An application of the closed graph theorem.

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SUMMARY

The discussion centers on demonstrating that a projection operator P on a Banach space X is bounded if and only if the kernel of P and the image P(X) are closed subspaces of X. The closed graph theorem is utilized to show that the graph G(P) is closed under these conditions. The user outlines their approach and refines their reasoning, ultimately concluding that if (x_n, y_n) converges to (x, y) with y_n = Px_n, then y must equal Px, confirming the boundedness of P.

PREREQUISITES
  • Understanding of Banach spaces
  • Familiarity with linear operators and projections
  • Knowledge of the closed graph theorem
  • Concept of closed subspaces in functional analysis
NEXT STEPS
  • Study the implications of the closed graph theorem in functional analysis
  • Explore properties of bounded linear operators on Banach spaces
  • Investigate the relationship between kernel and image of linear operators
  • Learn about convergence in Banach spaces and its applications
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Mathematicians, particularly those specializing in functional analysis, graduate students studying operator theory, and researchers exploring properties of Banach spaces.

Hjensen
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So I have to show that a projection P (i.e a linear operator with P=P²) on a Banach space X is bounded if and only if \ker (P) and P(X) are closed subspaces of X.My idea was to boil it down, using the closed graph theorem. What's left for me now is to show that the graph G(P):=\{(x,y)\in X\times X: y=Px\} is closed if \ker(P) and P(X) are closed. I don't quite know how this can be achieved though. Does anyone know how this could be done? Or am I simply taking the wrong approach by using the closed graph theorem?
 
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Edit: I wrote a nonsense. Thinking ...
 
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Assume ker(P) and P(X) closed. Let (x_n,y_n)\rightarrow (x,y),\, y_n=Px_n. Then (y_n-x_n)\in \ker(P) and so y=x+x',\, x'\in\ker (P). From (I-P)y_n=0 it follows (I-P)y=0, so y=Py=P(x+x')=Px.
 

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