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We have $L: \ell^{2} \to \mathbb{K},$ where $\mathbb{K}$ denotes real or complex space. And $L(x) = \sum_{n=1}^{\infty}\frac{x_{n}}{n^{q}}.$ What I've tried so far is the following.

$ \|L(x)\| = \| \sum_{n=1}^{\infty}\frac{x_{n}}{n^{}} \| \leq \sum_{n=1}^{\infty} \| \frac{x_{n}}{n^{}} \| = \sum_{n=1}^{\infty} \lvert \frac{1}{n^{}} \rvert \lvert x_{n}\rvert.$ But I'm not sure how to proceed from here. Any help is appreciated.

1 Answers1

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I think you are nearly there and just need to use the Cauchy Schwarz inequality. In particular, $$||L(x)||\leq \sum_{n=1}^\infty \left|\frac{1}{n}\right|\, |x_n|\leq ||x||_{l^2} \left(\sum_{n=1}^\infty \left|\frac{1}{n}\right|^2\right)^{1/2}=||x||_{l^2}\left(\sum_{n=1}^\infty\frac{1}{n^2}\right)^{1/2}<\infty.$$ Since $x\in l^2$ and $\sum_{n=1}^\infty 1/n^2$ converges by the $p$ test!

JayP
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  • Cauchy Schwarz is Holder's inequality with $p=q=2$. In this case $\sum_{n=1}^\infty |1/n||x_n|\leq \left(\sum_{n=1}^\infty |1/n|^2\right)^{1/2} \left(\sum_{n=1}^\infty |x_n|^2\right)^{1/2}=||x||{l^2}||y||{l^2}$ where $y=(1/n)_{n=1}^\infty$. Since $x,y\in l^2$ indeed $L$ is bounded when $q=1$. Are you asking for the result for arbitrary $q$? A similar argument should work. – JayP Feb 05 '23 at 03:18
  • But we're assuming that $q=1$; that's what's confusing me. So how can we set $q=2$ and apply Cauchy-Schwarz? –  Feb 05 '23 at 03:23
  • Sorry by previous comment was confusing. If we are only considering the case that $q=1$ then don't worry about Holder's inequality but rather look at https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality. Holder's inequality is a generalisation of this. This is what I have been applying in the solution. – JayP Feb 05 '23 at 03:26
  • Yes, Cauchy-Schwarz makes more sense here. Thank you. –  Feb 05 '23 at 03:56