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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:

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

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:

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

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