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.

  • $\begingroup$ Your question is confusing. The way it is stated you cannot choose $X$ and $Y$; they are given to you. $\endgroup$ Jan 7, 2019 at 11:54
  • $\begingroup$ Oh i am sorry. What I mean is choose $X$ and $Y$ and give a map $T$ toch is subjective bounded but not open $\endgroup$
    – user408856
    Jan 7, 2019 at 11:56
  • $\begingroup$ Thanks for the hint in your last sentence, so I could add part (ii) to my answer. $\endgroup$ Jan 7, 2019 at 15:32

1 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.

  • $\begingroup$ BTW...I saw an example in Amer. Math. Monthly of a complex normed linear space $X$ with a continuous linear bijection $h:X\to X$ such that $h^{-1}$ is discontinuous. $\endgroup$ Jan 7, 2019 at 15:19
  • $\begingroup$ The construction in part (ii) partly came from musing on the Closed Graph Theorem, applied to $g$. $\endgroup$ Jan 7, 2019 at 15:37
  • $\begingroup$ How do we know that $X$ is not complete? $\endgroup$
    – user408856
    Jan 7, 2019 at 18:24
  • $\begingroup$ Or is it due to the unboundedness of $g$? $\endgroup$
    – user408856
    Jan 7, 2019 at 18:42
  • 1
    $\begingroup$ Yes. First, $f^{-1}$ is a function because $f$ is a bijection. Second, $f$ is a continuous bijection, so if $f$ were open then $f^{-1}$ would be continuous. Third, if $f^{-1}$ were continuous then $f^{-1}$ would be bounded $\endgroup$ Jan 9, 2019 at 4:02

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