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Let $\vert \vert \vert \cdot \vert \vert \vert$ be a norm on $\ell^{1}$ with the following properties:

$1.$ $(\ell^{1},\vert \vert \vert \cdot \vert \vert \vert)$ is Banach Space

$2.$ for all $x \in \ell^{1}$: $\vert\vert x \vert \vert_{\infty}\leq \vert \vert \vert x \vert \vert \vert$

Show, using the closed graph theorem that $\vert \vert \vert \cdot \vert \vert \vert$ is equivalent to $\vert\vert \cdot \vert\vert_{1}$.

My idea:

Define $J: (\ell^{1}, \vert \vert \vert \cdot \vert \vert \vert) \to (\ell^{1}, \vert \vert \cdot \vert \vert_{1}), x \mapsto x$

I need to show that $J$ is a closed operator, as $(\ell^{1}, \vert \vert \vert \cdot \vert \vert \vert)$ and $(\ell^{1}, \vert \vert \cdot \vert \vert_{1})$ are already Banach.

So, let $(x^{n})_{n} \subseteq (\ell^{1}, \vert \vert \vert \cdot \vert \vert \vert)$ where $x^{n} \xrightarrow{n \to \infty} x$ and $\exists y \in (\ell^{1}, \vert \vert \cdot \vert \vert_{1})$ so that $Tx^{n}\xrightarrow{n \to \infty} y$. Now how can I show that $Tx=y$

And then how do I go on to show that $(\ell^{1}, \vert \vert \vert \cdot \vert \vert \vert)$ is closed?

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  • $\begingroup$ By definition, $x^n$ converges to $x$ for $||| \cdot|||$, thus for $\|\cdot\|_{\infty}$. Now, $x^n=Jx^n$ converges to $y$ for $\|\cdot\|_1$, thus for $\|\cdot\|_{\infty}$. Thus $x=Jx=y$. $\endgroup$ – Mindlack Jul 14 at 21:36
  • $\begingroup$ Why does convergence in $\vert \vert \cdot \vert \vert_{1} \Rightarrow$ convergence in $\vert \vert \cdot \vert \vert_{\infty}$. And how do I know $\operatorname{dom}(J)$ is closed? $\endgroup$ – SABOY Jul 15 at 5:26
  • $\begingroup$ By definition, $J$ has full domain. For your other question, note that $\|\cdot\|_1 \geq \|\cdot\|_{\infty}.$ $\endgroup$ – Mindlack Jul 15 at 9:21
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We have $x^n \to x$ in $|||\cdot|||$ so $x^n \to x$ in $\|\cdot\|_\infty$. In particular $x^n \to x$ pointwise.

Also $x^n=Jx^n \to y$ in $\|\cdot\|_1$ so in particular $x^n \to y$ pointwise. Therefore $Jx = x = y$.

Hence $J$ has closed graph so $\exists C > 0$ such that $\|x\|_1 = \|Jx\|_1 \le C|||x|||$ for all $x \in \ell^1$.

On the other hand, $J$ is a bounded operator which is bijective, where the inverse is the identity $J^{-1} : (\ell^1, \|\cdot\|_1) \to (\ell^1, |||\cdot|||)$. By the Bounded Inverse theorem, $J^{-1}$ is also bounded so there exists $D > 0$ such that $$|||x|||= |||J^{-1}x||| \le D\|x\|_1$$ It follows $$\frac1D |||x||| \le \|x\|_1 \le C|||x|||$$

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