completeness and the open mapping theorem

Question

Let $X,Y$ be normed vector spaces, I want to show that the open mapping theorem requires completeness of both spaces. So my question consists of two parts:

$\textit{i)}$ Let $X$ be a Banach space and $Y$ a normed space and find a bounded surjective linear operator which is not open.

$\textit{ii)}$ Let $X$ be a normed space and $Y$ a Banach space. Find a surjective linear operator which is not open.

For $\textit{i)}$, I chose $X=(\ell^1(\mathbb{N}),||\cdot||_1)$ and $Y=(\ell^1(\mathbb{N},||\cdot||_\infty)$ and $T:X\to Y$, $x\mapsto x$. This map is clearly surjective and bounded, but how can I show this is not open? I wanted to check the image of the open unit ball $B$ in $X$, so see whether $T(B)$ is open but I got stuck.

For $\textit{ii)}$, could anyone give me a hint which spaces I can use? I have not yet learend about Hamel basis, but I am allowed to use that there exists an unbounded linear functional on every infinite dimensional normed space.

Answer 1

For your example for (i), $T$ is continuous because $T$ is bounded and linear. A continuous open bijection is a homeomorphism. So if $T$ is open then $T^{-1}$ is continuous, and linear, and therefore bounded.... But $T^{-1}$ is NOT bounded. E.g. for $n,j\in \Bbb N$ let $x_{n,j}=1$ for $j\leq n$ and $x_{n,j}=0$ for $j>n$, and let $x(n)=(x_{n,j})_{j\in \Bbb N}$. Then $\|x(n)\|_{\infty}=1$ and $\|T^{-1}(x(n))\|_1=\|x(n)\|_1=n.$

(Recall that any linear map from any normed linear space to another normed linear space is continuous iff it is bounded.)

For (ii), let $Y$ be an infinite-dimensional real Banach space and let $g:Y\to \Bbb R$ be linear and unbounded. Let $X=\{(y,g(y)): y\in Y\}$ and let $\|(y,g(y))\|_X=\|y\|_Y+|g(y)|.$ Now let $f((y,g(y))=y.$ If $f$ were open then $f^{-1}$ would be continuous and hence $f^{-1}$ would be bounded . But $$\sup_{0\ne y\in Y}\frac {\|f^{-1}(y)\|_X}{\|y\|_Y}=\sup_{0\ne y\in Y}\frac {\|y\|_Y+|g(y)|}{\|y\|_Y}=\infty$$ because $g$ is unbounded.