Problem

Show f(x) = $\frac{1}{2}x^T Q x$ with Q symmetric positive definite, is a convex function on $\mathbb{R}^n$ with the formal definition of convexity:

$$f(y + (\alpha(x-y)) - \alpha f(x) - (1-\alpha)f(y) < 0$$

Solution

1. Plug in f(x) into the equation

$$(\frac{1}{2}(y + (\alpha(x-y))^T Q (y + (\alpha(x-y))) - \alpha (\frac{1}{2}x^T Q x) - (1-\alpha)(\frac{1}{2}y^T Q y) < 0$$

1. Distribute

$$(\frac{1}{2}(y^T + \alpha x^T - \alpha y^T) Q (y + \alpha x - \alpha y) - \alpha (\frac{1}{2} x^T Q x) - (1-\alpha)(\frac{1}{2}y^T Q y) < 0$$

$$(\frac{1}{2}(y^T Q + \alpha x^T Q - \alpha y^T Q) (y + \alpha x - \alpha y) - \alpha (\frac{1}{2} x^T Q x) - (1-\alpha)(\frac{1}{2}y^T Q y) < 0$$