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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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You are right @user1736.

If $0<a\leq1$ then $$\left(\sum |a_n|\right)^a\leq \sum|a_n|^a.\tag{1}$$

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


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 it 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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  • $\begingroup$ Where did you learn the first inequality? $\endgroup$ – newbie Jun 13 '13 at 9:36
  • $\begingroup$ @newbie I really don't remember, but it is not that difficult to prove. Added some steps. $\endgroup$ – AD. Jun 13 '13 at 12:40
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    $\begingroup$ @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). $\endgroup$ – AD. Jun 13 '13 at 14:50
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    $\begingroup$ Can you explain why to prove (2) it suffices to prove (3)? I can't see it. $\endgroup$ – DrHAL Sep 10 '16 at 23:25
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    $\begingroup$ @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 $y/x<1$. $\endgroup$ – AD. Sep 12 '16 at 7:40
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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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  • $\begingroup$ That makes sense. Thanks a lot! Is the inequality that I wrote down true though? $\endgroup$ – user1736 Sep 5 '10 at 20:24
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    $\begingroup$ Arturo, it is not too strong, as you can see in my answer. $\endgroup$ – AD. Jan 21 '11 at 12:40
  • $\begingroup$ This logic is easy to think, so useful. $\endgroup$ – Dutta Feb 9 '14 at 11:29
  • $\begingroup$ @Dutta Could you explain how to proceed using these hints? I do not see how to prove it using this. $\endgroup$ – Soap Mar 29 '17 at 20:03
  • $\begingroup$ @Simoes I am out of touch with functional analysis nowadays. You may go through a standard undergraduate textbook on functional analysis. $\endgroup$ – Dutta Mar 30 '17 at 5:25

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