An estimate for the lower Riemann sum for the derivative of a differentiable function

Question

I have been looking at this question for the past couple days and keep walking away from it as I have no clue where to begin.

Would anyone be able to point me in the right direction or give me any advice? Much appreciated!

Answer 1

Note that $$ m_j\le f'(t), \quad\text{for all}\,\, t\in[x_{j-1},x_j], $$ and hence, due to Mean Value Theorem, there exist points $\xi_j\in(t_{j-1},t_j)$, such that $$ f(x_j)-f(x_{j-1})=(x_{j}-x_{j-1})\,f'(\xi_j) \ge(x_j-x_{j-1})\,m_j=m_j\Delta x_j. $$ Thus $$ f(b)-f(a)=\sum_{j=1}^n \big(f(x_j)-f(x_{j-1})\big)\ge\sum_{j=1}^nm_j\Delta x_j=L(\,f',P). $$

Answer 2

Technical hint: Start by proving it for the special case where $n=1$ -- you'll use the Mean Value Theorem. Then, in the general case, if you already have the inequality for each interval in the partition, they all telescope into the desired result.

More high-level advice: It is difficult to see from your question if the problem you're hitting is that you don't intuitively understand what the claim you're asked to prove says, or that you're not convinced it's true, or that you can understand intuitively that it must be true but can't figure out how to express that understanding as a proof. You would get better answers if you give enough detail in your questions to allow the reader to understand where you're stuck.

Answer 3

This is just the mean value theorem. Since there exist points $c_j \in [x_{j-1},x_j]$ satisfying $f'(c_j)(x_j - x_{j-1}) = f(x_j) - f(x_{j-1})$ you obtain $$\sum_{j=1}^n m_j \Delta x_j \le \sum_{j=1}^n f'(c_j) \Delta x_j = \sum_{j=1}^n (f(x_j) - f(x_{j-1})) = f(x_n) - f(x_0) = f(b) - f(a).$$