# Help me tie up loose ends in operator theory

Science Advisor
Homework Helper
Dear knowledgeable person,

please, help me with the following problems:

Let $A$ be a closed operator in the Banach space $(B, ||.||)$.
Let D(A) be its domain in B. Prove that D(A) endowed with the graph norm is a Banach space.

The graph norm is defined as:

let 'a' be a vector in D(A). Then $||a||_{\mathbb{graph}} := ||a||+ ||A a||$.

I can easily show that the graph norm makes D(A) a pre-Banach space, but what about completeness wrt the graph norm ? I'm sure it has to do with the closedness of A, but how ?

Also, how does one prove the quivalence of the 2 known definitions of a closed operator in a banach space ? Is it really trivial, as everyone claims ?

See also here
http://planetmath.org/?method=l2h&from=objects&name=ClosedOperator&op=getobj

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## Answers and Replies

Are you aware that closed subsets of complete metric spaces are themselves complete? That should give you what you want.

Landau
Science Advisor
It seems trivial. If (a_n) is a Cauchy sequence w.r.t. the graph norm, then both (a_n) and (Aa_n) is a Cauchy sequence w.r.t. the usual norm. Hence they converge to some a and b, respectively. As A is closed, we get that a is in D(A) and b=Aa.