# 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

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.