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I’m trying very hard to solve the following problem

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

I’m practically making no progress. Anyway here’s my approach.

My attempt: There are three equivalent definition of continuity. (1)$\epsilon - \delta$ Definition, (2) $\lim$ of $f(x)$ is equal to $f(p)$ as $x$ tends to $p$, and (3) $f^{-1}(V)$ is open in $(a,b)$ where $V$ is an open subset of $ \mathbb{R}$.

When I try to use first definition. The most natural approach is to start from $|f(p)-f(x)| \leq$ some expression of $x$. So that $|f(p)-f(x)| \lt \epsilon$, for some $x$. Since there is not much information about $f$ given in the hypothesis of problem, to simply $|f(p)-f(x)|$, other than $f$ is convex. We have to somehow use the fact, that $f$ is convex. Since, $p,x \in (a,b)$, we have $f(\lambda p + (1-\lambda)x) \leq \lambda f(p) + (1-\lambda)f(x)= \lambda (f(p)-f(x)) + f(x)$. So $\frac{f(\lambda p + (1-\lambda)x) - f(x)} {\lambda} \leq f(p)-f(x)$. Now, I’m stuck.

When I try to use third definition. I can’t find a relation between $p$ and $f(p)$. If I could, then maybe I could construct an open ball around $p \in f^{-1}(V)$ of some radius(depends on the radius of ball around $f(p)$) that contains in $f^{-1}(V)$.

Question: I don’t want to look at the solution straight away. I want you to give multiple hints, like hint 1, hint 2 .... hint n, with each hint one is more likely to solve the problem. You are essential telling in which direction to go.

There are already solution of this problem exist on SE website. Link: Prob. 23, Chap. 4 in Baby Rudin: Every convex function is continuous and every increasing convex function of a convex function is convex . I’m little bit suspicious of his proof. Though I haven’t reviewed his proof. I have skim through it. His proof seems to be using the “exact” same notation($M, \eta , c , d,$etc) as the solution posted by University of Wisconsin, link: https://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwjChtS2k6L0AhVHuksFHZOyDaIQFnoECAgQAQ&url=https%3A%2F%2Fminds.wisconsin.edu%2Fhandle%2F1793%2F67009&usg=AOvVaw37kH6RefrxdcL6MZl2u-An .

user264745
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  • When I tried to solve this problem, I used the second definition. But I could only show that $\limsup f(x_n)\leq f(x)$. If you want to use this approach, you should at first fix an $x$, then choose $p,q$ with $p<x<q$, and finally choose any "one-sided" sequence in $(p,x)$ and consider the relation between $p,x_n,x$ and $x_n,x,q$. –  Nov 19 '21 at 02:18
  • @user912011 I will definitely try this approach. – user264745 Nov 19 '21 at 15:31

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