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I have a little problem to understand a Hilbert Space property.

Do we need the completeness of Hilbert space for the Riesz-Frechet representation to hold true ? Justify

I suppose we need it (intuition) I tried by taking a Cauchy sequence such that it is not convergent. Then I suppose I have to use the continuity of scalar product but I am stuck. Though this question will definitely help me to see why we suppose completeness in the definition of an Hilbert Space.

Good afternoon, Herosix

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  • $\begingroup$ I would suggest reading the wikipedia page for this representation. It shows than an N-dimesnional space can be built to arbitrary metrics provided they are completed to each other (connected) meaning each dimension is projected correctly (normalised to all dependent intervals). So the hilbert space remains isomorphic to a banach space. $\endgroup$ – McTaffy May 5 '17 at 12:29
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    $\begingroup$ @Sam Your comment is very unclear. $\endgroup$ – reuns May 5 '17 at 13:43
  • $\begingroup$ This question has already been asked, in particular here (mathoverflow.net/questions/35840/…) $\endgroup$ – Jean Marie May 5 '17 at 14:08
  • $\begingroup$ "MathOverflow is a question and answer site for professional mathematicians". I'm not a pro, and so the question has never been asked on this website. Anyway, thanks for the link it's useful in an global understanding $\endgroup$ – Holog May 5 '17 at 14:48
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    $\begingroup$ Fischer Riesz states that the scalar products gives you a canonical (anti-linear) isometry $H\to H^*$. If you know that dual spaces (the space of continuous linear functionals) of normed vector spaces are always complete, then you can see that the statement cannot be true if your original space is not complete. $\endgroup$ – s.harp May 5 '17 at 15:02
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Completeness is necessary. Consider the space of continuous functions $C([0,1])$ with the $L^2$-scalar product. The functional $f$ defined by $$ f(x):= \int_{1/2}^1 x(s)ds $$ cannot be represented as the scalar product with a continuous function.

Assume the contrary, i.e., there exists continuous $y$ such that $f(x)=\int_0^1 x(s)y(s)ds $ for all $x\in C([0,1])$. Then taking smooth test functions $x$ it follows that $y(x)=0$ almost everywhere on $(0,1/2)$ and $y(x)=1$ almost everywhere on $(1/2,1)$. Such a function $y$ cannot be continuous. Contradiction.

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    $\begingroup$ Scalar or dot product is for $\mathbb{R}^n$ otherwise we say inner product $\endgroup$ – reuns May 5 '17 at 13:44
  • $\begingroup$ I can't see why it is not possible to represent it as an inner product with a continuous function. Does this functional have nice properties saying the converse? $\endgroup$ – Holog May 5 '17 at 14:50
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    $\begingroup$ @user1952009 Seems that I do not belong to your 'we'. $\endgroup$ – daw May 5 '17 at 14:50
  • $\begingroup$ A side question: is completeness sufficient and necessary? $\endgroup$ – user66081 May 5 '17 at 15:06
  • $\begingroup$ @user66081 Sufficiecy = the Riesz-Frechet theorem, necessary -> see my answer $\endgroup$ – daw May 5 '17 at 15:56

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