I wish to prove that if $f$ is a convex function on $(a,b)$ with $-\infty \le a < b \le +\infty$, then $f$ is the sup of all affine functions majorised by $f$.
In order to prove it, I tried to prove first that
For every $a$ there esists an affine function $g_a$ such that:
- $g_a(a) = f(a)$
- for all $x$ in $(a,b)$, $g_a(x) \le f(x)$. For this, I used the definition of convexity, fixing two numbers $x$ and $y$ and writing $$ f(ty + (1-t)x) \le tf(x) +(1-t)f(y) = t[f(y) - f(x)] + f(x)$$ to deduce that $$ \frac{f(ty + (1-t)x) - f(x)}{t} \le f(y) - f(x) $$ And then, setting $ty + (1-t)x = t(y-x)+x = x + h$, $$ \frac{f(x+h) - f(x)}{h} \le \frac{f(y) - f(x)}{y - x} $$ But then I get stuck, because the LHS of the previous inequality not necessarily converges to the derivative of $f$, for $f$ is not necessarily differentiable.
Is there any more efficient way to prove this theorem?
The result cannot hold. Suppose $f$ is defined on the reals.
If $a,b$ are affine functions and $a \le b$ then we must have $b-a$ is a constant since the only affine function that is bounded above is a constant.
Hence if $\phi_n$ is an increasing sequence of affine functions, we must have $\phi_n = \phi_1+c_n$ where $c_n $ is a non decreasing sequence.
Then we have $f = \phi_1 + \lim_n c_n$ which is either affine or $+\infty$.