Show bounded and convex function on ℝ is constant (pt 2) Continuing from: Show bounded and convex function on $\mathbb R$ is constant
My thoughts: Assume a line through $(a, f(a))$ and $(c,f(c))$ which must lie above (or at) $f$ for some $b\in [a,c]$
$f(c) \ge f(a) + (c-a)\frac{f(b)-f(a)}{b-a}$
If $c \to +\infty$ then
$\displaystyle \lim_{c\to +\infty} f(c) \ge \lim_{c\to +\infty} \left( f(a) + (c-a)\frac{f(b)-f(a)}{b-a} \right)$
But since $f$ is bounded, $\lim_{c\to +\infty} f(c)$ should equal some constant $Q$. So we can see that in 
$\displaystyle \lim_{c\to +\infty} f(a) + (c-a)\frac{f(b)-f(a)}{b-a}$
$(c-a)$ has to cancel somehow. My thought was to first let $c$ go to positive infinity and then $a$ to negative infinity. If then we could somehow lock $f(c)$ between two values which are equal to each other we would have shown $f$ is constant. But I am stuck, any tips?
 A: Since $f$ is convex, we have
$$
f(\,\lambda x+ (1-\lambda)y\,)\le\lambda f(x)+(1-\lambda)f(y) \;\;  \forall x, y\in\mathbb{R} ,\;\forall \lambda\in[0,1].
$$
Let $S<\infty$ be the supremum of $f$ (that is, $S=\sup\{f(x),\; x\in\mathbb{R}\}$). 
Claim: $f(x)\equiv S$.
Fix $x\in \mathbb{R}$. Given $\varepsilon >0$ there exists $z\in\mathbb{R}$ such that $f(z)+\varepsilon>S$. Then 
$$
S<\varepsilon +f(z)\le \varepsilon +\lambda f(x) + (1-\lambda)f(y)\le \varepsilon +\lambda f(x)+(1-\lambda)S
$$
where $y$ is such that $z=\lambda x+(1-\lambda)y$.
Then
$$
\lambda S <\lambda f(x) +\varepsilon
$$
Since $\varepsilon$ is choosen arbitrarly small, we conclude that $S\le f(x)$.
A: If $f$ is not constant, then you can find $ a < b $ such that $ f(a) \neq f(b)$.
If $f(b) > f(a)$, then for any $c > b$, convexity implies that $f(c) \geq f(b) + (c-b)\frac{f(b) - f(a)}{b-a} $, and so it is clear that $\underset{{c \to \infty}}{\lim} f(c) = \infty$.
Likewise, if $f(b) < f(a)$, then for any $c < a$, convexity gives $f(c) \geq f(a) + (c-a)\frac{f(a) - f(b)}{a-b}$, and so $\underset{{c \to -\infty}}{\lim} f(c) = \infty$.
