# Bounded Inverse Theorem-Another proof

I am looking for another proof of BIT which doesn't use the Open mapping theorem. Actually, I want to use BIT in the proof of Open mapping theorem, so I cannot take that proof as it will be a paradox. I have looked at many texts but almost every proof is the same.

Kindly Help Thanks & regards in advance

• What is BIT? I have heard of it. – Kavi Rama Murthy Jul 9 '18 at 7:39
• Wikipedia states it is equivalent to the closed graph theorem, so you can try and use that to prove it – Bernard W Jul 9 '18 at 7:40
• Inverse of a bijective Bdd linear Operator is also bounded @KaviRamaMurthy – Devendra Singh Rana Jul 9 '18 at 9:12
• I think BIT, like Open Mapping Theorem, is an application of BCT. So I think you can prove BIT by imitating the proof of OMT but no simpler proof is possible. – Kavi Rama Murthy Jul 9 '18 at 9:32
• Why -1 please give reason?? – Devendra Singh Rana Jul 9 '18 at 12:23

## 1 Answer

Suppose that $T:X \to Y$ is an invertible bounded linear map where $X,Y$ are Banach spaces. We will prove that the inverse $T^{-1}$ is bounded by applying the closed graph theorem. We will use the condition from the functional analysis section from the relevant wikipedia page

Suppose $y_n$ is a sequence which converges to $y\in Y$ where each $y_n\in Y$. We want to show that the sequence $T^{-1}y_n$ converges to $T^{-1}y$

Let $x_n=T^{-1}y_n$ and note that by the closed graph theorem the sequence $(x_n,Tx_n) = (x_n , y_n)$ converges to an element $(x,y)$ on the graph of $T$, hence $\lim x_n=\lim T^{-1}y_n =x$ and $Tx = y$ which in turn implies $T^{-1}y= x$. Putting this together gives $\lim T^{-1}y_n = T^{-1}y.$

We conclude that $T^{-1}$ is continuous, hence bounded.

• This only shows that the Closed Graph Theorem implies the Bounded Inverse Theorem. Since it is well known that OMT, BIT and CGT are equivalent, this hardly answers the OP’s question. – M. Mueger Jan 16 at 21:43