0

I'm reading Elementary Analysis by Ross and don't understand the usage of $\frac{\alpha}{2}$ in the proof of the theorem below. Why not just use $\alpha (d - c)$ in the third inequality?

Proof:

Otherwise, since $g$ is continuous, there is a nonempty interval $(c,d) \subseteq [a,b]$ and $\alpha > 0$ satisfying $g(x) \geq \frac{\alpha}{2}$ for $x \in (c,d)$. Then $\int_{a}^{b} g \geq \int_{c}^{d} g \geq \frac{\alpha}{2}(d - c) >0$.

This post doesn't answer my question because I would still wonder why let $\epsilon = \frac{\alpha}{2}$?

m4th3m4t1cs
  • 53

1 Answers1

0

We have $g(x) \ge \frac{\alpha}2$ for all $x \in (c,d)$.

Integrating both sides, gives us the result.

We do not know that $g(x) \ge \alpha$. Hence you can't drop the $\frac12$.

We could have modified the proof by saying that $g(x) \ge \alpha$ directly, then you can just state $\int_c^d g(x)\, dx \ge \alpha (d-c) $ directly.

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
  • This is a duplicate. Maybe you should do a little search for that. – Mittens Sep 17 '21 at 03:23
  • @siong thank you! I just wonder why he did not use $g(x) \geq \alpha $. I thought there must be some reason to work with the more complicated $\frac{\alpha}{2}$. – m4th3m4t1cs Sep 17 '21 at 03:32
  • some context might be required, is it clear to you why we can state $g(x) \ge \alpha$? If not, one might work from the definition of continuity and obtain some constant multiplier. – Siong Thye Goh Sep 17 '21 at 03:54
  • Why care about $\alpha$ or $\alpha /2$? @m4th3m4t1cs You do not know what is $\alpha$ after all. – Arctic Char Sep 17 '21 at 08:59