Task
Show that $$\rho(x) = (\sqrt{|\xi_{1}|}+\sqrt{|\xi_{2}|})^2$$ does not define a norm on the vector space of all ordered pairs $x = (\xi_1> \xi_2)$, ... of real numbers
Attempt
First lets say that $X$ is vector space
My Idea is that we could have convex set $$M = \{x \in X | \lVert x\rVert \leq 1 \} \subset A \subset X$$
We can show that A is a convex set by simply
Let $x_{1}, x_{2} \in B(0;1)$ $$\lVert \lambda x_{1} + (1-\lambda) x_{2} \rVert \leq \lvert \lambda \rvert \lVert x_{1} \rVert + \lvert 1-\lambda \rvert \lVert x_{2} \rVert$$
And from here we could check that the statement is valid for the max value $x=1$
$$\lvert \lambda \rvert \lVert x_{1} \rVert + \lvert 1-\lambda \rvert \lVert x_{2} \rVert \leq \lambda + 1-\lambda = 1$$
Question:
We prove that the statement is correct because it's valid even in it's limit conditions?
Real task
We could assign $$P = \{ x \in X | \rho (x) = 1 \} \subset B(0;1)$$
Now we have a subspace of a norm convex vector space, we should disapprove some of it's norm axioms, the 1st 2 are obviously true:
$$\lVert x \rVert \geq 0$$ $$\lVert x \rVert = 0 \iff x = (0,0)$$
Got stuck in
Maybe, I have to contradict something in the other 2 axioms but don't see how. It's obvious that $P$ is not a convex set.
Question: Can you explain me how could I prove this statement?
Answer:
Let $\alpha=0.5$
$$\lVert (1,0)\rVert=\lVert (0,-1)\rVert=1$$
$$∥λx_1+(1−λ)x_2∥= \lvert 0.5 \rvert \lVert [(1, 0) + (0, -1)] \rVert = 0.5 \lVert [(1, -1)] \rVert = 2 ≤|λ|∥x_1∥+|1−λ|∥x_2∥ = 1$$
Note that the unit ball defined by $\rho$ is not convex. Indeed $\rho((1,0)) = \rho((0,-1)) =1$, but $$\rho(\frac 12 (1,0)+\frac 12 (0,-1)) = 2>1$$ Hence $\rho$ is not a norm.