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Show that $\rho(x)=\left( \sum_{i=1}^n|x_i|^{1/2} \right)^2$ is not a norm

It clearly satisfies the first two properties, that it be nonnegative and $\rho(ax)=|a|\rho(x)$. So it must not satisfy the triangle inequality: $\rho(x+y)\le \rho(x)+\rho(y)$ must fail for some x, y.

Choose $\begin{pmatrix}x_1\\x_2\end{pmatrix},\begin{pmatrix}y_1\\y_2\end{pmatrix}$ as some vectors and plug it into the formula.

$$\begin{split}\rho(x+y)&?\rho(x)+\rho(y)\\ [(x_1+y_2)^{1/2}+(x_2+y_2)^{1/2}]^2&?[x_1^{1/2}+x_2^{1/2}]^2+[y_1^{1/2}+y_2^{1/2}]^2\\ x_1+y_1+x_2+y_2+2\sqrt{(x_1+y_1)(x_2+y_2)}&?x_1+x_2+2\sqrt{x_1x_2}+y_1+y_2+2\sqrt{y_1+y_2}\end{split}$$

Compare the parts that are different:

$$\begin{split}2\sqrt{(x_1+y_1)(x_2+y_2)}&?2\sqrt{x_1x_2}+2\sqrt{y_1y_2}\\ 2\sqrt{x_1x_2+x_1y_2+y_1x_2+y_1y_2}&?2(\sqrt{x_1x_2}+\sqrt{y_1y_2})\\ x_1x_2+x_1y_2+y_1x_2+y_1y_2&?x_1x_2+y_1y_2+2\sqrt{x_1x_2y_1y_2}\\ x_1y_2+y_1x_2&?2\sqrt{x_1x_2y_1y_2}\\ x_1^2y_2^2+y_1^2x_2^2+2x_1y_1x_2y_2&?4x_1x_2y_1y_2\end{split}$$

If $(x_1,x_2)=(1,0)$ and $(y_1,y_2)=(0,1)$ we get $1$ on the LHS and $0$ on the RHS thus the triangle equality fails and this is not a norm because we have found a counterexample.

Vons
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  • What is your question? – Martin R Jun 18 '22 at 19:44
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    You really don't need to do all that work just compute directly $\rho((1,1)^T), \rho((1,0)^T)$ and $\rho((0,1)^T).$ – Thomas Andrews Jun 18 '22 at 19:45
  • See also https://math.stackexchange.com/q/102385/42969 – Martin R Jun 18 '22 at 19:46
  • I wanted to verify the solution and am glad I did because of Thomas's amazing insight for a noob like me. Also, thanks for the link. – Vons Jun 18 '22 at 19:52
  • If you know the first two properties, then the triangle inequality is equivalent to the statement that ${x\mid \rho(x)<1}$ is convex. That givens an intuitively look at a function $\rho$ as a norm. You can look at a shape of the ${x\mid \rho(x)=1}.$ – Thomas Andrews Jun 18 '22 at 19:53

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