Prove that convex function on $[a,b]$ is absolutely continuous

Question

In the book Roberts, Varberg, Convex functions, on pp.9-10 is proved that

If $f:(a,b)\rightarrow \mathbb R$ is continuous and convex then $f$ is absolutely continuous on each $[c,d]\subset (a,b)$.

How can this be extended to the case when the domain of $f$ is instead $[a,b]$?

Edit

I wish to prove the following theorem:

If $f:[a,b] \rightarrow \mathbb R$ is continuous and convex then $f(b)-f(a)=\int_a^b f'(t)dt$.

Thanks

First of all, if $f$ is convex on $(a,b)$ then it is automatically continuous. http://math.stackexchange.com/questions/258511/proof-of-every-convex-function-is-continuous — Tomasz Kania, Oct 05 '14 at 12:28
Yes, I know, but some authors by convex function mean Jensen convex function, which need not be continuous. — Richard, Oct 05 '14 at 12:31
If you explicitly require the continuity on $[a,b]$, it is okay, the absolute continuity of $f$ on $[a,b]$ follows. If not, $f$ could be discontinuous at $a$ and/or $b$. — Daniel Fischer, Oct 05 '14 at 13:16

score 3 · Answer 1 · answered Oct 05 '14 at 12:43

3

Suppose that $f\colon (a,b)\to \mathbb{R}$ is convex. Consider the function

$$F(x,t) = \frac{f(x+t) - f(x)}{t}\quad(t\neq 0).$$

Then $F$ is increasing in each variable. So $F$ is uniformly bounded on each compact subinterval $[c,d]\subset (a,b)$. In particular, $f$ is Lipschitz on $[c,d]$ hence absolutely continuous.

answered Oct 05 '14 at 12:43

Tomasz Kania

16,996

Thanks, but I need absolutely continuity also on the whole $[a,b]$. – Richard Oct 05 '14 at 12:45
@Richard: But $f$ is not even defined on the whole $[a,b]$! – Eric Stucky Oct 05 '14 at 12:46
In my question $f:[a,b] \rightarrow \mathbb R$. – Richard Oct 05 '14 at 12:48

score 2 · Accepted Answer · answered Jan 05 '15 at 13:33

2

Hint. It should be enough to use the case from the book, together with:

If $\phi$ is nondecreasing, then $$ \int_a^b \phi(t)\,dt = \lim_{\alpha\searrow a, \beta \nearrow b} \int_\alpha^\beta \phi(t),dt $$

answered Jan 05 '15 at 13:33

GEdgar

117,296

Does there always exist integral $\int_a^bf'(t)dt$ for convex $f$ ? In general the values $f'(a)$, $f'(b)$ can be in finite for convex continuous $f$. – Richard Jan 05 '15 at 13:50
1

But the limit of $f(\beta) - f(\alpha)$ is finite, right? – GEdgar Jan 05 '15 at 13:55
Ok, this limit is $f(b)-f(a)$, what is needed. Many thanks for explanation. – Richard Jan 05 '15 at 14:03

Prove that convex function on $[a,b]$ is absolutely continuous

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