# Graph norm of a closed operator

Let's say we have a separable Banach space ($$B$$, $$\Vert \cdot \Vert$$). $$A: D(A) \to B$$ is a closed operator defined on a linear subspace $$D(A)$$ of $$B$$. Define the graph norm on $$D(A)$$ by $$\Vert x \Vert_{D(A)}= \Vert x \Vert + \Vert Ax \Vert.$$ It is well-known that by closed graph theorem that $$(D(A), \Vert \cdot \Vert_{D(A)})$$ is a also Banach space. I'm asking if this Banach space necessarily separable?

$$\newcommand{nrm}{\left\lVert{#1}\right\rVert}\newcommand{norm}{\nrm{\bullet}}$$Let $$\norm':B\times B\to [0,\infty)$$ be the norm $$\nrm{(x,y)}'=\nrm x+\nrm y$$. This norm metrizes the topological space $$(B,\norm)\times (B,\norm)$$, which is separable because $$B$$ is.
The map $$T:\left(D(A),\norm_{D(A)}\right)\to (B\times B, \norm')\\ Tx=(x,Ax)$$ is an isometry, and therefore it is a homeomorphism onto its image. Said image is a separable metric space because it is subset of a separable metric space.
• Could you provide more detail? First, $D(A)$ is a subset of $B$ but not $B \times B$. Second, the norm here is $\Vert x \Vert + \Vert Ax \Vert$ but not $\Vert x \Vert + \Vert y \Vert$. – lye012 Apr 25 at 17:35