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A real valued function $f$ defined in $(a,b)$ is said to be convex if $$f(\lambda x+(1-\lambda)y)\le \lambda f(x)+(1-\lambda)f(y)$$ whenever $a < x < b,\; a < y < b,\; 0< \lambda <1$.
Prove that every convex function is continuous.

Usually it uses the fact:
If $a < s < t < u < b$ then $$\frac{f(t)-f(s)}{t-s}\le \frac{f(u)-f(s)}{u-s}\le\frac{f(u)-f(t)}{u-t}.$$

I wonder whether any other version of this proof exists or not?

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    All proofs I have seen boil down to something similar. The above fact is useful in that it shows that right- and left-hand derivatives exist at each point, and hence it is locally Lipschitz. This is true in $\mathbb{R}^n$ as well.2012-12-14
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    Your title is a bit misleading. It is *not* the case that every convex function is continuous. What is true is that every function that is finite and convex *on an open interval* is continuous on that interval (including $\mathbb{R}^n$). But for instance, a function $f$ defined as $f(x)=-\sqrt{x}$ for $x>0$ and $f(0)=1$ is convex on $[0,1)$, but not continuous.2014-08-15
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    Furthermore, in convex analysis we frequently refer to so-called "extended valued functions" defined on the extended real line $[-\infty,+\infty]$. Continuing my example above, for instance, we could define $f(x)=+\infty$ for $x<0$. If we define the secant rule above carefully, using sensible conventions for arithmetic on infinities, you will find that it holds for any points $(a,b)\in\mathbb{R}^n$---even $a,b<0$!2014-08-15
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    Ha ha! I did not notice that this question is almost two years old! Well. I think the clarifications are still important.2014-08-15
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    What is the "usual proof" that uses that fact?2015-03-08
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    It also doesn't hold if we are dealing with infinite-dimensional spaces.2016-11-18
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    See also: https://math.stackexchange.com/questions/24676/convex-function-in-open-interval-is-continuous2018-01-22

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