I can't seem to work out the inequality $(\sum |x_n|^q)^{1/q} \leq (\sum |x_n|^p)^{1/p}$ for $p \leq q$ (which I'm assuming is the way to go about it).
How do you show that $l_p \subset l_q$ for $p \leq q$?
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analysis
functional-analysis
inequality
lp-spaces
2 Answers
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You are right @user1736.
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0Where did you learn the first inequality? – 2013-06-13
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0@newbie I really don't remember, but it is not that difficult to prove. Added some steps. – 2013-06-13
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0@newbie If you are interested you might search for "locally bounded spaces" and "quasi-normed linear spaces". – 2013-06-13
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0Thanks for your clarification. Do you mean that $\ell^p$ with $p\in(0,1)$ is quasi-normed? – 2013-06-13
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1@newbie Some references are Stefan Rolewicz *Metric Linear Spaces* and Wiesław Żelazko *Metric generalizations of Banach algebras*. An other common name is $p$-norm, while quasi-norm is something like $\|x + y\|\leq K(\|x\| +\|y\|)$ which is *equivalent* to a $p$-norm. So, "Yes" - is a sense - at least there is a strong connection. (Very interesting algebras, if I may say so, many things that works for $p\geq1$ are not true while others are but needs different kind of proofs). – 2013-06-13
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1Can you explain why to prove (2) it suffices to prove (3)? I can't see it. – 2016-09-10
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0I also cant see that – 2016-09-11
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5@DrHAL Suppose we knew (3), and suppose $x>y$, then $$(x+y)^a=x^a(1+(y/x))^a \leq x^a(1 + (y/x)^a) = x^a + y^a$$ because $y/x<1$. – 2016-09-12
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0@LuizFernando Is this better? – 2016-09-12
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I don't think you need to prove the inequality you have in the question; that's a bit too strong. Note that $\{x_n\}\in\ell_p$ if and only if $\left(\sum|x_n|^p\right)^{1/p}$ is finite, if and only if $\sum|x_n|^p\lt\infty$. So you really just need to show that if $\sum|x_n|^p$ is finite, then $\sum|x_n|^q$ is finite, assuming $p\leq q$.
You want to remember is two things:
- if $p\leq q$, then for $|x|>1$ you have $|x|^p\leq|x|^q$, but if $|x|<1$, then $|x|^p \geq |x|^q$.
- $\sum_{n=1}^{\infty}a_n$ converges if and only if for every $m\geq 1$, $\sum_{n=m}^{\infty}a_n$ converges.
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0That makes sense. Thanks a lot! Is the inequality that I wrote down true though? – 2010-09-05
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1Arturo, it is not too strong, as you can see in my answer. – 2011-01-21
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0This logic is easy to think, so useful. – 2014-02-09
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0@Dutta Could you explain how to proceed using these hints? I do not see how to prove it using this. – 2017-03-29
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0@Simoes I am out of touch with functional analysis nowadays. You may go through a standard undergraduate textbook on functional analysis. – 2017-03-30