Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've made proof by induction over $n$ for triangle inequality : $\left \| x+y \right \|_{e}\leq \left \| x \right \|_{e}+\left \| y \right \|_{e}$ ,where $\left \| x \right \|_{e}=\sqrt{\sum_{i=1}^{n}x_{i}^{2}}$ for $x\in \mathbb{R}^{n}$. Is that proof also valid for triangle inequality for $\left \| x \right \|=\sqrt{\sum_{i=1}^{\infty}x_{i}^{2}}$ where $x\in \ell^2, \ell^2=\left\{{x\in\mathbb{R}^{\infty}}:\left \|x\right\|^{2}<\infty\right\}$ ? Maybe I should write down that proof, but I don't believe that induction proof could be also valid in case of infinite spaces.

$\left \| x+y \right \|_{e}\leq \left \| x \right \|_{e}+\left \| y \right \|_{e}$

$\sum_{i=1}^{n}(x_{i}+y_{i})^{2}\leq \sum_{i=1}^{n}(x_{i})^{2}+\sum_{i=1}^{n}(y_{i})^{2}+2\sqrt{\sum_{i=1}^{n}(x_{i})^{2}\sum_{i=1}^{n}(y_{i})^{2}}$

$\sum_{i=1}^{n}x_{i}y_{i}\leq \sqrt{\sum_{i=1}^{n}(x_{i})^{2}\sum_{i=1}^{n}(y_{i})^{2}}$

this is true when the true is that :

$\sum_{i=1}^{n}\left |x_{i}y_{i}\right |\leq \sqrt{\sum_{i=1}^{n}(x_{i})^{2}\sum_{i=1}^{n}(y_{i})^{2}}$

above inequality is true for $n=1$ and we assume that it's true for $n$.

For $n+1$ we get : $\sum_{i=1}^{n}\left |x_{i}y_{i}\right |+\left |x_{n+1}y_{n+1}\right |\leq \sqrt{\sum_{i=1}^{n}(x_{i})^{2}\sum_{i=1}^{n}(y_{i})^{2}+x_{n+1}^{2}\sum_{i=1}^{n}(y_{i})^{2}+y_{n+1}^{2}\sum_{i=1}^{n}(x_{i})^{2}+x_{n+1}^{2}y_{n+1}^{2}}$

using induction assumption we get :

$\sum_{i=1}^{n}\left |x_{i}y_{i}\right |+\left |x_{n+1}y_{n+1}\right |\leq \sqrt{\sum_{i=1}^{n}(x_{i})^{2}\sum_{i=1}^{n}(y_{i})^{2}}+\left |x_{n+1}y_{n+1}\right |\leq \sqrt{\sum_{i=1}^{n}(x_{i})^{2}\sum_{i=1}^{n}(y_{i})^{2}+x_{n+1}^{2}\sum_{i=1}^{n}(y_{i})^{2}+y_{n+1}^{2}\sum_{i=1}^{n}(x_{i})^{2}+x_{n+1}^{2}y_{n+1}^{2}}$

now we take cube of both sides and annihilate what we can, in result we get that 0=<(...+...)^2 so inequality is hold.

So is this proof valide for infite spaces ?

share|improve this question
    
Did you mean square ? –  checkmath May 7 '12 at 22:11
    
You need to do a last inequality saying that the geometric average is lesser than or equal to the Arithmetic average. –  checkmath May 7 '12 at 22:14

1 Answer 1

up vote 0 down vote accepted

Well you result is true for all $n$ natural so the inequality must hold for the limits! That is what you want.

$$\sqrt{\sum_{i=1}^{n}(x_i+y_i)^2}\leq \sqrt{\sum_{i=1}^{n}x_i^2} +\sqrt{\sum_{i=1}^{n}y_i^2}$$

All the sequence here are increasing so taking the limits when $n \to \infty$ we get the result desired.

share|improve this answer
    
so information about that we are in l2 space is worth nothing ? so it's also true for Hilbert cube Q ? –  Qbik May 7 '12 at 11:49
    
Meditate about when x or y aren't in $\ell^2$ –  checkmath May 7 '12 at 11:52
    
You may also see this equation by noticing that $\langle x\pm y,x\pm y\rangle \geq 0$. If you mean for the Hilbert cube the supremo norm, yeah that is true. But $\ell^{\infty}$ is a totally different space that cannot be said Hilbert. –  checkmath May 7 '12 at 11:55
    
You are using induction wrong because you have to suppose the result true for $n$ and the prove it for $n+1$. –  checkmath May 7 '12 at 12:24
    
you look again on my proof ? –  Qbik May 7 '12 at 14:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.