Subset of $\ell_2$ with the norm $\sup n|x_n|<\infty$ is a Banach space

Question

Consider the function $\lVert \cdot \rVert: \ell_2 \to [0,\infty]$ defined by $$\lVert x \rVert=\sup_{n\geq 1} n\lvert x_n\rvert$$ for every $x=\{x_n \}_{n=1}^{\infty}\in \ell_2$. Let $$X=\{ x\in \ell_2 \mid \lVert x\rVert <\infty \} $$ Show that $X$ with the said norm is a Banach space.

I have shown that the function defined is indeed a norm. I can think of two ways to prove a space is Banach: one is to get a Cauchy sequence and show that it is convergent, and two is to show that every absolutely convergent series is convergent. None of them has led me far, but I think the first one is more useful if I can relate the norm with the usual $\ell_2$ norm $\lVert \cdot \rVert_2$ in some way. Let's say if we had $c \lVert x \rVert<\lVert x \rVert_2 < C\lVert x \rVert$, then a Cauchy sequence in $X$ with the said norm yields a Cauchy sequence in $\ell_2$ by the right side of the inequality, and since $\ell_2$ is complete, the sequence will converge in $\ell_2$, but again using the left side of the equality we can say that it converges in X as well.

Answer 1

Let $(x^{(n)})$ be a Cauchy sequence in this space. Then $\sup_k k|x^{(n)}(k)-x^{(m)}(k)|\to 0$ as $n, m \to \infty$. In particular, $\sup_k |x^{(n)}(k)-x^{(m)}(k)|\to 0$ so $x_k=\lim x^{(n)}(k)$ exists for each $k$. If we choose $N$ such that $\sup_k k|x^{(n)}(k)-x^{(m)}(k)|<\epsilon$ for $n, m >N$ then $\sup_k k|x^{(n)}(k)-x_k|\le \epsilon$ for $n>N$ so $x^{(n)} \to x$ in the given norm.

The fact that $\sup_k k|x^{(n)}(k)-x_k|\le \epsilon$ for some $n$ implies that $x^{(n)} -x\in \ell^{2}$ and hence, $x \in \ell^{2}$.

Answer 2

The problem can be generalized. For $a_n>0$ let $$X=\{x\in \ell^2\,:\, \sup_na_n|x_n|<\infty\}$$ Then the space $X$ is complete with respect to the norm $$\|x\|=\sup_na_n|x_n|$$ if and only if $\{a_n^{-1}\}_{n=1}^\infty\in \ell^2.$ In particular the space $X$ is complete for $a_n=n.$

Indeed assume $\{a_n^{-1}\}_{n=1}^\infty\in \ell^2.$ The operator $$T(x)=\{a_nx_n\}_{n=1}^\infty $$ is an isometry from $X$ into $\ell^\infty.$ On the other hand the operator $$S(x)=\{a_n^{-1}x_n\}_{n=1}^\infty$$ from $\ell^\infty$ to $X$ is well defined since $\{a_n^{-1}\}_{n=1}^\infty\in \ell^2.$ Moreover $ST=I_{X}$ and $TS=I_{\ell^\infty}.$ Hence the spaces $X$ and $\ell^\infty$ are isometrically isomorphic, i.e. $X$ is complete.

Assume now that $\{a_n^{-1}\}_{n=1}^\infty\notin \ell^2.$ Then there exists a sequence $b_n>0$ such that $b_n\to 0$ and $$\sum_{n=1}^\infty {b^2_n\over a_n^2}=\infty \quad (*)$$ Consider the sequence of elements $x^{(k)}$ in $X$ defined by $$x^{(k)}_n=\begin{cases}{b_n\over a_n}& 1\le n\le k\\ 0 & n>k\end{cases}$$ The elements $x^{(k)}$ form a Cauchy sequence in $X$ as for $k<l$ we have $$\|x^{(l)}-x^{(k)}\|=\sup_{k<n\le l}b_n\underset{k\to \infty}{\longrightarrow}0 $$ However the sequence $x^{(k)}$ is not convergent in $X.$ Indeed, $x^{(k)}$ is convergent entrywise to the sequence $\{b_n/a_n\}_{n=1}^\infty$, but this sequence is not square summable.

Example For $a_n=n^{1/2}$ the sequence $a_n^{-1}$ is not square summable. In this case the sequence $b_n=[\log(n+1)]^{-1/2}$ satisfies (*).