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While studying for my optimization class, I came across the following problem:

Problem: Prove that if a function $f$ defined on an open interval $I \subseteq \mathbb{R}$ is convex, then $f$ is continuous. Must a convex function on an arbitrary interval be continuous?

I have seen proofs regarding this fact before, like this one: Is every convex function on an open interval continuous? which is fine and I was able to grasp.

However, the book I am using ("A Problem book in Real Analysis"), gives a slightly different proof containing an inequality which is bothering me a lot:

Proof:

Let $I=(a, b)$ be an open interval and assume $f$ is convex on $(a, b)$. Let $c \in(a, b)$. Suppose $a<s<c<d<t<b$. From the geometric interpretation of convexity, we know that the point $(c, f(c))$ lies below the line through the points $(s, f(s))$ and $(d, f(d))$. That is, $$ f(c) \leq f(s)+\frac{f(d)-f(s)}{d-s}(c-s) . $$ Now consider the line through the points $(c, f(c))$ and $(t, f(t))$. Again from convexity we have $$ f(d) \leq f(c)+\frac{f(t)-f(c)}{t-c}(d-c) . $$

The author continues and states that:

From the above inequalities we can obtain $$ f(s)+\frac{f(c)-f(s)}{c-s}(d-s) \leq f(d) \leq f(c)+\frac{f(t)-f(c)}{t-c}(d-c) . $$

Which I just cannot see why the LHS should be true, that is, why $$f(s)+\frac{f(c)-f(s)}{c-s}(d-s) \leq f(d)$$ holds.

Attempt:

Apparently, one should be able to conclude the above inequality from the previous two, as the authors says:

From the above inequalities one can obtain...

Having that in mind, I tried summing the inequalities, but ended up with $$ f(d) \leqslant f(s)+\frac{f(d)-f(s)}{d-s} \cdot(c-s)+\frac{f(t)-f(c)}{t-c} \cdot(d-c) $$

which is no good. I also tried to reverse engineer the inequality, but it did not go so well. I also tried using the geometric interpretation of convexity, but the endpoints in the LHS inequality do not seem to agree.

Can someone help?

Thanks in advance, Lucas

Lucas
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    Work with just the first inequality ($f(c) \leq \cdots$). Subtract $f(s)$ from both sides of the inequality, then multiply both sides by $(d-s)/(c-s)$, then add $f(s)$ to both sides. –  May 21 '21 at 22:55
  • Omg, that was super nice, how were you able to figure it out? – Lucas May 21 '21 at 23:01
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    Note that $1-{c-s\over d-s} = {d -c \over d-s}$. – copper.hat May 21 '21 at 23:06
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    Just asked myself a few questions: how can I get $f(c) - f(s)$ on one side, how can I get the ratio $(c-s)/(d-s)$, etc. That made it pretty straightforward. Sometimes you see the right manipulations immediately and other times you don't. In the latter case, try starting with small steps that get you closer to the desired form, and maybe with luck after a few such steps you get there. Noticing that the second inequality didn't have any of the stuff I needed led me to focus on the first one. –  May 21 '21 at 23:08

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