Let $X$ be Banach space and $Y$ be a normed vector space and suppose that I find a linear map $T \in L(X,Y)$. With the aid of $T$, I wonder under what condition I can conclude that $Y$ is also a Banach space.
Is $T$ a linear isomorphism enough? I encounter this question when think of an example like $\ell^p$-space equipped with norm $$ \|a\|_p := \left(\sum_{n=1}^\infty |a_n|^p\right)^{1/p} $$ and $\ell^p(w)$, a weighted $\ell^p$-space, equipped with norm $$ \|a\|_{p,w}:=\left(\sum_{n=1}^\infty |a_n|^p w(n)\right)^{1/p} $$ for $p \geq 1$ and $w: \mathbb{N} \to [0,\infty)$. It seems to me these two guys are isomorphic, and I was trying to argue $\ell^p(w)$-space is Bananch space without using the standard Cauchy sequence argument. Does this thought make sense? or it is completely off-track? Any comment or feedback is appreciated.
Edit What if $w(n) > 0$ is forced. Does the idea above alive?
If $X$ is a Banach space, $Y$ is a linear space and $T:X\to Y$ is a linear isomorphism, then $Y$ is a Banach space. To show this let $(y_n)$ be a Cauchy sequence in $Y$, then the sequence $(T^{-1}y_n)$ is Cauchy in $X$, so it converges to some $x$, and then $y_n\to Tx$. Therefore $Y$ is complete.
Now, with the definition of $\ell^p(w)$, choose $w(n)=\frac{1}{n}$. Then $T:\ell^2\to\ell^2(w)$ sending $x$ to $x$ is well defined, injective and continuous. However, $T$ is not surjective: note that $y=\left(1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},\dots \frac{1}{\sqrt{n}},\dots\right)\in \ell^2(w)$, but there is no $x\in\ell^2$ such that $Tx=y$. Therefore $T$ cannot be an isomorphism.