Norm on $D(A)$ where $A$ is a unbounded linear operator

Asked Apr 17 '17 at 22:06

Active Apr 17 '17 at 22:06

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Let $E,F$ Banach spaces. Brezis calls a operator $A: D(A)\subset E\to F$, where $D(A)$ is a subspace such that $\overline{D(A)} = E$ of unbounded linear operator.

In some exercise he gives the following hint: Consider $D(A)$ with the graph norm. In this case, if $A$ is closed, then $D(A)$ with such norm is a Banach space.

So my point is, what does mean to consider $D(A)$ with the graph norm? This is weird since $D(A)\subset E$. Is he meaning that $A$ can be view as the operator $A : (u,v) \in F^{\ast}\times E\mapsto \langle u,Av\rangle$ and $\|A(u,v)\| := \|u\| + \|Av\|?,$ where $D(A)$ is thought as a subespace of $F^{\ast}\times E?$

asked Apr 17 '17 at 22:06

L.F. Cavenaghi

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This might help: Closed operators – wdacda Apr 17 '17 at 22:30
THANK YOU SO MUCH! This helped more than you think. – L.F. Cavenaghi Apr 17 '17 at 22:34
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$(\mathcal{D}(A),|\cdot|_A)$ is a normed space, where $|x|_A = |x|_E+|Ax|_F$. This space is a Banach space if $A$ is closed. This norm is the norm induced by $x \mapsto \langle x,Ax\rangle \in \mathcal{G}(A)\subset E\times F$. – Disintegrating By Parts Apr 19 '17 at 16:14

Norm on $D(A)$ where $A$ is a unbounded linear operator

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