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Can someone help me with this one?

What is the norm of this functional on $l^2$?

$$f(x) = \sum_{n=1}^\infty \frac{3^n\cdot x_n}{\sqrt{(n+1)!}}, \qquad x=(x_1,x_2,\ldots)\in l^2. $$

It is easy to see this functional is bounded ($\|f(x)\|\leq \sum_{n=1}^\infty \frac{3^n}{\sqrt{(n+1)!}} \|x_n\|$ and the series converges).

But I can't compute the exact norm. Any help appreciated.

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What are your spaces? $\|f\| = \sup \{ |f(x)|: \|x\|=1\}.$ – user29999 Dec 11 '12 at 17:28
Is your functional $f: l² \rigtharrow l²?$ – user29999 Dec 11 '12 at 17:31
I changed $||x||$ to $\|x\|$. – Michael Hardy Dec 11 '12 at 19:54
@MichaelHardy, could you please take a look at… Nobody really knows what to do with this guy. From your "activity" you may have been asleep. – Will Jagy Dec 11 '12 at 21:57
up vote 1 down vote accepted

This functional is of the form $f(x) = \langle u,x \rangle$, where $u \in \ell^2$. What is $u$? What happens if you take $x = u$?

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Thank you. This lead me to the solution. – Uroš Dec 11 '12 at 19:07

As dual of $ l² $ is $ l² $, for this make sense we have that $ \Bigl(\frac{3^n}{\sqrt{(n+1)!}}\Bigr)_{n=1}^{\infty} \in l^2 $. Then the norm of this functional is $$ \| \Bigl(\frac{3^n}{\sqrt{(n+1)!}}\Bigr)_{n=1}^{\infty} \|_{l²} = \sum_{n=1}^{\infty} \frac{3^{2n}}{(n+1)!} = \sum_{n=1}^{\infty} \frac{9^{n}}{(n+1)!}. $$

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but this sum is greater than the number that is surely upper bound for this norm... – Uroš Dec 11 '12 at 18:11
I think the norm is square root of your sum. Do you agree? – Uroš Dec 11 '12 at 19:11
How can you see easily that $\|f(x)\|\leq \sum_{n=1}^\infty \frac{3^n}{\sqrt{(n+1)!}} \|x_n\|$? What are the norms $\| \cdot \|$? If your norm for $ x = (x_1, x_2, \cdots ) \in l^2 $ is $ \|x\|_{l^2} $ your estimate is wrong. if your functional is from $ l² $ then your funcional is $ f(x) = \langle x,\Bigl(\frac{3^n}{\sqrt{(n+1)!}}\Bigr)_{n=1}^{\infty} \rangle_{l²} $ and its norm is $ \| \Bigl(\frac{3^n}{\sqrt{(n+1)!}}\Bigr)_{n=1}^{\infty} \|_{l²} $. – user29999 Dec 12 '12 at 16:15

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