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I have a normed space: $$l_p = \{ (x_n)_{n = 1}^{\infty}: \sum\limits_{n = 1}^{\infty}|x_n|^{p} < \infty \}$$ With norm: $$||x|| = (\sum\limits_{n = 1}^{\infty} |x_n|^p)^{1/p}$$ where $p = \dfrac{5}{4}$

And I want to find norm of following functional:

$$f(x) = -x_{1} + x_{100} + \sum\limits_{n = 1}^\infty \dfrac{x_{2n}}{n!}$$

I was try apply Hölder's inequality, but not shure it was right:

$$|f(x)| \le ||x||_p ||g||_q, q = 5$$ and $g$ looks like:

$$g(x) = (-1,\dfrac{x_2}{1!}, 0, \dfrac{x_4}{2!}, 0, \ldots, \dfrac{(50! + 1) x_{100}}{50!}, 0, \dfrac{x_{101}}{51!}, \ldots)$$

But I have no idea how to move forward.

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1 Answer 1

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Mimicing proof from this answer you can show that the norm of functional $f$ equals to the $q$-norm of the sequence $$ a=\left(-1,\frac{1}{1!},0,\frac{1}{2!},0,\ldots,0,\frac{1}{49!},0,1,0,\frac{1}{51!},0,\ldots,0,\frac{1}{n!},0\ldots\right) $$ In other words $$ \Vert f\Vert=\left(|-1|^5 + \left|\frac{1}{1!}\right|^5+\left|\frac{1}{2!}\right|^5+\ldots+\left|\frac{1}{49!}\right|^5+1^5+\left|\frac{1}{51!}\right|^5+\ldots\right)^{1/5}\\ =\left(2-\frac{1}{(50!)^5}+\sum\limits_{n=1}^\infty\frac{1}{(n!)^5}\right)^{1/5} $$

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  • $\begingroup$ It works. Thanks a lot! $\endgroup$
    – piu
    Nov 19, 2013 at 14:05

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