# Proof by induction of triangle inequality in Hilbert space.

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 ?

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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

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.
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