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Let $f:mathbb{R} to mathbb{R}$ and $g: mathbb{R} to mathbb{R}$ be two strictly convex function on an open interval $(a,b)$. Assume $f$ and $g$ are not identical. Can we find an upper bound on how many times these two functions can intersect on $(a,b)$?

We can assume that both functions are analytic on $(a,b)$. This assumption makes the problem non-trivial.

There is a related question here, but it is about $mathbb{R}$ and not $(a,b)$.

Edit: Furthermore, can an assumption that one of the function is strictly increasing help?

An uncountably infinite number of times.

Consider the interval $[-1,1]$. Let $f(x)=x^2$.

And let
$$g(x)=left{begin{eqnarray} x^2 & text{ if } x < 0 \ 2x^2 & text{ otherwise }end{eqnarray}right..$$

These functions intersect an uncountably infinite number of times on the interval $[-1,0]$.

Restricting the example given in the answer in the related question to some interval $(a,b)$ that intersects $(-1,1)$ but is not contained in it. For any other $(a,b)$, just rescale and shift (which doesn't break analycity).

However, adding the condition of analycity does, indeed, change the answer (assuming that you're requiring $f$ and $g$ to not be the same function: as analytic continuations are unique, if $f = g$ on some subinterval $(c,d)$, then $f = g$ everywhere (cf here). Thus, this counterexample can't work. Further, it's not hard to see that this is the only way that we can have uncountably many points of intersection.

[NOTE: It's very possible there's something wrong at the end of the following paragraph. I'd be much obliged if someone who knows more than me about real analytic functions could check it and let me know how to fix it: I'm keeping this here because the rest of the answer is correct, and I've got a pretty strong impression that there is some way to fix this bit]

You also can't have countably infinitely many points where $f$ and $g$ coincide in a bounded interval $(a,b)$: if you do, then there's some convergent sequence $(a_n)to c$ of such point (as any enumeration of the points gives a bounded sequence, which has a convergent subsequence). Either every such $(a_n)$ is eventually constant, or not. In the former case, there must be some $delta > 0$ such that for all points $xneq y$ such that $f(x) = g(x)$ and $f(y) = g(y)$, we have $|x - y| > delta$, but there can't be infinitely many such points in a bounded interval, a contradiction. Thus, we have some convergent subsequence that is not eventually constant, hence ${x in (a,b) | f(x) = g(x)}$ has some accumulation points. [NOTE: This is where I went wrong in my initial attempt at writing this]. I feel like there's now a way to conclude that $f = g$, but I'm not sure how exactly.

You can still have any finite number that you like, though: take $f(x) = x^2 + sin(x)$ and $g(x) = x^2 - sin(x)$. Then both $f$ and $g$ are analytic and convex, but $f(x) = g(x)$ whenever $x = npi$ for any integer $n$. Thus, $(0,(n+1)pi)$ contains $n$ points at which $f = g$, and this can be mapped by something that doesn't break convexity or analyticy to any $(a,b)$.