Help me tie up loose ends in operator theory

[dextercioby]

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

 

[some_dude]

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

 

[Landau]

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.