1

I'm struggling to understand the lemma 3.5 and corollary 3.6, but some points make me difficult to fully understand. Before start, let me upload the lemma and corollary. There are 2 questions in total.

enter image description here enter image description here

Here is the first question about them.

(1) The last sentence of corollary 3.6 says $G(a_k) \le G(b_k)$ when $a = a_k$, but this isn't clear for me, because the previous lemma says $G(a_k) = G(b_k)$ where $E = \cup_k (a_k, b_k)$. The only difference between the lemma and this corollary is the support of the function $G$: $G$ is continuous on all $\mathbb R$ in lemma 3.5, while the $G$ of corollary 3.6 is defined on $[a,b]$. Then, what makes the relationship between $G(a_k)$ and $G(b_k)$ different? I proved the lemma by showing that $G(a_k) < G(b_k)$ doesn't occur due to the fact that $a_k \notin E$, so the fact that inequality $G(a_k) < G(b_k)$ holds in corollary 3.6 makes me confusing.

======================================================

Before asking the second question, let me post the theorem 3.4 and its proof. Although the proof is quite long, it is necessary in this question because I can show you where I'm stuck on.

enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here

This is the end of proof of theorem 3.4

(2) The proof says $E_\gamma \subset \cup_{k}(a_k, b_k)$ where $F(b_k) - F(a_k) \ge \gamma(b_k - a_k)$. I'm just assuming this is related to the last sentence of corollary 3.6 ($G(a_k) \le G(b_k)$), but I'm not sure how exactly (mathematically) this leads to the previous inequallity $E_\gamma \subset \cup_{k}(a_k, b_k)$. Also, there exist $G(x) = -F(-x) + rx$, $G(x) = F(x) - Rx$ in the latter part of the proof, as well as their corresponding inequalities $F(b_k) - F(a_k) \le r(b_k - a_k)$ and $F(b_{k,j} - F(a_{k,j}) \ge R(b_{k,j} - a_{k,j})$. I guess this is also associated with the previous definition of $G(x) = F(x) - \gamma x$ , but I'm not sure. I want to clarify how the inequalities $F(b_k) - F(a_k) \le r(b_k - a_k)$ and $F(b_{k,j} - F(a_{k,j}) \ge R(b_{k,j} - a_{k,j})$ are induced from each definition of $G(x)$. (maybe duplicated questions)

Any help would be greatly appreciated. Thank you.

jason 1
  • 727
  • For your first question, note in 3.6, $E$ is defined as a subset of $(a,b)$. This is the difference between 3.5 and 3.6 you may have missed when comparing them. – Brian Moehring Feb 18 '23 at 10:15
  • I do know the difference of E (similar to the difference of the function G), but I still don't know why. I think the logic of lemma 3.5 can be equally applied to the corollary 3.6, which will lead to same conclusition $G(a_k) = G(b_k)$ not $G(a_k) \ge G(b_k)$, since $a_k =a$ is not an element of E. – jason 1 Feb 18 '23 at 11:07
  • It sounds like you're still misunderstanding. $E$ is defined as a subset of $(a,b)$, not as a subset of $[a,b]$, so very little can be deduced from the fact $a \not\in E$, as $a \in E$ is disallowed regardless of the value of $G(a)$. – Brian Moehring Feb 18 '23 at 18:33
  • As an extreme example, consider $G(x) = x$. Then $G(a) < G(b)$ but still $a \not\in E$ just because we disallow $a \in E$. – Brian Moehring Feb 18 '23 at 20:36

0 Answers0