Prove that $U(P,f)-L(P,f)\le\lambda\alpha(b-a)$

Question

Let $\left \{ P \right \}_{i=1}^{\infty}$ be a partition of the interval $[a,b]$ with a norm $\lambda$.

And Let $f$ be a contionous on $[a,b]$ and derivative on $(a,b)$ such as $|f'|\le\alpha$

Prove that $U(P,f)-L(P,f)\le\lambda\alpha(b-a)$

My attempt:

Using MVT in every interval $[x_{i-1},x_i]$:

$\frac{f(x_i)-f(x_{i-1})}{\lambda}\le \frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}=f'(c_i)\le\alpha$

$U(P,f)-L(P,f)=\sum_{i=1}^{n}(M_i-m_i)\Delta x_i\le\sum_{i=1}^{n}(M_i-m_i)\lambda$

But I don't know how to continue from here.

Answer 1

Choose points $a_i, b_i \in [x_{i-1},x_i]$ satisfying $f(a_i) = M_i$ and $f(b_i) = m_i$. Either

• $a_i \not= b_i$, in which case there exists a point $c_i$ in between $a_i$ and $b_i$ satisfying $M_i - m_i = f'(c_i) (a_i - b_i)$, and since $M_i \ge m_i$, $$M_i - m_i = |f'(c_i)||a_i - b_i| \le \alpha \Delta x_i \le \alpha \lambda,$$ or
• $a_i = b_i$ in which $f$ is constant on $[x_{i-1},x_i]$ so that $M_i = m_i$ and you still have $$M_i - m_i \le \alpha \lambda.$$

Rearrange: $$ M_i \le m_i + \alpha \lambda$$ multiply by $\Delta x_i$: $$ M_i \Delta x_i \le m_i \Delta x_i + \alpha \lambda \Delta x_i$$ and sum, using the fact that the sum of the $\Delta x_i$ is just $b-a$.