Show that the norm of multiplicative linear operator $M_x :l^2\to l^2$ is $||x||_{\infty}$

Question

For each $x\in l^{\infty}$, define $M_x :l^2\to l^2$ by $M_x(y)(k)=x(k)y(k)$, $k=1,2,...$ for $y\in l^2$.Show that $||M_x||=||x||_{\infty}$

I know how to do the easy part: $||M_X||\leq ||x||_{\infty}$, but have no idea how to prove $||M_X||\geq ||x||_{\infty}$, could you please give me some hints? Thank you.

See http://math.stackexchange.com/questions/1337896/norm-of-multiplication-operator-in-ell2-bbb-n-is-x-infty?rq=1 — , Feb 22 '17 at 12:10

score 3 · Accepted Answer · answered Feb 22 '17 at 12:13

3

Let $u_1,u_2,...$ the usual orthonormal basis of $l^2$.

Then $M_x(u_k)=(0,...,0,x_k,0,...)$, hence $||M_x(u_k)||=|x_k|$, therefore

$$ \sup \{||M_x(u_k)||: k \in \mathbb N\}=||x||_{\infty},$$

thus

$$||x||_{\infty} \le \sup \{||M_x(y)||: y \in l^2, ||y ||_2=1\}=||M_x||$$

answered Feb 22 '17 at 12:13

Fred

78,422

Why $||x||_{\infty} \le \sup {||M_x(y)||: y \in l^2, ||y ||_2=1}=||M_x||$ ? – Chan Shue Long Feb 22 '17 at 13:21
$\sup {||M_x(u_k)||: k \in \mathbb N} \le \sup {||M_x(y)||: y \in l^2, ||y ||_2=1}$ – Fred Feb 22 '17 at 13:25
${||M_x(u_k)||: k \in \mathbb N} \subset {||M_x(y)||: y \in l^2, ||y ||_2=1}$
wasn't it a $\geq$ instead of $\leq$?
– Chan Shue Long Feb 22 '17 at 13:30
If $A$ and $B$ subsets or $\mathbb R$, $A \subseteq B$, and if $B$ is bounded from above, then $A$ is bounded from above and $\sup A \le \sup B$. – Fred Feb 22 '17 at 13:34
oh yes, I made a stupid mistake. Thank you. – Chan Shue Long Feb 22 '17 at 13:36

Show that the norm of multiplicative linear operator $M_x :l^2\to l^2$ is $||x||_{\infty}$

1 Answers1