The intersections of a convex function and y=x Suppose $f(x)$ is a convex function on $\mathbb{R}$, with $f(0)\ge0$, $f(1)=1$ and $f'(1)\gt1$. Then, do we have a unique point $x$ on $[0,1)$ such that $f(x)=x$? 
My Solution
Firstly, let $g(x):=f(x)-x$. Based on the properties of $f(x)$ mentioned above, we also have $g(0)\ge0$, $g(1)=0$ and $g'(1)\gt0$. 
Since $f(x)$ is convex, then the first order derivative $f'(x)$ should be continuous and monotonically increasing. This indicates that there is an interval $[a, b]\subset[0,1)$ with $f'(x)\ge1$ for any $x\in[a, b]$. Then $g'(x)\ge0$ on $[a, b]$  and hence $g(x)$ is increasing on $[a, b]$, which also equivalently suggests that $g(x)\lt0$ on $[a,b]$ since $g(1)=0$ and $b\lt1$.
Finally, because $g(0)\ge0$ and $g(x)\lt0$ for $x\in[a,b]\subset[0,1)$, there must be a point $\alpha$ on $[0,1)$ such that $g(\alpha)= 0$ and hence $f(\alpha)=\alpha$.
I looked at some graphs of convex functions and I guess my solution is somewhat intuitive. Is it possible to put the proof into a more rigorous framework?
 A: Here's another way boiling down to a simple fact (or definition if you like):

A function $f$ is convex if and only if the set of points above the curve $y = f(x)$ is a convex set. In particular, for any pair of points $a < b$, the line segment from $(a, f(a))$ to $(b, f(b))$ lies on or above the curve $y = f(x)$.

Now let's suppose that $f(a) = a$ and $f(b) = b$ where $a < b < 1$. Then I claim that $f(x) = x$ for all $x \in (a, 1)$. We know that the line segment $(a,a)$ to $(1,1)$ lies on or above the curve $y = f(x)$, so we know that $f(x) \le x$ for all $x \in (a,1)$.
Suppose that $f(x) < x$ for some $x \in (a,1)$. First, if $x < b$ then we look at the line segment from $(x, f(x))$ to $(1,1)$. But notice that the point $(b,b)$ lies on the curve $y = f(x)$ and is also above the line segment$^\dagger$. If $x > b$ we do the same with the line segment from $(a,a)$ to $(x, f(x))$.
$^\dagger$This is intuitively obvious. If we were to show this rigorously, we first use the fact that $f(x) < x$ to show that the slope $m$ of the line segment is $> 1$. So the line segment is $y = 1 - m(1 - x)$ and therefore $$ y < 1 - (1 - x) = x$$ for all $x < 1$.
Now if $f(x) = x$ for all $x \in (a, 1)$, and $f$ is differentiable at $1$, we must have
$$ f'(1) = \lim_{x \nearrow 1} \frac{f(x) - f(1)}{x - 1} = 1,$$
which is a contradiction.
This shows that there can be at most one point. To show that there is at least one, note that if $f(x) \ge x$ for all $x \in (0,1)$ then
$$ f'(1) = \lim_{x \nearrow 1} \frac{f(x) - f(1)}{x - 1} \le \lim_{x \nearrow 1} \frac{x - f(1)}{x - 1} = 1. $$
So there must be some $x \in (0,1)$ with $f(x) < x$. Now we can use the intermediate value theorem.
