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

  • $\begingroup$ Part of the definition of that convergence is that $x \in \ell^1$? $\endgroup$ – Rhys Steele Jul 24 at 22:53
  • $\begingroup$ 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 $\vert\vert\vert x \vert \vert\vert=\infty$ $\endgroup$ – MinaThuma Jul 25 at 8:44
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    $\begingroup$ 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 $x \in \ell^1$. $\endgroup$ – Rhys Steele Jul 25 at 8:45

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