Say for example, we look at the sequence space $\ell^{1}$, and we define some strange norm for it, call it $\vert \vert \vert \cdot \vert \vert \vert$ (that is NOT $\vert \vert \cdot \vert \vert_{1}$)
Then we look at the operator $Id: (\ell^{1}, \vert \vert \vert \cdot \vert \vert \vert)\to (\ell^{1}, \vert \vert \cdot \vert \vert_{1})$
Can I assume that $(\ell^{1}, \vert \vert \vert \cdot \vert \vert \vert)$ is closed? Because in a proof I have seen pertaining to closedness of the graph of $Id$ it is automatically assumed that for $x_{n}\xrightarrow{\vert\vert\vert \cdot \vert\vert\vert}x$ it immediately follows that $x \in (\ell^{1}, \vert \vert \vert \cdot \vert \vert \vert)$. I would say this is certainly the case when looking at $(\ell^{1},\vert \vert \cdot \vert \vert_{1})$ but since the norm $\vert \vert \vert \cdot \vert \vert \vert$ can be defined in all sorts of way, I do not understand why $x \in (\ell^{1}, \vert \vert \vert \cdot \vert \vert \vert)$.