Question on equivalent definitions of a convex function

Question

As I was reading the proof of Jensen's inequality by Evans' book, I came across with this:

Since $\;f\;$ is a convex function, for each $\;p \in \mathbb R\;$ there exists $\;r\in \mathbb R\;$ such that $$f(q) \ge f(p)+r(q-p).\quad\forall q \in \mathbb R$$

However the definition of convex function as I know it is the following:

$\;f:\mathbb R^n \to \mathbb R\;$ is convex if its domain is convex set and for all $\;x,y\;$ in its domain, and all $\;λ \in [0,1]\;$ we have $$f(λx+(1-λ)y) \le λf(x)+(1-λ)f(y).$$

Thus my question is, why is the statement in Evans'book true?

I also found this Theorem:

which states that if a function is convex then its graph should lie above then tangent line for all $\;x\;$ in its domain.Although in my case $\;f\;$ is not necessary differentiable and hence I can't assume $\;r=f'(p)\;$.

Last but not least, I found this question here but these answers weren't that helpful for me.

I would really appreciate if somebody could explain to me all the above and save me from all this confusion!

Thanks in advance!

If you don't have derivative (a limit) just take $\liminf$ instead. — , Feb 09 '18 at 13:07
The gradient of the line $r$ should depend on the point $x_0$ (which is your $p$.) — Calvin Khor, Feb 09 '18 at 13:08
@CalvinKhor Maybe I'm missing something here. I get why it should depend on $;x_0;$ but I cannot see why $;r;$ in Evans book is equal to $;f'(p);$ since there is no differentiability. Hope I made my confusion here more clear.. — kaithkolesidou, Feb 09 '18 at 13:14
I made no claims that it should be of that form, just commenting on your sentence near the end "on the one hand, in that post the number $r$ depends on $x_0$" — Calvin Khor, Feb 09 '18 at 13:15
Maybe you'll find this useful https://en.wikipedia.org/wiki/Subderivative Evans is using the fact that for convex functions the subdifferential at a point is not empty (which gives you the r). — Uskebasi, Feb 09 '18 at 13:28
@Uskebasi I think this is extremely helpful right now. Thanks.. I had no idea about subdifferential — kaithkolesidou, Feb 09 '18 at 13:40

score 5 · Accepted Answer · answered Feb 09 '18 at 13:46

It is well known that one equivalent definition of convexity is the statement that the gradient of secant lines increase. One way to write this down is as follows: for all $x<y<z$,

$$ \frac{f(y)-f(x)}{y-x} ≤ \frac{f(z)-f(y)}{z-y}$$ Let $y$ be fixed. Take $r := \liminf_{z\downarrow y} \frac{f(z) - f(y)}{z-y}, $ so that $$ \frac{f(y) - f(x) }{y-x} ≤ r$$ Then $$f(x) ≥ f(y) + r(x-y), \quad x<y$$

Similarly if $r':= \limsup_{x\uparrow y}\frac{f(y)-f(x)}{y-x}$ one obtains $$f(z) ≥ f(y) + r'(z-y), \quad y<z$$ If $r'>r$ then $(r'-r)(x-y) ≤0 $ for $x<y$, so we can write $$ f(x) ≥ f(y) + r(x-y) + (r'-r)(x-y) = f(y) + r'(x-y) $$ and hence $$ f(x) ≥ f(y) + r'(x-y) \quad ∀ x$$

If instead $r'<r$ we can similarly unify both inequalities, hence the result. Also, if $f$ were differentiable at $y$ then $r=r'=f'(y)$.

Thanks a lot for your answer and your time! This is exactly what I needed... — kaithkolesidou, Feb 09 '18 at 14:00
Nice elementary proof. The only proof I knew used the support hyperplane theorem. — Uskebasi, Feb 09 '18 at 14:00
@Uskebasi If i recall correctly, something similar happens in that proof before applying Zorn's lemma. — Calvin Khor, Feb 09 '18 at 14:04

Question on equivalent definitions of a convex function

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