MHB Show $\|.\|$ is an Norm: Prove Triangle Inequality

[Julio1]

If $\|(x,y)\|=\sqrt{x^2+4y^2}.$ Show that $\|.\|$ is an norm.

Hello MHB! :)

The cases $\|(x,y)\|\ge 0$ and $\|\lambda (x,y)\|=|\lambda|\|(x,y)\|$ are obvious, but the case $\|(x,y)+(x',y')\|\le \|(x,y)\|+\|(x',y')\|$ I don't know how it do? Help me!

 

[Nono713]

Hint: you know that $\sqrt{x^2+y^2}$ is a norm, being the Euclidean metric. Proceed by using the relation below:
$$\sqrt{x^2+(2y)^2}=\sqrt{x^2+4y^2} = \|(x,y)\|$$

 

[Chris L T521]

If $\|(x,y)\|=\sqrt{x^2+4y^2}.$ Show that $\|.\|$ is an norm.

Hello MHB! :)

The cases $\|(x,y)\|\ge 0$ and $\|\lambda (x,y)\|=|\lambda|\|(x,y)\|$ are obvious, but the case $\|(x,y)+(x',y')\|\le \|(x,y)\|+\|(x',y')\|$ I don't know how it do? Help me!

If you can show instead that $\|(x,y)+(x^{\prime},y^{\prime})\|^{\color{red}2} \leq (\|(x,y)\|+\|(x^{\prime},y^{\prime})\|)^{\color{red}2}$, then it immediately follows that $\|(x,y)+(x^{\prime},y^{\prime})\| \leq \|(x,y)\|+\|(x^{\prime},y^{\prime})\|$. Do you know how to proceed from here?

 

[Julio1]

Hint: you know that $\sqrt{x^2+y^2}$ is a norm, being the Euclidean metric. Proceed by using the relation below:
$$\sqrt{x^2+(2y)^2}=\sqrt{x^2+4y^2} = \|(x,y)\|$$

Thanks Bacterius :)

But I don't understand which is the relation of the norm $\|.\|_2$ with the norm $\sqrt{x^2+4y^2}$ ?

If you can show instead that $\|(x,y)+(x^{\prime},y^{\prime})\|^{\color{red}2} \leq (\|(x,y)\|+\|(x^{\prime},y^{\prime})\|)^{\color{red}2}$, then it immediately follows that $\|(x,y)+(x^{\prime},y^{\prime})\| \leq \|(x,y)\|+\|(x^{\prime},y^{\prime})\|$. Do you know how to proceed from here?

Thanks Cris L T521.

Emm?, this it's good?

$\|(x,y)+(x',y')\|^2=\|(x+x',y+y')\|^2=\langle (x+x',y+y')\rangle =\langle (x,y)\rangle +\langle (x',y')\rangle +\langle (x',y)\rangle +\langle (x,y')\rangle=?$PD.: Sorry, my English is bad. :(

 

[Nono713]

Thanks Bacterius :)

But I don't understand which is the relation of the norm $\|.\|_2$ with the norm $\sqrt{x^2+4y^2}$ ?

Well, from the Euclidean metric we get that for all $(x, y)$ and $(x', y')$ we have:

$$\sqrt{(x + x')^2 + (y + y')^2} \leq \sqrt{x^2 + y^2} + \sqrt{x'^2 + y'^2}$$

So by direct substitution of $(u, 2v) \to (x, y)$ and $(u', 2v') \to (x', y')$ we have:

$$\sqrt{(u + u')^2 + (2v + 2v')^2} \leq \sqrt{u^2 + (2v)^2} + \sqrt{u'^2 + (2v')^2}$$

$$\sqrt{(u + u')^2 + (2(v + v'))^2} \leq \sqrt{u^2 + (2v)^2} + \sqrt{u'^2 + (2v')^2}$$

$$\sqrt{(u + u')^2 + 4(v + v')^2} \leq \sqrt{u^2 + 4v^2} + \sqrt{u'^2 + 4v'^2}$$

Which is what you wanted. Does that make sense?

 

[Chris L T521]

Thanks Cris L T521.

Emm?, this it's good?

$\|(x,y)+(x',y')\|^2=\|(x+x',y+y')\|^2=\langle (x+x',y+y')\rangle =\langle (x,y)\rangle +\langle (x',y')\rangle +\langle (x',y)\rangle +\langle (x,y')\rangle=?$PD.: Sorry, my English is bad. :(

Hm, it's not good. :-/

I would start off like this:

\[\begin{aligned}\|(x,y) + (x^{\prime},y^{\prime})\|^2 &= \langle (x,y) + (x^{\prime},y^{\prime}),(x,y) + (x^{\prime},y^{\prime})\rangle \\ &= \langle (x,y),(x,y)\rangle + \langle (x,y),(x^{\prime},y^{\prime})\rangle + \langle (x^{\prime},y^{\prime}), (x,y)\rangle + \langle (x^{\prime},y^{\prime}),(x^{\prime},y^{\prime})\rangle \\ &= \langle (x,y),(x,y)\rangle + 2 \langle (x,y),(x^{\prime},y^{\prime})\rangle + \langle (x^{\prime},y^{\prime}),(x^{\prime},y^{\prime})\rangle \\&= \|(x,y)\|^2 + 2\langle (x,y),(x^{\prime},y^{\prime})\rangle + \|(x^{\prime},y^{\prime})\|^2 \end{aligned}\]

To get the desired result, you'll need to recall that $\langle (x,y),(x^{\prime},y^{\prime})\rangle \leq \|(x,y)\|\|(x^{\prime},y^{\prime})\|$ by the Cauchy-Schwarz inequality.

Do you think you can wrap things up from here? (Bigsmile)