking the derivative lower bound $k=\inf \{|f'(x)| : a\leqslant x \leqslant b\}=1$, we have
$$\left| L(P,f) - \int_a^b f(x) \, dx\right| = \frac{1}{2} \leqslant 1 = k(b-a)^2$$