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Does Pythagorean law holds in Hilbert space or plane?

I know that Pythagorean law holds in Euclidean plane while Hilbert space is a generalisation of Euclidean space.

I know that finite orthonormal set $S$ in a hilbert space $H$, then Pythagorean law holds.

Can someone explain it elaborately whether Pythagorean law holds in Hilbert plane.

MAS
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1 Answers1

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Let $S= \{x_1, \ldots, x_n\}$ be a finite orthogonal set in an inner product space. We have $$\left\|\sum_{i=1}^n x_i\right\|^2 = \left\langle \sum_{i=1}^n x_i, \sum_{j=1}^n x_j\right\rangle = \sum_{i=1}^n \sum_{j=1}^n \langle x_i, x_j\rangle = \sum_{i=1}^n \sum_{j=1}^n \|x_i\|^2\delta_{ij} = \sum_{i=1}^n \|x_i\|^2$$ which is the Pythagorean identity.

mechanodroid
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  • Sir, I saw for finite orthonormal set it holds. My question whether Pythagorean law holds in Hilbert space or not. – MAS Mar 06 '19 at 12:31
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    @M.A.SARKAR It holds in arbitrary inner product spaces so in particular it holds in Hilbert spaces, yes. – mechanodroid Mar 06 '19 at 12:32
  • Sir, Ok, then why we are using the restriction of finite orthogonal set ? – MAS Mar 06 '19 at 15:36
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    @M.A.SARKAR Because for arbitrary orthogonal sets ${x_\lambda}{\lambda \in \Lambda}$ the sum $\sum{\lambda \in \Lambda} x_\lambda$ need not converge. If it does converge, then the same proof as above yields that $\sum_{\lambda \in \Lambda} |x_\lambda|^2 < +\infty$ and $$\left|\sum_{\lambda \in \Lambda} x_\lambda\right|^2 = \sum_{\lambda \in \Lambda} |x_\lambda|^2$$ You also need to use continuity of the inner product for this. – mechanodroid Mar 06 '19 at 16:24
  • Sir, great answer from you. I verified from Wikipadea also. Now my question is what is the difference between Hilbert space and Hilbert plane? Secondly, Does Pythagorean theorem holds in Hilbert plane also ? – MAS Mar 07 '19 at 07:10
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    @M.A.SARKAR If I'm not mistaken, Hilbert plane is simply the standard Euclidean plane founded by David Hilbert's axioms from $1899$. So the Pythagorean theorem certainly holds there. On the other hand, a Hilbert space is a complete inner product space, which is completely abstract. But, as we showed above, what we call the Pythagorean theorem still holds. – mechanodroid Mar 07 '19 at 11:39
  • Sir, ok nice. So Hilbert;s axioms is included in Hilbert space. Is not it? – MAS Mar 08 '19 at 05:47
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    @M.A.SARKAR Not really, a Hilbert space is something much more abstract and general. Besides, a Hilbert space doesn't have to be two-dimensional like the Euclidean plane. And it can be over the field $\mathbb{C}$, not only $\mathbb{R}$. – mechanodroid Mar 08 '19 at 07:35
  • very nice sir. I got it. My last request to you- Can you give an example of Hilbert Space which does not satisfy Hilbert's axiom? For example $L[a,b]$, the set of all square integrable functions on $[a,b]$ is a Hilbert Space. How to show that this space does not hold Hilbert's axiom? or Any other counter-example. – MAS Mar 08 '19 at 09:28
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    @M.A.SARKAR Well, $L^2[a,b]$ is an infinite-dimensional space so in particular, it cannot be a plane. This question discusses that Hilbert spaces are actually direct generalizations of the Euclidean plane to $n$ dimensions, or even infinite dimensions. Otherwise, I'm not sure. – mechanodroid Mar 08 '19 at 10:33
  • ok , You helped a lot. If you in future think about a counter-example, then you may include it here. Again a lot thanks to help me – MAS Mar 08 '19 at 10:55