Norm and conjugate of operator on $l^p$

Question

I want to solve the following problem:

Show that for every sequence $(\alpha_n) \in l^{\infty}$ formula $$A(x_n) = (\alpha_n x_n), (x_n) \in l^p,$$ gives bounded linear operator on $l^p$, find its norm and operator $A^*: (l^{p})^* \to (l^p)^*$, such that $A^*f = fA, \forall f \in (l^{p})^*$.

I proved that it is bouned operator $$||A(x_n)||_p = (\sum_{n=1}^{\infty} |\alpha_n x_n|^p)^{1/p} \leq (sup|\alpha_n|^p)^{1/p} ||x||_p \leq ||(\alpha_n)||_{\infty}||x||_p < \infty,$$ so $||A|| \leq sup|\alpha_n|$. But also, $$ ||A|| \geq ||A(e_n)|| = |\alpha_n|,$$ where $e_n$ is sequence with 1 in $n$-th position, 0 otherwise, so $||A|| = sup|\alpha_n|$.

I don't know what to do for the third part, how to find $A^*$.

score 0 · Answer 1 · answered Mar 16 '23 at 23:18

0

If $A^{*}(x_n)=(y_n)$ then $\sum z_n\overline {y_n}= \langle (z_n), (y_n) \rangle=\langle A(z_n), (x_n) \rangle=\sum \alpha_nz_n\overline {x_n}$ for all $(z_n)$. This gives $y_n=\overline {\alpha_n} x_n$. Hence, $A^{*}(x_n)=(\overline {\alpha_n} x_n)$.

[If your sequences are real sequences then you can just remove the bars on top].

answered Mar 16 '23 at 23:18

Kavi Rama Murthy

359,332

How can you say $A^(x_n)$, when $A^$ is defined on $B(l^p, \mathbb{C})$? – Maria Mar 19 '23 at 11:08
@Maria $B(\ell^{p},\mathbb C)$ is identified with $\ell^{q}$. I am taking $(x_n) \in \ell^{q}$. – Kavi Rama Murthy Mar 19 '23 at 11:25
Oh, ok then. But I don't understand how this satisfies the condition that $A^f = fA, \forall f \in (l^p)^$. – Maria Mar 19 '23 at 11:38
That is the definition of the adjoint operator. – Kavi Rama Murthy Mar 19 '23 at 11:51

Norm and conjugate of operator on $l^p$

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