Upper and lower darboux sum

Question

I have recently been going through some questions on Riemann integration.I got stuck in one of those questions which happen to be a multiple-select question. It says that:

Let $f$ be a continuously differentiable real-valued function on $[a,b]$ such that $\left\lvert f'(x)\right\rvert\le k$ for all $x$ in $[a,b]$. Then for any partition $P$ of $[a,b]$, which of the following is/are correct?

1. $\left\lvert L(f,P)\right\rvert\le k(b-a) \le \left\lvert U(f,P)\right\rvert$
2. $U(f,P)-L(f,P) \le k(b-a)$
3. $U(f,P)-L(f,P) \le k \left\lVert P\right\rVert$ where $\left\lVert P\right\rVert$ is the norm of partition $P$
4. $U(f,P)-L(f,P )\le k\left\lVert P\right\rVert(b-a)$

This is what I have done so far:

Since $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ therefore by mean-value theorem there exists a $c$ in $(a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.

And then by the hypothesis we obtain that $\left\lvert f(b)-f(a)\right\rvert\le k(b-a)$.

Now how should I proceed?

Answer 1

Put $$P=(x_i)_{i=0,n}$$

then

$$U(f,P)-L(f,P)=\sum_{i=0}^{n-1}(M_i-m_i)(x_{i+1}-x_i)$$ where $$M_i=\sup_{[x_i,x_{i+1}]}f \; \; and $$ $$m_i=\inf_{[x_i,x_{i+1}] } f .$$ $f $ is continuous at $ [x_i,x_{i+1}] $, thus $$M_i=f(b_i), \; and \; m_i=f(a_i)$$

with $(a_i,b_i)\in [x_i,x_{i+1}]^2$.

Wlog, we assume that $a_i<b_i$.

$f $ is continuous at $[a_i,b_i]$ and differentiable at $ (a_i,b_i) $, then by MVT, $$M_i-m_i=f(b_i)-f(a_i)=(b_i-a_i)f'(c_i)$$

with $b_i-a_i\le x_{i+1}-x_i\le ||P||$. therefore

$$M_i-m_i\le ||P||k$$ and finally $$U(f,P)-L(f,P)\le k||P||(b-a)$$