MHB R^n as a normed space .... D&K Lemma 1.1.7 .... .... some inequalities ....

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 1: Continuity ... ...

I need help with an aspect of Lemma 1,1,7 (iv) ...

Duistermaat and Kolk"s Lemma 1.1.7 reads as follows:
View attachment 7877
https://www.physicsforums.com/attachments/7878

At the start of the proof of (iv) we read the following:

" ... ... $$\mid x_j \mid \le \left( \sum_{ 1 \le j \le n } \mid x_j \mid^2 \right)^{ \frac{ 1 }{ 2 } } = \| x \|$$ ... ... ... "
Suppose now we want to show, formally and rigorously that $$\mid x_j \mid \le \left( \sum_{ 1 \le j \le n } \mid x_j \mid^2 \right)^{ \frac{ 1 }{ 2 } }$$Maybe we could start with (obviously true ...)

$$\mid x_j \mid = ( \mid x_j \mid^2 )^{ \frac{ 1 }{ 2 } }$$ ... ... ... ... ... (1)

then we can write

$$\mid x_j \mid = ( \mid x_j \mid^2 )^{ \frac{ 1 }{ 2 }} \le ( \mid x_1 \mid^2 + \mid x_2 \mid^2 + \ ... \ ... \ + \mid x_j \mid^2 + \ ... \ ... \ + \mid x_n \mid^2 )^{ \frac{ 1 }{ 2 } } $$and we note that$$( \mid x_1 \mid^2 + \mid x_2 \mid^2 + \ ... \ ... \ + \mid x_j \mid^2 + \ ... \ ... \ + \mid x_n \mid^2 )^{ \frac{ 1 }{ 2 } } = ( x_1^2 + x_2^2 + \ ... \ ... \ + x_j^2 + \ ... \ ... \ + x_n^2 )^{ \frac{ 1 }{ 2 } } = \| x \|$$ ... ... ... ... (3) ... BUT ... I worry that (formally anyway) (1) is invalid ... or compromised at least ...

... for suppose for example $$x_j = -3$$ then ...... we have LHS of (1) = $$\mid x_j \mid = \mid -3 \mid = 3$$... BUT ...

RHS of (1)$$ = ( \mid x_j \mid^2 )^{ \frac{ 1 }{ 2 } } = ( \mid -3 \mid^2 )^{ \frac{ 1 }{ 2 } } = ( 3^2 )^{ \frac{ 1 }{ 2 } } = 9^{ \frac{ 1 }{ 2 } } = \pm 3 $$My question is as follows:

How do we deal with the above situation ... and

... how do we formally and rigorously demonstrate that $$\mid x_j \mid \le \left( \sum_{ 1 \le j \le n } \mid x_j \mid^2 \right)^{ \frac{ 1 }{ 2 } } $$Hope someone can help ...

Peter

 

[tkhunny]

RHS of (1)$$ = ( \mid x_j \mid^2 )^{ \frac{ 1 }{ 2 } } = ( \mid -3 \mid^2 )^{ \frac{ 1 }{ 2 } } = ( 3^2 )^{ \frac{ 1 }{ 2 } } = 9^{ \frac{ 1 }{ 2 } } = \pm 3 $$

How do we deal with the above situation ... and

$9^{1/2}\ne\pm 3\;rather\;9^{1/2} = 3$

You may be thinking about solutions to $x^2 = 9$. That's a different animal.
It is appropriate to say $\sqrt{x^{2}} = |x|$ but only if we know nothing of x.

... how do we formally and rigorously demonstrate that $$\mid x_j \mid \le \left( \sum_{ 1 \le j \le n } \mid x_j \mid^2 \right)^{ \frac{ 1 }{ 2 } } $$

Maybe start with this?
(x + y)^2 = x^2 + 2xy + y^2 = x^2 + y^2 + Stuff

(x + y + z)^2 = x^2 + y^2 + z^2 + Stuff

Now, we need to show only that Stuff >= 0.

Just an idea. Let's see where you go with it.

 

[Math Amateur]

$9^{1/2}\ne\pm 3\;rather\;9^{1/2} = 3$

You may be thinking about solutions to $x^2 = 9$. That's a different animal.
It is appropriate to say $\sqrt{x^{2}} = |x|$ but only if we know nothing of x.
Maybe start with this?
(x + y)^2 = x^2 + 2xy + y^2 = x^2 + y^2 + Stuff

(x + y + z)^2 = x^2 + y^2 + z^2 + Stuff

Now, we need to show only that Stuff >= 0.

Just an idea. Let's see where you go with it.

Hi tkhunny,

Thanks for the help ...

You're right of course ...

Thanks again ...

Peter