# Norms Abstract Analysis

I have a question relating to norms and have been giving functions and need to state whether they are norms or not...

which of the following are norms on $\mathbb{R}^2$? Give reasons for your answers. For $x=(x_1,x_2)$ let

1. $G_2 (x) = 7|x_1| + 3|x_2|$
2. $G_3 (x) = max\left\{|x_1|^2, |x_2|^2\right\}$

I know the criteria for it to be a norm...

$||x|| \geq 0$

$||x|| =0$ if $x =0$

$||\lambda x|| = |\lambda|\cdot||x||$

$||x+y||\leq ||x|| + ||y||$

I just cant work out how to use this to prove whether or not they're norms.

-
I did not know this sort of things are called abstract analysis :) – Hui Yu Feb 20 '13 at 15:52

## 2 Answers

To check whether a function is a valid norm, check the conditions for being a norm one by one. Construct a general case for each of them, and either prove that the condition is met or provide a counterexample. You can use $x = (x_1, x_2)$ and $y = (y_1, y_2)$ when two vectors are called for (and $\lambda$ is just $\lambda$, of course).

For $G_2(x)$:

$\|x\| = 7|x_1| + 3|x_2| = |7x_1| + |3x_2|$; the first condition is met, since the sum of two absolute values has to be nonnegative. The second is also met: if $x_1$ or $x_2$ is nonzero, its absolute value is positive, and thus the sum must also be positive.

Constructing a general case for the third property goes like this:

$\| \lambda x \| = \| (\lambda x_1, \lambda x_2) \| = 7|\lambda x_1| + 3|\lambda x_2| = 7|\lambda||x_1| + 3|\lambda||x_1| = |\lambda|(7|x_1| + 3|x_2|) = |\lambda| \|x\|.$

for the fourth one, the following holds: $\|x+y\| = \|(x_1 + y_1, x_2 + y_2)\| = 7|x_1 + y_1| + 3|x_2 + y_2|$, and since $|x_1 + x_2| \leq |x_1| + |x_2|$ (triangle inequality for real numbers), we have $\|x+y\| = 7|x_1 + y_1| + 3|x_2 + y_2| \leq 7(|x_1| + |y_1|) + 3(|x_2| + |y_2|) = (7|x_1| + 3|x_2|) + (7|y_1| + 3|y_2|) = \|x\| + \|y\|.$

For $G_3(x)$:

$\|\lambda x\| = \max{|\lambda x_1|^2, |\lambda x_2|^2} = \max{|\lambda|^2 |x_1|^2, |\lambda|^2 |x_2|^2} = |\lambda|^2 \max{|x_1|^2, |x_2|^2} = |\lambda|^2\|x\| \neq |\lambda| \|x\|$, so $G_3$ can't be a norm.

(You can check that the first and second conditions hold for $G_3$ if you wish, though it's not strictly necessary.)

-
How can I then use this for the next part of the question... Let w: [a,b] $\rightarrow$ $\mathbb{R}$ with w(x) $\geq$ c $\geq$ for some c $\in$ $\mathbb{R}^+$ and all x $\in$ [a,b]. Prove that ||f||$_w$ = $\displaystyle\int^b_a |f(t)|w(t)\ dt$ is a norm on C([a,b]) – Mathsstudent147 Feb 20 '13 at 15:36
Do you mean $0 < w(x) \leq c$? – alcedine Feb 20 '13 at 16:09
Otherwise the first two don't hold in the general case. As for the third and fourth, try substituting $\lambda f$ or $f + g$ in place of $f$ in the integral and see what you get. Remember that integration is linear. – alcedine Feb 20 '13 at 16:15

For $G_2(x)$, clearly it is nonnegative, and as each term is nonnegative, we know that it can be zero if and only if $x_1=x_2=0$, hence we have established the first property of norms. Now, let $\lambda\in\Bbb R$, then $$G_2(\lambda x)=7|\lambda x|+3|\lambda x|=|\lambda|\big(7|x_1|+3|x_2|\big)=|\lambda|G_2(x).$$ Now, for $x,y\in\Bbb R^2$, from the definition of $G_2(x)$ and the triangle inequality for $|\cdot |$ of the real numbers, it is easy to check that $$G_2(x+y)\leq G_2(x)+G_2(y).$$

For $G_3(x)$, however, we can check that $$G_3\big(2\dot(1,1)\big)=\max\{4,4\}=4\neq 2\max\{1,1\}=2G_3\big((1,1)\big).$$Therefore, $G_3$ cannot be a norm.

-