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Let $(x_n) \subset \ell_2$ and let operator $L:\ell_2\to \mathbb R$ be defined by:

$\displaystyle L((x_n)) := \sum_{n=1}^{\infty} \frac{x_n}{\sqrt{n(n+1)}}$.

Find the norm of L.

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    Sorry, i did try to work on it but got a bit stuck. It's not a homework, i'm just exercising before the test and am running out of time. Next time I'll try to show my work when asking a question $:$)2012-11-10

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Let $a\in\ell^2$ the sequence defined by $a(n):=\frac 1{\sqrt{n(n+1)}}$. Then $T(x)=\langle a,x\rangle_{\ell^2}$.

Cauchy-Schwarz inequality gives that $\lVert T\rVert=\lVert a\rVert_{\ell^2}= \sum_{n=1}^{+\infty}\frac{n+1-n}{n(n+1)}=\sum_{n=1}^{+\infty}\left(\frac 1n-\frac 1{n+1}\right)=1.$

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    It's easier in the case of$a$linear functional on a Hilbert space. Here, we just use the definition of $T$ to find the vector $a$ such that $T(x)=\langle a,x\rangle$.2012-11-21
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$L(x) = \left\langle x,\Bigl((n(n+1))^{-1/2}\Bigr)\right\rangle = \langle x,\xi\rangle.$

So by the Riesz representation theorem for Hilbert spaces we have

$\lVert L\rVert^2 = \lVert \xi\rVert_2^2 = \sum_{n=1}^\infty \left|\frac1{\sqrt{n(n+1)}}\right|^2 = \sum_{n=1}^\infty \frac1{n(n+1)}.$

You should be able to sum the series yourself.

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    Yes, I can :). Thanks a lot :)2012-11-10