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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:

\begin{equation} \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} \end{equation}

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.

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

2 Answers 2

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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 $y=-x$.

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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

$-||a.x+(1-a).y||^2 \geq a.(-||x||^2) + (1-a).(-||y||^2)$

Now, think about the equality. There are a few cases where the equality occurs. Thus, there are many counterexamples as shown in the comments.

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