0
$\begingroup$

My professor gives an example to show that the completion of the codomain is necessary for the Open Mapping Theorem.

First, as setup, my professor provided the following equivalent formulation of an open map:

Let $T \in \mathcal{L}(X,Y)$. $T$ is an open map if there exists $A < \infty$ such that for each $y \in Y$, we can find $x \in X$ such that $||x|| \leq A||y||$.

Now, for the example, he considers $X = \ell^\infty$ and the following subspace thereof: $Y = \{(x_n) \in \ell^\infty \mid \sup_{n \in \mathbb{N}} n|x_n| < \infty \}$.

Then, he considers $T$ to send a sequence $(x_n)$ to the sequence $(n^{-1}x_n)$.

First of all: Why is $Y$ not complete? Presumably, I can exhibit a sequence of sequences whose limit (in $\ell^\infty$) fails to lie in $Y$, but I'm not sure what that would be.

Second of all: Why does $T$ fail to satisfy the OMT?

$\endgroup$

1 Answer 1

0
$\begingroup$

Here are some preliminary thoughts. I'm a bit tired right now, so apologies if they're half-baked.

With regards to question 1), I'm currently thinking about the following sequence of sequences:

A preliminary thought is to choose the following sequence of sequences (upper index for the `outer' sequence):

$x^m_n = \begin{cases} \mathrm{ln}(e+\frac{1}{n}) - 1 & n \leq m \\\ 0 & n > m \end{cases}$

With regards to question 2, I'm considering the following sequences in $Y$ with norm equal to $1$:

$$(1,0,...), (1,1,0,...),...$$

Whose preimages are $$(1,0,...), (1,2,0,...)$$ and hence have norms $1, 2, 3, ...$

Therefore, you cannot exhibit any such $A$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .