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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

Richard
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    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
  • I guess it is not true on the whole domain $[a,b]$. – Siminore Oct 05 '14 at 13:08
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    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

2 Answers2

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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.

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

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

GEdgar
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