(Solved): a. State and prove the open mapping Theorem, and deduce the closed graph Theorem. (You may assume a ...

Expert Answer

---

a. Let's start by stating and proving the Open Mapping Theorem, and then we will deduce the Closed Graph Theorem from it.

Open Mapping Theorem:

Let    and    be Banach spaces, and let    be a continuous linear operator that is onto (surjective). Then    is an open mapping, i.e., it maps open sets in    onto open sets in   .


Proof of the Open Mapping Theorem:

To prove the Open Mapping Theorem, we will make use of the Baire Category Theorem, which we assume is given.

First, let's define some terms.
A set    in a topological space is called nowhere dense if the closure of    has an empty interior. A set    is called meager (or first category) if it can be expressed as the union of countably many nowhere dense sets. A set    is called comeager (or second category) if it is not meager.

Now, let's proceed with the proof:


Step 1: We will show that   is a comeager set in   , where   is the closed unit ball in   .
Consider the sets   , where    is the closed unit ball in   .
Since    is onto, for any    in   , there exists    in    such that   .
Thus, we can write   for any    in    (the set of natural numbers).
This implies that y belongs to    for any    in   .
Note that each    is the sum of two closed sets,    and   , and hence is closed.

Now, we claim that each    is nowhere dense.
To prove this, let's assume that    has a non-empty interior for some   .
Then, there exists    in    and   such that   , where    denotes the open ball centered at   with radius   .

Since    is onto, for any    in   , there exists    in    such that   .
But then, we have   , which contradicts the assumption that   .
Hence,    has an empty interior, and it is nowhere dense.

Now, we have expressed   as the countable union of nowhere dense sets, i.e.,   .
Since each    is closed and nowhere dense, it follows that   is meager.
Therefore, by the Baire Category Theorem, the complement of    denoted by   , is comeager.


Step 2: We will show that    is open for any open set    in   .
Let    be an open set in   .
We want to show that    is open in   .
Take any   .
We need to find an open ball centered at    that lies entirely in   .

Since    is onto, there exists    in    such that   .
Since    is open, there exists    such that   , where    denotes the open ball centered at    with radius   .

Consider the set   , where    is the closed unit ball in  . V is an open set in X.
Now, for any    we have    for some   
Hence,   , which implies that   
Since   is comeager (Step 1), its complement    is a dense set in Y.
Therefore, there exists z in    such that   .
Thus, we have shown that for any  , there exists an open ball   that lies entirely in   . Therefore,   is open in Y.


Hence, we have proved the Open Mapping Theorem.