# Confusion on when we can assume $\operatorname{dom}(T)$ is closed within itself

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 $$x \in (\ell^{1}, \vert \vert \vert \cdot \vert \vert \vert)$$.

• Part of the definition of that convergence is that $x \in \ell^1$? – Rhys Steele Jul 24 at 22:53
• I thought that the fact whether $x \in \ell^{1}$ depends on the norm we may be using (e.g. $\vert \vert \vert \cdot \vert \vert \vert$), it is clear that $x \in (\ell^{1}, \vert \vert \cdot \vert \vert_{1})$ but I do not see why it is neccessarily true that $x \in (\ell^{1}, \vert\vert \vert \cdot \vert \vert \vert)$. In othere words, for me it is clear that $x_{n} \xrightarrow{\vert\vert \cdot \vert \vert_{1}} x \Rightarrow \vert\vert x \vert \vert_{1}<\infty$ but $x_{n} \xrightarrow{\vert\vert\vert \cdot \vert \vert\vert} x$ could still mean $\vert\vert\vert x \vert \vert\vert=\infty$ – MinaThuma Jul 25 at 8:44
• The elements of the space do not change when you change the norm. $\ell^1$ is a fixed vector space on which you are putting different norms. By definition of $\vert\vert\vert \cdot \vert \vert\vert$ being a norm on $\ell^1$ you know that $\vert\vert\vert x \vert \vert\vert < \infty$ for every $x \in \ell^1$. – Rhys Steele Jul 25 at 8:45