My professor asked us for an example of a Bounded Operator with whose range is not closed, after some attempts I thought this, but I am not sure about it:
Consider $n\in \mathbb{N}$ and $\mathbb{R}^n$ with the $\max$ norm, i.e if $\mathbb{R}^n\ni x = (x_1,\dots,x_n)$ then $|| x|| = \displaystyle \max_{i=1,\dots,n}{x_i}. $ And the operator $T=\mathbb{R}^n\to \mathbb{R}^n$ such that, if we denote $||x||=a$ then $T:x\mapsto(a,\dots,a) \in \mathbb{R}^{n}$. This is all the non-negative multiples of the vector $(1,1,…,1)$.
Clearly $||T(x)||=||x||$, so $T$ is bounded, but I am not sure if the range is closed, for this I thought the following: We know that a set $A$ is closed iff $A^{c}$ is open. In this case $A= T(\mathbb{R}^n)$ so I can consider an element in the complement of the image of $T$, this is $y=(y_1,\dots,y_n)$ such that there are at least two $y_i$ that are different. Let's denote $b=||y||$ and consider an open ball $B(y,\epsilon)$, note that for every possible $\epsilon$ the element $T(y)$ is contained in $B(y,\epsilon)$, so there is no open ball in $A^c$ for $y$, so $A^c$ is not an open set, therefore $A$ is not a closed set.
Is this correct? I feel like I am missing something but I am not sure. If I am wrong, can you give me some other example of a bounded operator with no closed range? I saw a couple but most of the examples I found use the $L^p$ spaces that we haven't seen on class. Thanks.
Best Answer
Let X be a non reflexive Banach space and $Y$ be a reflexive Banach space. Suppose that $A \colon X \to Y$ is an injective bounded linear operator. I claim that the range of $A$, denoted by $R(A)$, can't be closed in $Y$.
Arguing by contradiction, suppose that $R(A)$ is closed and thus reflexive. Then, the operator $ A \colon X \to R(A) $ is a continuous bijection and thus (by virtue of the Open Mapping Theorem) is an isomorphism. It's not hard to check that $X$ has to be reflexive, since it's isomorphic to a reflexive Banach space, which leads to a contradiction.