# Proof of theorem 3.4 in Stein's real analysis

(Corollary 3.6.) Suppose $$G$$ is a real-valued and continuous on a closed interval $$[a,b]$$. If $$E$$ denotes the set of points $$x$$ in $$(a,b)$$ so that $$G(x+h)>G(x)$$ for some $$h>0$$, then $$E$$ is either empty or open. In the latter case, it is a disjoint union of countably many intervals $$(a_k,b_k)$$, and $$G(a_k) = G(b_k)$$, except possibly when $$a =a_k$$, in which case we only have $$G(a_k)\leq G(b_k)$$.

Let $$O\subset\Bbb R$$ be an open set. Then $$O$$ can be written as $$\bigcup I_n$$, with $$I_n$$ disjoint open intervals. Fix $$n$$ and apply Corollary 3.6 to the function $$G(x) = -F(-x)+rx$$ on the interval $$-I_n$$. Reflecting through the origin again yields an open set $$\bigcup_k(a_k,b_k)$$ contained in $$I_n$$, where the intervals $$(a_k,b_k)$$ are disjoint, with $$F(b_k)-F(a_k)\leq r(b_k-a_k).$$

I don't understand the meaning of Reflecting through the origin here. Using the notation of Corollary 3.6, $$E$$ of $$-I_n$$ can be written as $$E =\bigcup_k(\alpha_k,\beta_k)$$. Then $$G(\alpha_k)\leq G(\beta_k)\iff -F(-\alpha_k)+r\alpha_k\leq -F(-\beta_k)+r\beta_k$$. What should I do next from here?

• Hard to say without knowing what you are trying to do. Commented Feb 16, 2023 at 3:19
• @copper.hat I'm trying to understand why $F(b_k)-F(a_k)\leq r(b_k-a_k)$. Commented Feb 16, 2023 at 4:37

"Reflecting through origin" just means negating the interval: $$(u, v) \mapsto (-v, -u)$$. But there is a problem.
Letting $$G(x) = F(x) - rx$$ and applying Corollary 3.6 results in finding intervals satisfying $$F(a_k) - ra_k \le F(b_k) - rb_k$$, that is, $$r(b_k - a_k) \le F(b_k) - F(a_k)$$, exactly the opposite inequality than is needed.
This could have been solved by negating the function, $$G(x) = -(F(x) - rx)$$, so the intervals found are those where $$F(x) - rx$$ is lower at the right endpoint instead of higher. Or it could have been solved by reflecting the domain of the function: $$G(x) = F(-x) -r(-x)$$, so the intervals found are those where $$F(x) - rx$$ is higher at the left endpoint instead of at the right. Either would have worked to get the desired inequality.
Instead, the book went with $$G(x) = -F(-x) + rx = -(F(-x) + r(-x))$$. The intervals found by Corollary 3.6 then satisfy $$G(a_k) \le G(b_k)\\-F(-a_k) + ra_k \le -F(-b_k) + rb_k\\F(-b_k) - F(-a_k) \le r(b_k - a_k)$$ Now since $$a_k < b_k$$ for the intervals found, we have $$-b_k < -a_k$$, so by a relabelling $$\alpha_k = -b_k, \beta_k = -a_k$$, we have intervals $$(\alpha_k, \beta_k)$$ for which $$F(\alpha_k) - F(\beta_k) \le r(-\alpha_k -(-\beta_k))$$ or $$F(\beta_k) - F(\alpha_k) \ge r(\beta_k - \alpha_k)$$. This isn't the inequality they claimed.
Because they talk about reflecting through the origin and $$-I_n$$, I think they intended $$G(x) = F(-x) +rx$$, and the negative sign in front of $$F$$ was just a typo.