Checking if a functional $F(x)$ is a norm in $\mathbb{R}^2$

Question

There is a functional given: $$F(x) = \sqrt{2x_1^2 + 3x_2^2}$$ Of course $x\in\mathbb{R}^2 \rightarrow x = (x_1, x_2)$
It is easy to check that:
1) $\forall x \in \mathbb{R}^2$ $F(x) \ge 0$
2) $F(x) = 0 \iff x = 0$
3) $ F(\lambda x) = | \lambda | F(x)$
I failed however to show that $F(x+y) \le F(x) + F(y)$. How should it be proven?

Answer 1

You can just use the Cauchy-Schwartz inequality:

\begin{align} F(x+y)^2 &= 2(x_1+y_1)^2+3(x_2+y_2)^2 \\ &= \underbrace{2x_1^2 + 3x_2^2}_{F(x)^2} + \underbrace{2y_1^2+3y_2^2}_{F(x)^2}+2(2x_1y_1 + 3x_2y_2)\\ &\stackrel{CSB}{\le} F(x)^2 + F(y)^2 + 2\sqrt{2x_1^2+3x_2^2}\sqrt{2y_2^2+3y_2^2}\\ &= F(x)^2 + F(y)^2 + 2F(x)F(y)\\ &= (F(x) + F(y))^2 \end{align}

Answer 2

\begin{align*} (F(x)+F(y))^2 &= 2x_1^2 + 2y_1^2+3x_2^2 + 3y_2^2+2\sqrt{2x_1^2 + 3x_2^2}\sqrt{2y_1^2 + 3y_2^2} \\ F(x+y)^2 &= 2(x_1 + y_1)^2 +3(x_2 + y_2)^2 \end{align*} So you just have to verify that $$ (4x_1y_1 + 6x_2y_2)^2 \leq 4(2x_1^2 + 3x_2^2)(2y_1^2 + 3y_2^2)$$ which is equivalent after expending and simplification to $$ 0 \leq -2 x_1x_2y_1y_2 + x_1^2y_2^2+ x_2^2y_1^2 \quad \Leftrightarrow \quad 0\leq ( x_1y_2- x_2y_1)^2$$ which is always true.

Answer 3

Does your book prove that the standard Euclidean norm is an inner poduct? I would recommend repeating their proof:

Define an inner product $\langle v, w \rangle $ so that $\|v\| = \sqrt{\langle v, v\rangle}$. Show that $\langle , \rangle$ satisfies the Cauchy-Schwartz inequality. Then use that to conclude the triangle inequality. (This is what the other answer is suggesting, though more explicitly)

If you happen to be given that the Euclidean norm is actually a norm, then you might want to try reducing this norm to that norm.

Answer 4

General principle: if $f:E\longrightarrow E$ is a linear isomorphism and $\|\|$ is a norm in $E$, then $\|x\|_f = \|f(x)\|$ is a norm in $E$. Proof of subadditivity: $$ \|x + y\|_f = \|f(x + y)\| = \|f(x) + f(y)\|\le\|f(x)\| + \|f(y)\| =\|x\|_f + \|y\|_f. $$ In your case, $f(x_1,x_2) = (\sqrt2\,x_1,\sqrt3\,x_2)$.