(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?

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

1 Answer 1


"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.


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