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I want to show that the function $N: \mathbb{R} \rightarrow \mathbb{R^2}$ defined by $\sqrt{x_1^2 - x_1x_2 + 4x_2^2}$

I know that in order to do so, I have to show the 3 following properties:

  • $N(x) = 0 \iff x= 0$

  • $N(\lambda x) = |\lambda|N(x)$

  • $N(x+y) ≤ N(x) + N(y)$

I tried to show the frst property but I got stuck.

I don't know how to prove that $x_1^2 - x_1x_2 + 4x_2^2 = 0 \iff (x_1, x_2) = (0,0)$

For the third property I guess I'd just have to use the triangle inequality. But I don't know how to show the second property either.

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    For the first part, write $x_{1}^2 - x_1 x_2 + 4 x_{2}^{2} $ as $(x_1 - \frac{1}{2} x_2)^2 + \frac{15}{4}x_2^2$. For the second part, try squaring the inequality. Can you do the algebra to see what inequality one would have to prove in that case? – 2017-01-21
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    $N(\lambda x) = \sqrt{ (\lambda x_1)^2 - (\lambda x_1) (\lambda x_2) + 4 (\lambda x_2)^2}$. Play around with the expression on the right-hand side. – 2017-01-21

2 Answers 2

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One way: $$Q(x_1,x_2)=x_{1}^2 - x_1 x_2 + 4 x_{2}^{2}=\begin{pmatrix}x_1,\;x_2\end{pmatrix}\begin{pmatrix}{1}&{-1/2 }\\{-1/2 }&{4}\end{pmatrix}\begin{pmatrix}x_1\\{x_2}\end{pmatrix}$$ is a definite positive quadratic form, so $$\langle (x_1,x_2),(y_1,y_2) \rangle=\begin{pmatrix}x_1,\;x_2\end{pmatrix}\begin{pmatrix}{1}&{-1/2 }\\{-1/2 }&{4}\end{pmatrix}\begin{pmatrix}y_1\\{y_2}\end{pmatrix}$$ is an inner product. As a consequence, $$N(x_1,x_2)=\sqrt{\langle (x_1,x_2),(x_1,x_2) \rangle}$$ is a norm.

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Let $x=(a,b)$ and $y=(c,d)$.

Hence, $$N(x)+N(y)=\sqrt{a^2-ab+4b^2}+\sqrt{c^2-cd+4d^2}=$$ $$=\sqrt{\left(a-\frac{b}{2}\right)^2+\frac{15}{4}b^2}+\sqrt{\left(c-\frac{d}{2}\right)^2+\frac{15}{4}d^2}\geq\sqrt{\left(a+c-\frac{b+d}{2}\right)^2+\frac{15}{4}(b+d)^2}=$$ $$=\sqrt{(a+c)^2-(a+c)(b+d)+4(b+d)^2}=N(x+y)$$ The rest is obvious, I think.