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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).

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Is this homework? –  Aryabhata Sep 5 '10 at 20:11
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Nope, it's a side remark that I am trying to work out in a book that I am reading. –  user1736 Sep 5 '10 at 20:17
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up vote 12 down vote accepted

You are right @user1736.

If $a\leq1$ then $$\left(\sum |a_n|\right)^a\leq \sum|a_n|^a.\tag{1}$$

Hence if $p\leq q$ we have $p/q\leq1$ $$\left(\sum_n |x_n|^q\right)^{1/q}=\left(\sum_n |x_n|^q\right)^{p/qp}\leq \left(\sum_n |x_n|^{q(p/q)}\right)^{1/p}=\left(\sum|x_n|^p\right)^{1/p}$$


Edit: How do we prove (1) (for $0<a<1$)?

Step 1. It is sufficient to prove this for finite sequences because then we may take limits.

Step 2. To prove the statement for finite sequences it is sufficient to prove $$(x+y)^a\leq x^a+ y^a,\quad\text{ for $x,y>0$}\tag{2}$$ because the finite case is just iterations of (2).

Step 3. To prove (2) it suffice to prove $$(1+t)^a\leq 1+t^a,\quad\text{ where $0<t<1$} \tag{3}$$

Now, the derivative of the function $f(t)=1+t^a-(1+t)^a$ is given by $f'(t) = a(t^{a-1} - (1+t)^{a-1})$ and which is positive since $a>0$ and $t\mapsto t^b$ is decreasing for negative $b$. Hence, $f(t)\geq f(0)=0$ for $0<t<1$, which proves (3).

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Where did you learn the first inequality? –  newbie Jun 13 '13 at 9:36
    
@newbie I really don't remember, but it is not that difficult to prove. Added some steps. –  AD. Jun 13 '13 at 12:40
    
@newbie If you are interested you might search for "locally bounded spaces" and "quasi-normed linear spaces". –  AD. Jun 13 '13 at 13:38
    
Thanks for your clarification. Do you mean that $\ell^p$ with $p\in(0,1)$ is quasi-normed? –  newbie Jun 13 '13 at 13:48
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@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). –  AD. Jun 13 '13 at 14:50
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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:

  1. if $p\leq q$, then for $|x|>1$ you have $|x|^p\leq|x|^q$, but if $|x|<1$, then $|x|^p \geq |x|^q$.
  2. $\sum_{n=1}^{\infty}a_n$ converges if and only if for every $m\geq 1$, $\sum_{n=m}^{\infty}a_n$ converges.
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That makes sense. Thanks a lot! Is the inequality that I wrote down true though? –  user1736 Sep 5 '10 at 20:24
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Arturo, it is not too strong, as you can see in my answer. –  AD. Jan 21 '11 at 12:40
    
This logic is easy to think, so useful. –  Dutta Feb 9 at 11:29
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