Prove that convex real functions on [0,1] form a meet-semilattice Let C be the set of all continuous strictly convex real valued functions on the interval [0,1]. For f,g $\in$ C, define f $\leq$ g IFF f(x) $\leq$ g(x) for all x $\in$ [0,1]. Prove this is a meet-semmilattice, but not a join-semilattice.
My intuition at first was that this is meet-semilattice because a linear function, which is convex and concave, would be the GLB and there is not LUB for all subsets because I could define an infinite amount of functions to have a greater slope, so there would be a lattice with a linear function on the bottom with an infinite amount of functions above it growing infinitely upward. but the problem calls for strictly convex functions, so I'm not sure that linear functions are allowed. I don't really know how to approach this or even if I'm thinking about it correctly.
This is an exercise from Gratzer's Lattice Theory: First Concepts and Distributive Lattices in the first chapter. I'm working through it on my own, it's not a homework assignment or anything. If it wouldn't be too much of a burden, I'd rather someone gently point me in the right direction to think about the problem rather than outright solve it for me. 
Thank you!
 A: I think it is precisely the opposite, as it must follow from the reasoning below.
I'm using the Wikipedia definition of strictly convex function, that is, $f : X \to \mathbb R$ is strictly convex if, whenever $x_1, x_2 \in X$ are such that $x_1 \neq x_2$, and $0 < t < 1$, then
$$f(tx_1 + (1-t)x_2) < tf(x_1) + (1-t) f(x_2).$$
Now suppose $f,g : [0,1] \to [0,1]$ are strictly convex, and define $h:[0,1]\to[0,1]$ by making 
$$h(x) = \max\{f(x),g(x)\}.$$
For any $t \in (0,1)$,
\begin{align}
h(tx_1 + (1-t) x_2)
&= \max\{ f(tx_1 + (1-t) x_2), g(tx_1 + (1-t) x_2) \}\\
&< \max\{ tf(x_1)+(1-t)f(x_2), tg(x_1)+(1-t)g(x_2) \}\\
&\leq th(x_1) + (1-t) h(x_2),
\end{align}
so $h$ is strictly convex. 
So just define $f \vee g = h$, above.
On the other hand, $f(x) = x^2$ and $g(x) = (x-1)^2$ are both strictly convex, but their meet is not (take $x_1 = 0$, $x_2 = 1$ and $t=1/2$).

Added.
Notice that in the example above, with $f(x)=x^2$ and $g(x)=(x-1)^2$, there isn't even any strictly convex function in $C$ below $f$ and $g$, since if $h \in C$ is such that $h \leq f$ and $h \leq g$, then $h(0) \leq f(0) = 0$ and $h(1) \leq g(1) = 0$, whence $h(0)=h(1)=0$.
The only convex function $h$ in these conditions is the constant $h(x)=0$, which clearly is not strictly convex, and so doesn't belong to $C$.
Added 2.
After Rob Arthan's comment, I noticed that $C$ is the set of strictly convex functions from $[0,1]$ to $\mathbb R$, not just from $[0,1]$ to $[0,1]$, which makes the claim in the previous paragraph wrong, as is points out in his comment.
Here's an easy fix.
If $h \in C$ is such that $h \leq f, g$, then $h(0), h(1) \leq 0$, and since $h$ is strictly convex, $h(x)<0$, for $0<x<1$; define $h_1$ by making $h_1(x) = h(x)/2$.
It is an easy task to show that $h_1 \in C$, but $h < h_1$.
So there is no greatest lower bound of $f$ and $g$.
