# completeness and the open mapping theorem

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.

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

## 1 Answer

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.

• 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. – DanielWainfleet Jan 7 at 15:19
• The construction in part (ii) partly came from musing on the Closed Graph Theorem, applied to $g$. – DanielWainfleet Jan 7 at 15:37
• How do we know that $X$ is not complete? – user408856 Jan 7 at 18:24
• Or is it due to the unboundedness of $g$? – user408856 Jan 7 at 18:42
• 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 – DanielWainfleet Jan 9 at 4:02