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I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable" ... ...
I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...
I need some help with an aspect of the proof of Theorem 11.4.1 ...
Garling's statement and proof of Theorem 11.4.1 reads as follows:
View attachment 7921In the above proof by Garling we read the following:
" ... ... Let $$f_j = x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i$$. Since$$ x_j \notin W_{ j-1 }, f_j \neq 0$$.
Let $$e_j = \frac{ f_j }{ \| f_j \| } $$. Then $$\| e_j \| = 1$$ and
$$\text{ span } ( e_1, \ ... \ ... \ e_j ) = \text{ span } ( W_{ j - 1 } , e_j ) = \text{ span }( W_{ j - 1 } , x_j ) = W_j $$
... ... "
Can someone please demonstrate rigorously how/why $$f_j = x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i $$
and
$$e_j = \frac{ f_j }{ \| f_j \| }$$imply that $$\text{ span } ( e_1, \ ... \ ... \ e_j ) = \text{ span } ( W_{ j - 1 } , e_j ) = \text{ span }( W_{ j - 1 } , x_j ) = W_j$$
Help will be much appreciated ...
Peter
I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...
I need some help with an aspect of the proof of Theorem 11.4.1 ...
Garling's statement and proof of Theorem 11.4.1 reads as follows:
View attachment 7921In the above proof by Garling we read the following:
" ... ... Let $$f_j = x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i$$. Since$$ x_j \notin W_{ j-1 }, f_j \neq 0$$.
Let $$e_j = \frac{ f_j }{ \| f_j \| } $$. Then $$\| e_j \| = 1$$ and
$$\text{ span } ( e_1, \ ... \ ... \ e_j ) = \text{ span } ( W_{ j - 1 } , e_j ) = \text{ span }( W_{ j - 1 } , x_j ) = W_j $$
... ... "
Can someone please demonstrate rigorously how/why $$f_j = x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i $$
and
$$e_j = \frac{ f_j }{ \| f_j \| }$$imply that $$\text{ span } ( e_1, \ ... \ ... \ e_j ) = \text{ span } ( W_{ j - 1 } , e_j ) = \text{ span }( W_{ j - 1 } , x_j ) = W_j$$
Help will be much appreciated ...
Peter