Norm in R^3

[Lambda96]

Homework Statement

For which $\lambda$ does $\Phi$ have a norm in $\mathbb{R^3}$

Relevant Equations

none

Hi,

The task is as follows

In order for it to be a norm, the three properties must be fulfilled.

1. Positive definiteness
2. Absolute homogeneity
3. Triangle inequality

$\textbf{Positive definiteness}$
Since all three elements are given in absolute value, the result of $\max{}$ will always be positive, no matter what value the $\lambda$ will have

$\textbf{Absolute homogeneity}$
$\max\{|s \cdot x|,|s \cdot y|^{\lambda}, |s \cdot z|\} = |s| \cdot \max\{|x|,|y|^{\lambda},|z|\}$

$\textbf{Triangle inequality}$
$|\max\{|x_1|,|y_1|^{\lambda},|z_1|\} +\max\{|x_2|,|y_2|^{\lambda},|z_2|\}| \leq |\max\{|x_1|,|y_1|^{\lambda},|z_1|\}| + |\max\{|x_2|,|y_2|^{\lambda},|z_2|\}|$

Now, I'm not sure, but $\lambda$ can now take on all values without the triangle inequality not being fulfilled. At least I can't think of an example where the triangle inequality is not fulfilled.

 

[fresh_42]

Homework Statement: For which $\lambda$ does $\Phi$ have a norm in $\mathbb{R^3}$
Relevant Equations: none

Hi,

The task is as follows

View attachment 348485
In order for it to be a norm, the three properties must be fulfilled.

1. Positive definiteness
2. Absolute homogeneity
3. Triangle inequality

$\textbf{Positive definiteness}$
Since all three elements are given in absolute value, the result of $\max{}$ will always be positive, no matter what value the $\lambda$ will have

$\textbf{Absolute homogeneity}$
$\max\{|s \cdot x|,|s \cdot y|^{\lambda}, |s \cdot z|\} = |s| \cdot \max\{|x|,|y|^{\lambda},|z|\}$

$\textbf{Triangle inequality}$
$|\max\{|x_1|,|y_1|^{\lambda},|z_1|\} +\max\{|x_2|,|y_2|^{\lambda},|z_2|\}| \leq |\max\{|x_1|,|y_1|^{\lambda},|z_1|\}| + |\max\{|x_2|,|y_2|^{\lambda},|z_2|\}|$

Now, I'm not sure, but $\lambda$ can now take on all values without the triangle inequality not being fulfilled. At least I can't think of an example where the triangle inequality is not fulfilled.

What if $|y|^\lambda$ is the maximum? Then $\max\{|s \cdot x|,|s \cdot y|^{\lambda}, |s \cdot z|\} \neq |s| \cdot \max\{|x|,|y|^{\lambda},|z|\}$ for $s>1.$

You also have to check that $\phi(x,y,z)=0 \Longleftrightarrow x=y=z=0.$

 

[Lambda96]

Thank you fresh_42 for your help

$\textbf{Positive definiteness}$
If $\Phi(x,y,z)=0$ the following must apply $\max\{ |0|,|0|^{\lambda} , |0| \}$ from which follows $x=y=z=0$

$\textbf{Absolute homogeneity}$
Because of $\max\{|s \cdot x|,|s \cdot y|^{\lambda}, |s \cdot z|\} \neq |s| \cdot \max\{|x|,|y|^{\lambda},|z|\}$ for $s>1.$ it follows that $\lambda =1$, because then the following is true

$\max\{|s \cdot x|,|s \cdot y|^{\lambda}, |s \cdot z|\} = |s| \cdot \max\{|x|,|y|^{\lambda},|z|\}$ with $\lambda=1$

 

[fresh_42]

Thank you fresh_42 for your help

$\textbf{Positive definiteness}$
If $\Phi(x,y,z)=0$ the following must apply $\max\{ |0|,|0|^{\lambda} , |0| \}$ from which follows $x=y=z=0$

$\textbf{Absolute homogeneity}$
Because of $\max\{|s \cdot x|,|s \cdot y|^{\lambda}, |s \cdot z|\} \neq |s| \cdot \max\{|x|,|y|^{\lambda},|z|\}$ for $s>1.$ it follows that $\lambda =1$, because then the following is true

$\max\{|s \cdot x|,|s \cdot y|^{\lambda}, |s \cdot z|\} = |s| \cdot \max\{|x|,|y|^{\lambda},|z|\}$ with $\lambda=1$

Yes, absolute homogeneity requires $\lambda =1.$ Otherwise we would have $|sy|^\lambda=|s|^\lambda |y|^\lambda.$

But we have dropped the factor $2$ in the first component. What about this factor?

And we still need the triangle inequality:
\begin{align*}
\phi(x+x',y+y',z+z')&=\max\{2|x+x'|,|y+y'|,|z+z'|\}\\
&\leq \max\{2|x|,|y|,|z|\}+\max\{2|x'|,|y'|,|z'|\} \tag{1}\\
&=\phi(x,y,z) +\phi(x',y,'z')
\end{align*}
Why does the inequality $(1)$ hold? The triangle inequality holds in each of the three components. But could it be that the maximum switches from one component to another, e.g. because of the factor two?

 

[Lambda96]

Thank you fresh_42 for your help

Regarding the 2 in the absolute homogeneity, I would say that this has no influence, because the 2 acts only on one of the components and you could write the following $2|x|=|a|$ from which follows

$\max\{|s \cdot a|,|s \cdot y|, |s \cdot z|\} = |s| \cdot \max\{|a|,|y|,|z|\}$


Regarding the 2 in the triangle inequality. Let's say that $x$ and $x'$ are the maximum. Then if both terms are positive, both terms would be equal, so $2|x+x'|=2|x|+2|x'|$, but if one of the two terms were negative, the following would apply $2|x+x'|<2|x|+2|x'|$

The triangle inequality would therefore still hold, even with the factor 2

 

[fresh_42]

Thank you fresh_42 for your help

Regarding the 2 in the absolute homogeneity, I would say that this has no influence, because the 2 acts only on one of the components and you could write the following $2|x|=|a|$ from which follows

$\max\{|s \cdot a|,|s \cdot y|, |s \cdot z|\} = |s| \cdot \max\{|a|,|y|,|z|\}$


Regarding the 2 in the triangle inequality. Let's say that $x$ and $x'$ are the maximum. Then if both terms are positive, both terms would be equal, so $2|x+x'|=2|x|+2|x'|$, but if one of the two terms were negative, the following would apply $2|x+x'|<2|x|+2|x'|$

The triangle inequality would therefore still hold, even with the factor 2

Yes, correct. But you have to rule out the possibility that the maximum in the second component on the left cannot become the maximum in the first component on the right or vice versa. The triangle inequality holds component-wise, yes, but the maxima could possibly switch among the components. You need an argument that this cannot happen. It is no question between $y$ and $z$, so it is sufficient to consider the cases where $\max\{2|x+x'|,|y+y'|\}=2|x+x'|$ and $\max\{2|x+x'|,|y+y'|\}=|y+y'|.$ The right-hand sides $\max\{2|x|,|y|\}+\max\{2|x'|,|y'|\}$ could be messy. How can you rule out for instance
$$
\max\{2|x+x'|,|y+y'|\}=2|x+x'| \text{ and } \max\{2|x|,|y|\}+\max\{2|x'|,|y'|\}=|y|+2|x'|\;?
$$
Or if it cannot be ruled out, why $\leq$ still holds despite of it?

 

[pasmith]

Having shown that $\lambda =1$ (which you easily can do by restricting attention to the subspace $x =z= 0$), all you need do is prove that:

(1) If $T : V \to V$ is an injective linear map and $f$ is a norm on $V$, then $f \circ T$ is a norm on $V$.
(2) $T: \mathbb{R}^3 \to \mathbb{R}^3: (x,y,z) \mapsto (2x,y,z)$ is an injective linear map.
(3) $f: \mathbb{R}^3 \to [0, \infty) : (x,y,z) \mapsto \max \{ |x| ,|y|, |z|\}$ is a norm on $\mathbb{R}^3$.

EDIT: It may also help to regard $\mathbb{R}^3 = (\mathbb{R} \oplus \mathbb{R}) \oplus \mathbb{R}$ and use the following proposition twice:

If $V_1$ and $V_2$ are normed spaces with norms $\|\cdot\|_1$ and $\|\cdot\|_2$ respectively, then $\|(x,y)\| = \max(\|x\|_1,\|y\|_2)$ is a norm on $V = V_1 \oplus V_2$.