0
$\begingroup$

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.

  • 0
    See http://math.stackexchange.com/questions/1337896/norm-of-multiplication-operator-in-ell2-bbb-n-is-x-infty?rq=1 – 2017-02-22

1 Answers 1

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

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

  • 0
    Why $||x||_{\infty} \le \sup \{||M_x(y)||: y \in l^2, ||y ||_2=1\}=||M_x||$ ? – 2017-02-22
  • 0
    $\sup \{||M_x(u_k)||: k \in \mathbb N\} \le \sup \{||M_x(y)||: y \in l^2, ||y ||_2=1\}$ – 2017-02-22
  • 0
    $\{||M_x(u_k)||: k \in \mathbb N\} \subset \{||M_x(y)||: y \in l^2, ||y ||_2=1\}$ wasn't it a $\geq$ instead of $\leq$? – 2017-02-22
  • 0
    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 $\sup A \le \sup B$. – 2017-02-22
  • 0
    oh yes, I made a stupid mistake. Thank you. – 2017-02-22