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In order to prove the space is not complete w.r.t to the given norm, we have to find a Cauchy sequence in the space but it does not converge to an element in the given set.

My attempt was taking $(Z^n) = ( (1,0,0, ... ) , (1, 1/2, 0, 0, ... ), ....)$ I tried to prove this is Cauchy and ended up with,

here $m>n$,

$\lVert Z^n - Z^m \rVert \leq \dfrac {(m-n)^{1/2}}{n}$

how could I make this arbitrarily small? ($<\epsilon$)

here the $n$th term of the sequence would be $Z^n = (1,1/2,1/3,...,1/n,0,0,0...)$ So when $n \rightarrow \infty$, $z^n$ has infinitely many non zero terms, would that be enough to say the space is incomeplete?

Mauro ALLEGRANZA
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Lucifer
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2 Answers2

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It is enough because of the following argument: Convergence in 2-norm implies convergence of each coordinate. It follows that if your sequence converges in the norm them the limit has to be $(1,\frac 1 2,\frac 1 3, \frac 1 4,....)$, but this last sequence has infinitely many non-zero terms.

geetha290krm
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  • thank you for the comment. I got that. But I am not sure how exactly to say my original sequence is Cauchy... to make $<\epsilon$ – Lucifer Mar 11 '23 at 06:09
  • @Lucifer If $a_n>0$ and $\sum a_n <\infty$ the $ \sum\limits_{k=n}^{m}a_k \to 0$ as $m>n \to \infty$. Apply this basic fact to the series $\sum \frac 1 {n^{2}}$. – geetha290krm Mar 11 '23 at 06:11
  • ahhh thank you so much!! so not necessarily needed for $\epsilon$ here? – Lucifer Mar 11 '23 at 06:28
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As you say, if $m>n$, $$ Z^m-Z^n=\Big(0,\ldots,0,\frac1{n+1},\ldots,\frac1m,0,\ldots\Big). $$ So $$ \|Z^m-Z^n\|_2^2=\sum_{k=n+1}^m\frac1{k^2}\leq\int_n^\infty\frac1{x^2}\,dx=\frac1n. $$ So the sequence is Cauchy: given $\varepsilon>0$, choose $n_0$ with $n_0\geq\frac1{\sqrt \varepsilon}$. Then for any $n,m\geq n_0$ you have $\|Z^n-Z^m\|<\varepsilon$.

Your space embeds naturally in $\ell^2(\mathbb N)$. There it is easy to check that $Z^n\to Z$, where $Z=(1,\frac12,\frac13,\ldots)$. Since $Z$ is not in the space $c_{00}$ of finitely supported sequences and since limits are unique, the sequence $\{Z^n\}$ does not converge in $c_{00}$.

Martin Argerami
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