Is the inverse of a bounded operator always bounded , if yes how to prove it ?
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1In which kind of spaces are you working? If you are working in Banach spaces, I think this can be a useful topic – PhDing Jan 03 '16 at 11:12
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Related : https://math.stackexchange.com/questions/1580369 – Watson Apr 06 '16 at 10:08
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If You consider an invertible, i.e. bijective and bounded linear operator $A:X\rightarrow Y$, between two Banach-spaces (it´s important they are complete), then as a consequence of Baires category theorem A is open (open mapping theorem) and so $A^{-1}:Y \rightarrow X$ is continuous, i.e. bounded.
Peter Melech
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Let T be a bounded linear operator from a normed space X on to a normed space Y. If there exists a positive b such that $||Tx||\geq b ||x|| \ \forall x\in X,$ then $T^{-1}$ exists and is bounded.
Ravindra
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I See that if $T^{-1}$ exists it would be bounded, but could you give an argument for the existence of $T^{-1}$? – Aug 17 '21 at 15:49
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In fact consider the right shift $T:\mathcal{l}^2\to \mathcal{l}^2,(x_1,x_2,x_3,\dots)\mapsto (0,x_1,x_2,\dots)$. Then $\Vert Tx\Vert = \Vert x \Vert$, but $T$ is not surjective. – Aug 18 '21 at 07:41
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@user592521 https://www.math.cuhk.edu.hk/course_builder/1516/math4010/solution4.pdf there you go! – ARROW Feb 18 '24 at 11:54
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It depends on how you define "inverse" of an operator. In general, if you don't require it be bounded, then no. For example, consider the bijection from $\mathbb{R}$ onto $(-1,1)$. But sometimes we say an operator is invertable only if we can find a bounded operator such that the composition is identity.
John
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, thanks , i think i found it here http://planetmath.org/boundedinversetheorem – lucaluca Jan 03 '16 at 11:23
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2But operator means linear operator and there is no bounded linear operator from $\mathbb{R}$ to $(-1,1)$. So I don't understand this answer. – Caleb Stanford Jul 30 '16 at 09:27