- #1
Oxymoron
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Im in need of some guidance. No answers, just guidance.
Question.
Let [tex](x_m)[/tex] be a Cauchy sequence in an inner product space, show that
[tex]\left\{\|x_n\|:n=1,\dots,\infty\right\}[/tex]
is bounded.
[tex]proof[/tex]
From the definition we know that all convergent sequences are Cauchy sequences. But it is not true that the converse always holds. However, in normed linear spaces the converse is always true. That is our Cauchy sequence is convergent to a point [tex]x[/tex].
Since [tex](x_m)[/tex] is a convergent sequence we take an [tex]\epsilon[/tex] such that
[tex]\|x_m - x\| \leq \epsilon[/tex] if [tex]m > N[/tex].
Now we fix [tex]N[/tex] so that
[tex]\|x_m\| \leq \|x_m - x\| + \|x\| < \epsilon + \|x\|[/tex]
which holds for all [tex]m > N[/tex]. This is just a matter of adding [tex]\|x\|[/tex]
Now we define
[tex]M = \mbox{max}\{\|x_1\|,\|x_2\|,\dots,\|x_N\|, (1+\|x\|)\}[/tex]
Then [tex]\|x_m\| < M \, \forall \, m[/tex]
That is, the sequence is bounded.
Question.
Let [tex](x_m)[/tex] be a Cauchy sequence in an inner product space, show that
[tex]\left\{\|x_n\|:n=1,\dots,\infty\right\}[/tex]
is bounded.
[tex]proof[/tex]
From the definition we know that all convergent sequences are Cauchy sequences. But it is not true that the converse always holds. However, in normed linear spaces the converse is always true. That is our Cauchy sequence is convergent to a point [tex]x[/tex].
Since [tex](x_m)[/tex] is a convergent sequence we take an [tex]\epsilon[/tex] such that
[tex]\|x_m - x\| \leq \epsilon[/tex] if [tex]m > N[/tex].
Now we fix [tex]N[/tex] so that
[tex]\|x_m\| \leq \|x_m - x\| + \|x\| < \epsilon + \|x\|[/tex]
which holds for all [tex]m > N[/tex]. This is just a matter of adding [tex]\|x\|[/tex]
Now we define
[tex]M = \mbox{max}\{\|x_1\|,\|x_2\|,\dots,\|x_N\|, (1+\|x\|)\}[/tex]
Then [tex]\|x_m\| < M \, \forall \, m[/tex]
That is, the sequence is bounded.