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On this question, Part of @user147263's answer goes:

Lemma: If $f : \mathbb{R} \to \mathbb{R}$ is convex, then it is Lipschitz on any interval $[a, b]$ with Lipschitz constant $L$ such that $$ L \leq 2 \max_{[a-1, b+1]} |f| $$ Proof: Let $y = g(x)$ be the equation of some secant line to the graph of $f$, based on two points $x_1, x_2 \in [a, b]$. By convexity, $g(a-1) \leq f(a - 1)$ and $g(b+1) \leq f(b+1)$. Hence, the (constant) slope of $g$ satisfies $$ g' = \frac{g(b+1) - g (x_1)}{b+1-x_1} \leq \frac{f(b+1) - f(x_1)}{1} \leq 2 \max_{[a-1, b+1]} |f| \tag{1}\label{gprime} $$ and similarly for $g' \geq -2 \max_{[a-1, b+1]} |f|$.

One issue that comes to mind is that on the first inequality in (\ref{gprime}) he is tacitly assuming that $g(b+1) \geq g(x_1)$. But let's juts assume that this is the case, i.e, that $g$ is non-decreasing.

Now, as for that last part

and similarly for $g' \geq -2 \max_{[a-1, b+1]} |f|$.

I simply could not prove this. The inequalities simply don't seem to line up. I'd appreciate if someone could provide a proof.

Caio Lins
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  • How is it assumed that $g(b+1) \geq g(x_1)$? – Frank Seidl Jul 21 '22 at 16:35
  • @FrankSeidl If $d > c > 0$, then $\frac{a}{c} \geq \frac{a}{d}$ only if $a \geq 0$. So he is either assuming that $g(b+1) \geq g(x_1)$ or that $f(b+1) \geq f(x_1)$. – Caio Lins Jul 21 '22 at 16:40
  • What are $a,c,d$? – Frank Seidl Jul 21 '22 at 16:55
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    What is causing confusing is that, in his solution, the user executed two steps in the same passage, namely, he switched $g(b+1)$ by $f(b+1)$, which is greater, and he removed the positive value $b-x_1$ from the denominator. Supposing that he did the latter first, than the correspondence with my previous comment would be $a = g(b+1) - g(x_1)$, $c = 1$ and $d = b + 1 - x_1$. – Caio Lins Jul 21 '22 at 17:01
  • Ah, I see that now. So if the convex function $f$ is say, $-x$, then $f(b+1)=g(b+1)$, and so the inequality simply fails. I'm not sure there is an easy way to salvage the proof. – Frank Seidl Jul 21 '22 at 17:40

2 Answers2

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If the proof has some flaws it's maybe a good idea to try to find a better proof.

One can use the following

Fact If $p<q<r$ then $\displaystyle \frac{f(q)-f(p)}{q-p}\leq \frac{f(r)-f(q)}{r-q}$.

Proof It's equivalent to the inequality $\displaystyle f(q)\leq \frac{r-q}{r-p}f(p)+\frac{q-p}{r-p}f(r)$ and we have $\displaystyle q=\frac{r-q}{r-p}\cdot p+\frac{q-p}{r-p}\cdot r$.

Corollary If $p<q\leq r<s$ then $\displaystyle \frac{f(q)-f(p)}{q-p}\leq \frac{f(s)-f(r)}{s-r}$.

Now we are prepared to prove the lemma.

Proof Put $\displaystyle M:=\max_{x\in [a-1,b+1]}|f(x)|$.

Let $a\leq x<y\leq b$. Then, from the Corollary, we get $$\frac{f(a)-f(a-1)}{a-(a-1)}\leq \frac{f(y)-f(x)}{y-x}\leq \frac{f(b+1)-f(b)}{(b+1)-b},$$ so

$$-2M\leq f(a)-f(a-1)\leq \frac{f(y)-f(x)}{y-x}\leq f(b+1)-f(b)\leq 2M.$$

This shows that $|f(y)-f(x)|\leq 2M|y-x|$, which ends the proof.

Mateo
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The last part, $$ g' \geq -2\max_{[a-1,b+1]}|f|,$$ follows by symmetry after you've done the rest of the proof. Consider the function $h(x) = f(a+b-x)$, which is a left-right reflection of $f$ across the same interval. Hence if $g'$ is the slope of a secant of $f$, then $-g'$ is the slope of a secant of $h$. Thus from the earlier answer, we know

$$ -g' \leq 2\max_{[a-1,b+1]}|h| = 2\max_{[a-1,b+1]}|f|.$$

Multiplying by $-1$ gives the desired lower bound.

Frank Seidl
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