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Consider the normed vector space $c_{00}(\mathbb N ) = \{ x=(x_n) \in \mathbb{R} ^{\mathbb N} : \{ x_n \neq 0 \} \quad \text {is finite} \}$. Let $Id$ the identity operator. Let $ X =( c_{00} , \| \cdot\|_1)$ and $ Y= (c_{00} , \| \cdot\|_2)$, where $\|\cdot\|_p$ is the $\ell^p$-norm $p=1,2$; i.e. $\|(x_n)\|_p=\left(\sum |x_n|^p\right)^{1/p}$ $p=1,2$.

(a) Does $Id \in \mathcal{B}(X,Y)$ ?

(b) Does $ Id \in \mathcal{B}(Y,X)$ ?

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@DavideGiraudo: You mean to take $ x=(1,1,0,0, \cdots)$ and $||x||_1=2$ and $||x||_2 =\sqrt 2$. And from here? – passenger Feb 21 '12 at 20:32
up vote 7 down vote accepted

Denote by $e_k$ the sequence all of whose terms are $0$ except the term $n$ which is $1$. Let $x_n:=\sum_{k=1}^ne_k$. Then $||x_n||_2=\sqrt n$ and $||x_n||_1=n$, so $Id\colon Y\to X$ is not continuous.

For $a=(a_1,\ldots,a_n,0,\ldots,)$, we have $$||a||_1^2=\left(\sum_{k=1}^n|a_k|\right)^2=\sum_{k=1}^na_k^2+\sum_{i\neq j}|a_i|\cdot |a_j|\geq \sum_{k=1}^na_k^2=||a||^2_2,$$ which gives the continuity of $Id$ from $X$ to $Y$.

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So $ Id : X \to Y $ is bounded with $ ||Id|| \leq 1$. Thank you for your time. – passenger Feb 21 '12 at 21:05
You're welcome. In fact you can check that the norm is in fact $1$, taking $x=e_1$ for example. – Davide Giraudo Feb 21 '12 at 21:06

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