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A real-valued function $f$ defined in $(a,b)$ is said to be convex if $$f(\lambda x+(1-\lambda)y)\leqslant \lambda f(x)+(1-\lambda)f(y)$$ whenever $a<x<b, \quad a<y<b,\quad 0<\lambda<1$.

Prove that every convex function is continuous.

I also proved that if $f$ is convex in $(a,b)$ and $a<s<t<u<b$ then $$\dfrac{f(t)-f(s)}{t-s}\leqslant\dfrac{f(u)-f(s)}{u-s}\leqslant\dfrac{f(u)-f(t)}{u-t}.$$ But how to use this inequality in proving of continuity?

Please can anyone show not hard proof of continuity without using derivatives.

RFZ
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    There is $u_1$ such that $f(t)-f(s)\le\frac{f(u_1)-f(t)}{u_1-t}(t-s)=L_1(t-s)$. Similarly there is $u_2$ such that $f(t)-f(s)\ge\frac{f(t)-f(u_2)}{t-u_2}(t-s)=L_2(t-s)$. So $|f(t)-f(s)|\le L|t-s|$ for some $L$. – Ziyuan Oct 02 '15 at 09:34
  • @ziyuang, Why such $u_1$ exists? Can you explain it in detail please? – RFZ Oct 02 '15 at 09:39
  • What is $t,s$? any point from interval? – RFZ Oct 02 '15 at 09:40
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    See here: http://math.stackexchange.com/questions/148599/proving-that-a-convex-function-is-lipschitz – Math1000 Oct 02 '15 at 09:41
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    http://math.stackexchange.com/questions/258511/proof-of-every-convex-function-is-continuous?rq=1 here's the proof – user300 Oct 02 '15 at 09:44
  • @SaikatPanja, what proof do yo mean? – RFZ Oct 02 '15 at 09:47
  • @RFZ that every convex function is continuous – user300 Oct 02 '15 at 09:55
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    @Math1000, Thanks a lot for helpful link. Our function $f$ is convex in $(a,b)$ then it would be a Lipschitz at any segment $[c,d]$ such that $a<c<d<b$. Hence it would be continuous at $[c,d]$. So $f$ is also continuous at $(a,b)$. Am I right? – RFZ Oct 02 '15 at 10:53
  • @RFZ any $t<u_1<b$ and $a<u_2<s$ are OK. – Ziyuan Oct 02 '15 at 13:03

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