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How do I find the norm of the following operator i.e. how to find $\lVert T_z\rVert$ and $\lVert l\rVert$?

1) Let $z\in \ell^\infty$ and $T_z\colon \ell^p\to\ell^p$ with $$(T_zx)(n)=z(n)\cdot x(n).$$ What my thoughts were to use Banach-Steinhaus theorem but it seems straight forward and I don't know if I am right.

$\lVert T_z\rVert _p \leqslant\lVert z\lVert \cdot n\cdot\lVert x\rVert_p n=n^2\lVert x\rVert _p$ so if I choose $x=1$ then I get $\lVert T_z\rVert =n^2$.

2) Let $0\leqslant t_1\leqslant\cdots\leqslant t_n=1$ and $\alpha_1,\dots,\alpha_n \in K$ , $l\colon C([0,1])\to K$ with $l(x)=\sum_{i=1}^n \alpha_i x(t_i)$.

How to I find operator norm in this case as well? I am quite sure I am not right. I would be glad if I could get some help. Definitely some hints would be great! Thanks in advance.

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  • $\begingroup$ I'm sorry, could you elaborate where you applied Banach-Steinhaus? And also how you get the other direction, $\|T_z\| \geq n^2$? $\endgroup$ Nov 12, 2012 at 22:19
  • $\begingroup$ @MattN. : But that means the operator is not bounded isn't it ? I am not exactly getting what i should be looking for in this case . $\endgroup$
    – Theorem
    Nov 12, 2012 at 22:20
  • $\begingroup$ @MattN. : That was just a thought but i don't know if i should be thinking in that direction . $\endgroup$
    – Theorem
    Nov 12, 2012 at 22:23
  • $\begingroup$ Well, "finding the operator norm" mean you have to show $\|T\|=K$ for some $K$, so you have to show $\geq$ and $\leq$. $\endgroup$ Nov 13, 2012 at 6:35

1 Answer 1

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  1. As for each $n$, $|z(n)x(n)|^p\leqslant \lVert a\rVert_{\infty}|x(n)|^p$, then we certainly have $\lVert T_z\rVert\geqslant \lVert a\rVert_{\infty}$. To get the other inequality, fix $\delta$ and pick $k$ such that $|a(k)|\geqslant \lVert a\rVert_{\infty}-\delta$ (the case $a=0$ is obvious).

  2. We assume $t_j$ distinct. Let $f_j$ a continuous map such that $f_j(t_j)=e^{i\theta_j}$, where $e^{i\theta_j}\alpha_j=|\alpha_j|$ and $f_j(t_k)=0$ if $k\neq j$. We can choose the $f_j$'s such that $\lVert \sum_{j=1}^nf_j\rVert=1$.

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  • $\begingroup$ Davide I am confused with the fact that when i see it , it looks to me that the operator acts on $x$ and $n$ so ie there are lots of variables involved and applying direct definition seems quite confusing . Can you give me some insight ? $\endgroup$
    – Theorem
    Nov 12, 2012 at 23:02
  • $\begingroup$ The operator $T_z$ acts on sequence: it transforms a sequence into a sequence. To define it, we have to say how each term of the sequence is transforms. In the OP, your bounds are too large to conclude. $\endgroup$ Nov 12, 2012 at 23:03
  • $\begingroup$ that means $z(n)$ and $x(n)$ are actually sequences right ? ie in this case $T_z$ is simply taking $x(n)\to z(n)x(n)$ where $z(n)$ is some sequence of functions. is my understanding correct about the map ? $\endgroup$
    – Theorem
    Nov 12, 2012 at 23:45
  • $\begingroup$ $z(n)$ is a real number, $\{z(n)\}_{n=1}^{+\infty}$ is a sequence. $\endgroup$ Nov 13, 2012 at 10:11
  • $\begingroup$ I have tried to understand but i didn't get your proof :( . Can you make it a bit more elaborate . Thank you . $\endgroup$
    – Theorem
    Nov 13, 2012 at 13:28

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