Vtyjkyuk

$f:I to mathbb R$ is a convex function on an open interval $I$. I know that if $f$ is differentiable on $I$, then $forall x_0,x in I , f(x_0)+f'(x_0)(x-x_0) le f(x)$

I read somewhere that something similar true even if $f$ is not differentiable, i.e. $exists a in mathbb R: forall x_0, x in I, f(x) ge a(x-x_0)+f(x_0)$.

Does anybody know a proof for that? It would e.g. help to show that $forall x_0 in I$ there is a convex differentiable function $g:I to mathbb R, g(x_0)=f(x_0)$ and $forall x in I, g(x) le f(x)$

It is well-known that a convex function $f : I to mathbbR$ defined on an open interval is continuous and has a left derivative $f'_-(x_0) = lim_x to x_0^- fracf(x) - f(x_0)x-x_0$ and a right derivative $f'_+(x_0) = lim_x to x_0^+ fracf(x) - f(x_0)x-x_0$ at each $x_0 in I$ such that for $x_- < x_0 < x_+$

$$fracf(x_-) - f(x_0)x_- - x_0 le f'_-(x_0) le f'_+(x_0) le fracf(x_+) - f(x_0)x_+ - x_0 .$$

Then for any $a in [f'_-(x_0),f'_+(x_0)] $ and $x ne x_0$ we get

$$a(x - x_0) le f(x) - f(x_0) .$$

If $f$ is differentiable at $x_0$, this yields

$$f'(x_0)(x - x_0) le f(x) - f(x_0) .$$

Added on request:

A function $f : I to mathbbR$ is convex if for all $x_1, x_2 in I$ and all $t in [0,1]$ one has $f(tx_1 + (1-t)x_2) le tf(x_1) + (1-t)f(x_2)$. Here is an alternative characterization of convex functions:

Lemma. For a function $f : I to mathbbR$ the following are equivalent:

(1) $f$ is convex.

(2) For all $a,b,x in I$ such that $a le x le b$, $f(x) le f(a) + fracf(b) - f(a)b-a(x - a)$.

(3) The function $Delta_f : I backslash x_0 to mathbbR, Delta_f(x) = fracf(x) - f(x_0)x - x_0$ is monotonically increasing for all $x_0 in I$.

Proof. (1) $Rightarrow$ (2) : Define $t = fracb-xb-a in [0,1]$. Then $x = ta + (1-t)b$ and $1-t = fracx-ab-a$ so that $f(x) le tf(a) + (1-t)f(b) = f(a) + (1-t)(f(b) - f(a)) = f(a) + fracf(b) - f(a)b-a(x - a)$.

(2) $Rightarrow$ (1) : It suffices to consider the case $x_1 < x_2$. Apply (2) for $a = x_1, b = x_2, x = tx_1 + (1-t)x_2$ and use $x - x_1 = (1-t)(x_2-x_1)$.

(2) $Rightarrow$ (3) : Let $x_1, x_2 in I backslash x_0 $ such that $x_1 < x_2$. Case 1: $x_0 < x_1$. Let $a = x_0, b = x_2, x = x_1$. Then $f(x_1) le f(x_0) + fracf(x_2) - f(x_0)x_2 - x_0(x_1 -x_0)$, hence $fracf(x_1) - f(x_0)x_1 - x_0 le fracf(x_2) - f(x_0)x_2 - x_0$. Case 2: $x_2 < x_0$. $a = x_1, b = x_0, x = x_2$. Case 3: $x_1 < x_0 < x_2$. $a = x_1, b = x_2, x = x_0$.

(3) $Rightarrow$ (2) : $x_1 = a,x_2 =b, x_0 = x$.

The lemma shows that $f'_-(x_0) = lim_x to x_0^-Delta_f(x)$ exists because $Delta_f(x)$ is bounded from above for $x < x_0$. Similarly $f'_+(x_0) = lim_x to x_0^+Delta_f(x)$ exists because $Delta_f(x)$ is bounded from below for $x > x_0$ (by $f'_-(x_0)$ which proves the second inequality).