Splitting Function Across Sum With Inequality Given Convex Function
Theorem
If \(f : X \to \mathbb{R}\) is a convex function where \(X\) as a convex subset of \(\mathbb{R}\) and \(f(0) \leq 0\) then for all \(a, b \in X\)
\[ f(a + b) \geq f(a) + f(b).\]
Proof
Let \(f : X \to \mathbb{R}\) be a convex function as above. Then consider for any \(a, b \in X\) that because \(f\) is convex,
\[ \frac{a}{a + b} f(0) + \frac{b}{a + b}f(a + b) \geq f\left(\frac{a}{a + b}(0) + \frac{b}{a + b}(a + b)\right) = f(b)\]
given that
\[ \frac{a}{a + b} + \frac{b}{a + b} = \frac{a + b}{a + b} = 1.\]
Because \(f(0) \leq 0\), we then have the inequality
\[ \frac{b}{a + b}f(a + b) \geq f(b).\]
Interchanging the role of \(a\) and \(b\) because of symmetry we get
\[ \frac{a}{a + b}f(a + b) \geq f(a)\]
and then adding these two inequalities we acquire
\[ f(a + b) = \frac{a}{a + b}f(a + b) + \frac{b}{a + b}f(a + b) \geq f(a) + f(b).\]
Theorem
If \(f : X \to \mathbb{R}\) is a concave function where \(X\) is a convex subset of \(\mathbb{R}\) and \(f(0) \geq 0\) then for all \(a, b \in X\)
\[ f(a + b) \leq f(a) + f(b).\]