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I am struggling to find the norm of the following linear operator: $A:l_2 \rightarrow l_2, \ A(x_1,x_2,...,x_n,...) = (x_1, \frac{1}{2}x_2,...,\frac{1}{2^{n-1}}x_n,...)$

I considered $$||Ax||^2_{l_2} = \sum_{k=1}^{\infty} |\frac{1}{2^{k-1}} x_k|^2 = \sum_{k=1}^{\infty} (\frac{1}{2^{k-1}})^2| x_k|^2, \tag{1}$$ but I don't know what to do next. Bounding $(\frac{1}{2^{k-1}})^2$ above with 1 we get $$||Ax||^2_{l_2} \leq \sum_{k=1}^{\infty} |x_k|^2 = ||x||^2_{l_2},$$ which means that $||A|| \leq 1$, but this is too crude because it's easy to see that if we consider $x_0 = (\frac{1}{2}, \frac{1}{4}, ..., \frac{1}{2^n})$ we get $$||Ax_0||_{l_2}=\frac{2}{3}|x_0|_{l_2},$$ suggesting that $||A|| \leq \frac{2}{3}$. But I'm missing the way to derive that from (1). Any tips?

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2 Answers 2

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$\|Ax_0\|=\frac 2 3 \|x_0\|$ does not tell you that $\|A\| \leq \frac 2 3$. (We may even have $\|Ax_0\|=0(\|x_0\|)$ for some non-zero $x_0$ and we cannot conclude that $\|A\| \leq 0$ right?)

In general if $A(x_n)=(a_nx_n)$ then $\|Ax\| \leq \sup_n |a_n|\|x\|$ so $\|A\|\leq \sup_n |a_n|$. Also $Ae_n=a_ne_n$ so $\|A\| \geq |a_n|$ for each $n$. Hence $\|A\| \geq \sup_n |a_n|$ proving that $\|A\|=\sup_n |a_n|$. In the present case the norm is $1$.

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  • $\begingroup$ I managed to show that $||A|| \leq \frac{2}{\sqrt{3}}$, but plugging in $x=a$ yields $||Ax|| = \sqrt\frac{16}{15} = \frac{4}{\sqrt15} = \frac{2}{\sqrt3} * \frac{2}{\sqrt5} \neq \frac{2}{\sqrt3}||x||.$ Am I missing something? $\endgroup$
    – toss
    Commented Apr 22, 2021 at 1:04
  • $\begingroup$ @toss I have edited my answer. $\endgroup$ Commented Apr 22, 2021 at 5:38
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Take $e_1=(1,0,0,0,...)$. Clearly, $\lVert e_1\rVert = 1$, and $Ae_1=e_1$, so...?

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