Let $X:=\{ x \in \ell^{1}:\vert\vert\vert x \vert \vert \vert< \infty\}$, and that $\vert\vert\vert x \vert \vert \vert=\sum\limits_{j=1}^{\infty}j\vert x_{j}\vert$
$a.$ Show that $(X, \vert\vert\vert \cdot \vert \vert \vert)$ is a Banach space
$b.$ Show that $(X, \vert\vert \cdot \vert \vert_{1})$ is not a Banach space.
Ideas:
$a.$ Let $(x^{n})_{n\in\mathbb N}\subset X$ be a cauchy sequence of sequences. This means for an arbitrary $\epsilon > 0$ there exists $N \in \mathbb N$ so that for any $n> m \geq N$: $\vert\vert\vert x^{n}-x^{m} \vert \vert \vert=\sum\limits_{i=1}^{\infty}i\vert x_{i}^{n}-x_{i}^{m}\vert<\epsilon$
thus the $\vert x_{i}^{n}-x_{i}^{m}\vert < \frac{\epsilon}{i}\leq\epsilon$,thus
we have a Cauchy sequence $(x_{i}^{n})_{n \in \mathbb N}$ for a arbitrary, but fixed $i \in \mathbb N$. Now my thought (and only one at that), is to use that fact that $(x_{i}^{n})_{n \in \mathbb N}$ is a cauchy sequence oa a complete space (when equipped with the euclidean norm), therefore a pointwise limit $x_{i}:=\lim\limits_{n \to \infty}x_{i}^{n}$ exists so that we have an ideal candidate for a limit in $(X, \vert\vert\vert \cdot \vert \vert \vert)$, namely $x:=(x_{i})_{i \in \mathbb N}$
But the more I think about it, the less it makes sense that $\vert\vert\vert x \vert \vert \vert<\infty$ yet $\vert \vert x \vert \vert_{1}=\infty$ given the definitions of the norms, and that $j\vert x_{j}\vert\geq\vert x_{j}\vert$ for $j \in \mathbb N$
Any help is greatly appreciated
Best Answer
For (a), note that $\ell_1\to X$, $(a_n)\mapsto (n^{-1}a_n)$ is an isometry $(\ell_1,\lVert-\rVert_1)\to(X,\lVert\!\lvert-\rvert\!\rVert)$ makes the question rather trivial, but let's copy the proof of $\ell_1$ being complete here.
To show that $(x_i)\in X$, given $\varepsilon>0$ arbitrary, for every $N\in\mathbb{N}$, we have $$\require{color} \sum_{j=1}^N j\lvert x_j\rvert\leq\underbrace{\color{red}\sum_{j=1}^Nj\lvert x_j-x_j^{(m)}\rvert}_{<\varepsilon} + \underbrace{\color{blue}\sum_{j=1}^N j\lvert x_j^{(m)}\rvert}_{\leq M} $$ where $M=\sup\{\lVert\!\lvert x^{(m)}\rvert\!\rVert\colon m\in\mathbb{N}\}<\infty$ (since $(x^{(m)})$ is Cauchy in $X$) and the red term is $<\varepsilon$ for sufficiently large $m$ since we are only taking finitely many componentwise limit $j=1,2,\dots,N$. Therefore, $$ \sum_{j=1}^Nj\lvert x_j\rvert\leq M+\varepsilon $$ for every $N$. Now take $N\to\infty$ shows $x\in X$.
Of course, finally we need to show $x^{(n)}\to x$. Can you do that?
For (b), note that the space of eventually zero sequences $c_{00}$ is contained in $X$. So $$ x^{(n)}_j=\begin{cases} j^{-2} & j<n\\ 0 & \text{otherwise} \end{cases} $$ is a sequence in $X$. This sequence clearly converges to the sequence $(j^{-2})$ in $\ell_1$ since $\sum_j j^{-2}<\infty$, but this sequence is not in $X$.