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I'm learning about Hilbert spaces and operators theory, from some book. I came across the following question - enter image description here

And the books' answer:

enter image description here

What I don't understand in the proof -

Why can we understand that each sequence is Cauchy?

Moreover, why the following inequallity is true?

enter image description here

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    $\begingroup$ Every sequence is not Cauchy. Completeness is a condition on Cauchy sequences, that is why the proof only needs to work with those that are so. $\endgroup$ – user647486 Apr 12 at 17:24
  • $\begingroup$ @user647486 I meant, every sequence in $x^{(k)}$ is Cauchy. Why? $\endgroup$ – ChikChak Apr 12 at 17:28
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    $\begingroup$ In the chain of inequalities, the first one is triangle inequality $|x_j|=|x_j-x_j^{(k)}+x_j^{(k)}|\leq |x_j-x_j^{(k)}| + |x_j^{(k)}|$. The second comes from choosing $\epsilon <1$. The last one is from the definition of $\|x^{(k)}\|_{\infty}$, which is the suppremum of $|x_j^{(k)}|$ for all $j$. $\endgroup$ – user647486 Apr 12 at 17:28
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If $\sup_n|x_n^{(k)}-x_n^{(m)}|\leq\epsilon$ then it follows that for each $n\in\mathbb{N}$ we have $|x_n^{(k)}-x_n^{(m)}|\leq\epsilon$, because this expression is not bigger than the supremum on $n$. It follows that $(x_n^{(k)})_{k=1}^\infty$ is Cauchy for each $n$. (Cauchy with respect to the usual metric in $\mathbb{C}$).

As for the second question: the first inequality is the triangle inequality, the second follows from the fact that $x_j^{(k)}\to x_j$, the third follows from the definition of supremum norm.

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  • $\begingroup$ Well why can we assume that $\sup_n|x_n^{(k)}-x_n^{(m)}|\leq\epsilon$? We are taking some arbitrary Cauchy sequence in $l_\infty$ i.e., each element of $x_k$ is a sequence in $l_\infty$. So how can we know that $\sup_n|x_n^{(k)}-x_n^{(m)}|\leq\epsilon$? $\endgroup$ – ChikChak Apr 12 at 17:45
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    $\begingroup$ We take a Cauchy sequence $(x^{(k)})_{k=1}^\infty$ of elements in $l_\infty$. This is the only thing we assume. Now, an element in $l_\infty$ is a sequence of complex numbers. So for each $k\in\mathbb{N}$ we have that $x^{(k)}$ is a complex sequence $(x_n^{(k)})_{n=1}^\infty$. Now what does it mean that $(x^{(k)})_{k=1}^\infty$ is Cauchy? It means that for each $\epsilon>0$ there is $N\in\mathbb{N}$ such that for all $k,m>N$ we have $||x^{(k)}-x^{(m)}||_\infty<\epsilon$. And now just use the definition of the infinity norm. For a sequence $x=(x_n)$ the definition is $||x||_\infty=\sup_n|x_n|$. $\endgroup$ – Mark Apr 12 at 17:53

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