# How can I prove a a negative norm is not convex?

I want to prove that $$f(x) = -\lVert x \rVert^2$$ is not convex.

I know that $$f(x) = \lVert x \rVert^2$$ is convex by the following proof:

$$$$\label{eq1} \begin{split} \lVert \alpha x + (1 - \alpha )y \rVert^2 & \leq \lVert \alpha x\rVert^2 + \lVert (1 - \alpha)y\rVert^2 \\ & = \alpha^2\lVert x\rVert^2 + (1 - \alpha)^2\lVert y\rVert^2 \\ & \leq \alpha \lVert x\rVert^2 + (1 - \alpha) \lVert y\rVert^2 \end{split}$$$$

How do I prove that the negative version of this is not convex? Intuitively I know that it must be concave, but I don't know how I can prove.

• In the zero space $-\|x\|^2$ is convex. If the space has an $x\neq0$, then you have that $\frac{(-\|x\|^2)+(-\|-x\|^2)}{2}=-\|x\|^2<0=\left\|\frac{x+(-x)}{2}\right\|$ Dec 15, 2019 at 0:55
• $\lVert \alpha x\rVert^2 + \lVert (1 - \alpha)y\rVert^2 = \alpha\lVert x\rVert^2 + (1 - \alpha)\lVert y\rVert^2$? I do not think so. $\lVert \alpha x\rVert^2=\alpha^2\lVert x\rVert^2$, not $\alpha\lVert x\rVert^2$. Dec 15, 2019 at 1:52
• Why not? They are just scalars. Dec 15, 2019 at 1:54

To prove $$f$$ is not convex, all you need is to find an example where $$f(\alpha x + (1-\alpha) y) > \alpha f(x) + (1-\alpha) f(y)$$ with $$0 \le \alpha \le 1$$. Try $$\alpha = 1/2$$ and $$y=-x$$.
You have $$||a.x+(1-a).y||^2 \leq a.||x||^2 + (1-a).||y||^2$$ for $$0\leq a\leq 1$$. Multiplying both sides by -1, you have
$$-||a.x+(1-a).y||^2 \geq a.(-||x||^2) + (1-a).(-||y||^2)$$