# Find the norm of an operator on $\ell_2$

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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Are you assigning homework to us? Please don't post in the imperative and without showing any effort of your own. Some people find that kind of post rather rude. –  kahen Nov 10 '12 at 11:19
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 :) –  hncas Nov 10 '12 at 11:40

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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Ok, I see :). Should've seen that :D. Thanks a lot :) –  hncas Nov 10 '12 at 11:41
Quick question: writing $T(x) = <a,x>_{l^2}$ is using writing it in the Riesz Representation? Looking at the norm of a is always the way of finding the operator norm? –  Lost1 Nov 21 '12 at 22:04
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 $T(x)=\langle a,x\rangle$. –  Davide Giraudo Nov 21 '12 at 22:09
$$L(x) = \left\langle x,\Bigl((n(n+1))^{-1/2}\Bigr)\right\rangle = \langle x,\xi\rangle.$$
$$\lVert L\rVert^2 = \lVert \xi\rVert_2^2 = \sum_{n=1}^\infty \left|\frac1{\sqrt{n(n+1)}}\right|^2 = \sum_{n=1}^\infty \frac1{n(n+1)}.$$