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We know all isometries of $\mathbb R^n $ are composition of transfer by orthogonal linear functions.

How to find all surjective isometries of Hilbert space?

Is there similarity?

Daejvad
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    The unitary operators on $H$ form a group under composition, which is the isometry group of $H$ (see http://en.wikipedia.org/wiki/Hilbert_space). – Dietrich Burde Aug 07 '14 at 20:21
  • @Dietrich: That is the group of bijective, linear isometries. Daejvad has not indicated that bijectiveness is assumed, and from context appears to be not only considering linear transformations. – Jonas Meyer Aug 07 '14 at 20:28
  • @Daejvad: The accepted answer appears to not answer your question. Could you please clarify how it does? – Jonas Meyer Aug 07 '14 at 20:41
  • I was searching similarity. indeed the question was for all surjective isometries of Hilbert – Daejvad Aug 07 '14 at 21:57
  • @Daejvad: When I posted that comment the answer did not answer even that question, and it had been pointed out in comments, but then was accepted anyway. Hence my question to you. Subsequently, you may see that after 6 or so iterations it has gotten to an answer to the question with the additional assumption of surjectiveness, before you had said that you were assuming surjectiveness. – Jonas Meyer Aug 07 '14 at 22:00
  • @JonasMeyer: my homework was asking onto isometries.I thought each one to one is surjective in this case , so i didn't mentioned it in the question. I accepted the answer immediately without thinking for it's correctness. – Daejvad Aug 07 '14 at 22:16

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norm induced by inner product is strictly subadditive.

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By mazur-ulam theorem every surjective isometry of hilbert space is composition of a transfer and a linear function. So every surjective isometry of hilbert spaces are determined.

Source: Functional Analysis of Peter.Lax page 47 for theorem 9 and page 61 for theorem 10

Fin8ish
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    Daejvad appears to be not asking about linear functions only. This is also not correct for linear operators, but if that were the question, the answer could be corrected by changing the last word to "set". (Isometries on infinite dimensional Hilbert space need not be surjective.) – Jonas Meyer Aug 07 '14 at 20:26
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    masoud: OK, I'll reiterate with direct evidence this time. (1) This is false even for linear operators, e.g., because there are nonsurjective isometries. If ${e_n}{n=1}^\infty$ is an onb, then the operator $\sum_n a_n e_n\to \sum_n a_ne{n+1}$ is a linear isometry, but it sends the onb ${e_n}{n=1}^\infty$ to the orthonormal set ${e_n}{n=2}^\infty$, which is not complete, hence not an onb. (2) Sillier counterexamples for the nonlinear case include $x\mapsto x+a$ for fixed $a$ in the space. So it is not clear to me what you mean to say. [this was in response to a deleted comment] – Jonas Meyer Aug 07 '14 at 20:35
  • @JonasMeyer: sry, I corrected own answer. – Fin8ish Aug 07 '14 at 20:47
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    I do not see any change in meaning after your edit. The first part is still only true under the assumption that $f$ is linear, and the second part is still not true even for linear functions in general, unless you are assuming surjectiveness. – Jonas Meyer Aug 07 '14 at 20:49
  • Now that edit helps, but it is still assuming surjectiveness, which was never an assumption in Daejvad's question. – Jonas Meyer Aug 07 '14 at 20:59
  • Now that edit helps even further, but you're still missing the point that a linear isometry need not be surjective. – Jonas Meyer Aug 07 '14 at 21:04
  • @JonasMeyer: you are right, but theorem 10 above says it is surjective automatically. – Fin8ish Aug 07 '14 at 21:12
  • masoud: No, theorem 10 says that every linear isometry is obtained in this way if it is assumed to be surjective. I already posted an explicit example in the second comment above showing that linear isometries need not be surjective. – Jonas Meyer Aug 07 '14 at 21:14
  • @JonasMeyer: oh sry , I was in a bad mistake. – Fin8ish Aug 07 '14 at 21:16
  • What is the source of the images you have posted here? – Jonas Meyer Aug 07 '14 at 21:36
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    @Jonas Meyer: Functional Analysis of Peter.Lax page 47 for theorem 9 and page 61 for theorem 10. – Fin8ish Aug 07 '14 at 21:39
  • @masoud: you should add the source of the images to your answer. This avoids some legal issues. – robjohn Aug 08 '14 at 09:16