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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?

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$\newcommand{nrm}[1]{\left\lVert{#1}\right\rVert}\newcommand{norm}{\nrm{\bullet}}$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.

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  • $\begingroup$ 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 $\Vert x \Vert + \Vert y \Vert$. $\endgroup$ – lye012 Apr 25 at 17:35
  • $\begingroup$ @lye012 Fair enough, thank you for pointing that out. I'll edit. $\endgroup$ – Saucy O'Path Apr 25 at 17:39
  • $\begingroup$ Thanks! It's clear now. $\endgroup$ – lye012 Apr 25 at 18:50

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