Today Wongting raised the question in the real analysis prelims at UW about proving that the Closed Graph Theorem (CGT) implies the Open Mapping Theorem (OMT). This is a standard exercise in the first course of functional analysis.

Let us recall the three basic theorems. We assume are Banach spaces. Recall that a linear operator is bounded if and only if it is continuous on .

**OMT**: If be a bounded surjective linear operator, then is an open mapping.

**Inverse Mapping Theorem (IMT)**: If is a bounded bijective linear operator, then the inverse is bounded.

**CGT**: Suppose is a linear operator. Then is bounded if and only if its graph is closed in .

Standard texts tell us that OMT implies IMT and IMT implies CGT. Once we proved CGT implies OMT, we see that the above three theorems are indeed equivalent.

**Theorem.*** CGT implies OMT. *

** Proof.** Assume CGT is true. Let be a bounded surjective linear operator. We want to show (wlog, by linearity) for some (Notation: is the open ball centered at with radius ).

Intuitively (if you have learnt some general topology), that a map is open is related to the continuity of its inverse. Therefore we want to construct the inverse of , but may not be invertible. One way to do is to consider equivalence classes, or more precisely, quotient spaces.

Define an operator (Notation: is the null space of ; Note now is a closed subspace of ) by where . It is easy to see this map is well defined and linear. Then:

Claim 1: The graph of is closed.

Indeed, if , then , and . Thus by passing to subsequence, for some , and so , also . Since , has a limit and thus Cauchy, has a convergent subsequence, so and . We get .

By CGT, is bounded. Then, by linearity, we may assume there exists such that . After proving the following second claim, we are done.

Claim 2: .

Indeed, let . Then and thus . Therefore there exists such that . It follows that .

The work is very helpful . Thanks. I need some help in homology group

Nai ana yeh exams mein

Thanks