# Show that $(X, \vert\vert\vert \cdot \vert \vert \vert)$ is a Banach Space and that $(X, \vert \vert \cdot\vert \vert_{1})$ is not)

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

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 $$x^{(n)}_j=\begin{cases} j^{-2} & j is a sequence in $$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$$.