# Show that $\vert \vert \vert \cdot \vert \vert \vert$ is equivalent to $\vert \vert \cdot \vert \vert_{1}$

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?

• 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 $x=Jx=y$. – Mindlack Jul 14 at 21:36
• 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? – SABOY Jul 15 at 5:26
• By definition, $J$ has full domain. For your other question, note that $\|\cdot\|_1 \geq \|\cdot\|_{\infty}.$ – Mindlack Jul 15 at 9:21

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 $$\frac1D |||x||| \le \|x\|_1 \le C|||x|||$$