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I have a small confusion,i know how to prove $l^2(\mathbb R)$ is a vector space but i am not getting any idea to prove $l^3(\mathbb R)$ is a vector space.

Vector space is defined as

Vector space is Defined as

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  • $\begingroup$ How do you define this vector space? $\endgroup$
    – Ian Coley
    Feb 17, 2014 at 20:47
  • $\begingroup$ @IanColey The definition is in the question. $\endgroup$ Feb 17, 2014 at 21:09
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    $\begingroup$ Since it is a subset of a vector space - the space of all real sequences - you need to show that 1. it is nonempty ($0$ pretty obviously belongs to $l^3(\mathbb{R})$), 2. scalar multiples of sequences in $l^3(\mathbb{R})$ belong to $l^3(\mathbb{R})$, this one is pretty obvious too, and finally, that the sum of two elements of $l^3(\mathbb{R}$ belongs to $l^3(\mathbb{R})$. The last one is not too difficult either, but not trivial. $\endgroup$ Feb 17, 2014 at 21:09
  • $\begingroup$ @DanielFischer I know this axioms,Just in case how can i use Triangular inequality here ?! $\endgroup$
    – Nithish
    Feb 17, 2014 at 21:19
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    $\begingroup$ @Jp when I commented, the definition wasn't there, so I'm glad it is now. $\endgroup$
    – Ian Coley
    Feb 17, 2014 at 22:27

2 Answers 2

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Non-negativity and other norm properties are trivial to prove. The one which may be difficult is the triangle inequality.

To prove the triangle inequality for $\ell^p$ for $1<p<\infty$ (the cases for $p=1,\infty$ are trivial), you should know the following:

  1. Young's inequality: For $a,b>0$, $1<p,q<\infty$ where $\frac{1}{p}+\frac{1}{q}=1$, $$ab \leq \frac{a^p}{p} + \frac{b^q}{q}$$.
  2. Holder's inequality: for $1 \leq p , q \leq \infty$, $\frac{1}{p}+\frac{1}{q}=1$, $x \in \ell^p$, $y \in \ell^q$, $$ \sum_i |x_i y_i| \leq ||x||_p ||y||_q$$ (this is proved by a simple application of Young's inequality. Note for $p=q=2$, you get Cauchy-Schwarz. For $p=1$, this is a straightforward result since $q=\infty$)

Now, you can prove the triangle inquality in $\ell^p$ for $1<p<\infty$ (try $p=1,p=\infty$ on your own): $\sum_i |x_i+y_i|^p = \sum_i |x_i+y_i|^{p-1} |x_i+y_i| \leq \sum_i (|x_i+y_i|^{p-1} |x_i| + |x_i+y_i|^{p-1} |y_i|)$.

Now, apply Holder's inequality to both terms to get the upper bound $(\sum_i |x_i+y_i|^{(p-1)q})^{1/q} (\sum_i |x_i|^p)^{1/p} + (\sum_i |x_i+y_i|^{(p-1)q})^{1/q} (\sum_i |y_i|^p)^{1/p}$. Now, divide through by $(\sum_i |x_i+y_i|^p)^{1/q}$ and note that $(p-1)q=p$ and $1-\frac{1}{q}=\frac{1}{p}$ to get the result that $(\sum_i |x_i|^p)^{1/p}$ is indeed a norm.

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  • $\begingroup$ Note that when you say "divide through by $(\sum_i |x_i+y_i|^p)^{1/q}$" you are assuming that such sum is finite, which is precisely what you want to prove. Because of that is that in the proof of Minkowski's Inequality you first need to do the step I did, and then you prove the inequality as you did. $\endgroup$ Feb 18, 2014 at 0:10
  • $\begingroup$ Yes, this answers (b), while you answered (a). $\endgroup$
    – Batman
    Feb 18, 2014 at 0:40
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Suppose that $\sum_n|x_n|^3<\infty$, $\sum_n|y_n|^3<\infty$. The key inequality is $$ |a+b|^3\leq(|a|+|b|)^3\leq(2\max\{|a|,|b|\})^3=8\max\{|a|^3,|b|^3\}\leq 8(|a|^3+|b|^3). $$

Then $$ \sum_n|x_n+y_n|^3\leq 8\,\sum_n|x_n|^3+8\,\sum_n|y_n|^3<\infty. $$

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