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We defined the convexity of an $f: \mathbb{R} \to \mathbb{R}$ function like this:

$f$ is convex on the interval $I$ if $\forall a,b\in I$, $a<b$, $\forall x \in I$ $$f(x)\leqslant f(a)+\frac{f(b)-f(a)}{b-a}(x-a)$$

And then we had the following theorem:

The following statements are equivalent:
(i) $f$ is convex on $I$
(ii) $\forall a \in I$, $m_a(x)=\frac{f(x)-f(a)}{x-a}$ is monotone increasing on $I\setminus\{a\}$

I tried to prove (i)$\implies$(ii) like this: Let $a,x,y \in I$. If $a<x<y$, then $$f(x) \leqslant f(a) + \frac{f(y)-f(a)}{y-a}(x-a)$$ $$\frac{f(x)-f(a)}{x-a} \leqslant \frac{f(y)-f(a)}{y-a}$$ $$m_a(x) \leqslant m_a(y)$$ Which is fine, but when $x<a<y$, we have that $x-a \leqslant 0$, so $$\frac{f(x)-f(a)}{x-a} \geqslant \frac{f(y)-f(a)}{y-a}$$ $$m_a(x) \geqslant m_a(y)$$ And it's not good. Did I miss something or is the theorem false as it is stated?

Botond
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    To me, the original definition implies the additional constraint that x also be an element of the interval I (although it might be more precise to make the constraint explicit in the definition statement). Your counter example violates that constraint. – John Polcari Dec 20 '18 at 13:22
  • @JohnPolcari You are right, I did miss that out. But I don't see why did my counterexample violate that constraint? – Botond Dec 20 '18 at 13:26
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    Because there is nothing in your counter example that prevents x from being outside I (think about the case where a is the left edge of I). Actually, as I think about it more, I suspect that the original definition should require x to actually be between a and b. – John Polcari Dec 20 '18 at 13:33
  • @JohnPolcari Yes, you are right. It was not mentioned in the formal definition, just before it, that's why I didn't see it. Thank you! – Botond Dec 20 '18 at 13:40
  • Actually, it looks like the definition can be further simplified so that a and b are specifically defined as the edges of I rather than some arbitrary pair of points in I. – John Polcari Dec 20 '18 at 14:00
  • @JohnPolcaribI see that the $\forall a<b \in I \dots$ will imply that it will be true for the endpoints as well, but how can I see the other direction? – Botond Dec 20 '18 at 14:23
  • Never mind. I see where the proposed simplification will get you into trouble... Sorry for the confusion. – John Polcari Dec 20 '18 at 14:37

2 Answers2

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If $x<a<y$, one should use $$ f(a) \leqslant f(x) + \frac{f(y)-f(x)}{y-x}(a-x)\tag{*} $$ instead of $$ f(x) \leqslant f(a) + \frac{f(y)-f(a)}{y-a}(x-a), $$ which holds for $a<x<y$. Then one gets $$ m_a(x)\leq m_x(y). $$ One can also show that $$ m_x(y) \leq m_a(y) $$ which is equivalent to $(*)$.

Botond
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Myunghyun Song
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  • But why isn't the $f(x) \leqslant \dots$ working? – Botond Dec 20 '18 at 13:27
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    I think the definition is wrong. It should require that $ x\in [a,b]$, rather than $x\in I$. – Myunghyun Song Dec 20 '18 at 13:31
  • You are right. It was stated in the very first "definition": The graph of $f$ is under the line passing through $(a,f(a))$ and $(b,f(b))$ on $[a,b]$, but somehow It was missing from the formal definition. Thank you! – Botond Dec 20 '18 at 13:37
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You want to look up the secant lemma, as that is what is likely being a alluded too in the second part of their definition. The problem is, they wrote it incorrectly. For example, by their current definition, $x$ is convex on $I=[a,b]$ since $\frac{d^2}{(dx)^2}(x)\geq 0$, however, letting $f(x)=x$ in the second “equivalent” definition, we find it simplifies to $1$ which is not monotonic.

The issue comes from the actual secant lemma being much more general, and on you mixing strict vs non strict inequalities in both parts.

R.Jackson
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  • I think $f(x)=x$ is both convex and concave. Also, $m_a(x)=\frac{x-a}{x-a}=1$, which is increasing, because $1=m_a(x)\geq m_a(y)=1$. So it looks ok for me. – Botond Dec 20 '18 at 13:31