Answer to Question #272582 in Functional Analysis for Prathibha Rose

Question #272582

Show that the self adjoint operator is continuous map


1
Expert's answer
2021-12-01T17:11:25-0500

A linear map T : X → Y is continuous if and only if its operator norm is finite


To prove that T is continuous, prove that it is bounded. From Cauchy-Schwarz-Bunyakowsky

Ty2=Ty,TyX=y,TTyYyTTyyTTy|T^*y|^2 = |\langle T^* y, T ^∗ y\rangle_X| = |\langle y, TT ^∗ y\rangle_Y| ≤ |y| · |T T^ ∗ y| ≤ |y| · |T| · |T ^∗ y|

where |T| is the operator norm. For Ty0T ^∗y \neq 0 , divide by it to find

TyyT|T^ ∗ y| ≤ |y| · |T|

Thus,TT|T^ ∗ | ≤ |T| . In particular, TT^ ∗ is bounded.


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