6

If $H$ is a Hilbert space then $H^*$ is isomorphic with $H$. I am asking if we have a vector space H equipped with inner product (,) and $H^*$ is isomorphic with $H$ is it true to say that $H$ is Hilbert? Edit:I am also interested for cases that the norm of H is not the ordinary norm given from inner product

Arctic Char
  • 16,972
George Giatilis
  • 317
  • Yes, because the dual space is complete – Calvin Khor Jul 26 '20 at 04:29
  • yes the norm is the ordinary given from inner product – George Giatilis Jul 26 '20 at 04:35
  • One can ask if a Banach space $X$ isometrically isomorphic to its dual is a Hilbert space. – Kavi Rama Murthy Jul 26 '20 at 04:46
  • 1
    The vector space with inner product need not be $H$? – user10354138 Jul 26 '20 at 04:46
  • @CalvinKhor If H is a general normed linear space then N* is complete(even if the case that H is not complete;).If thats right i am also thinking of cases that H is a Banach space but is not Hilbert(because H does not come from inner product).For example if you consider l2 with his ordinary norm its a Hilber space but if we consider an equivalent norm with the || ||2 such as ||x||=||x||2+||x||00(notice that ||x||2<=||x||<=2||x||2 so l2 is Banach with the || || norm) – George Giatilis Jul 26 '20 at 04:52
  • but its not Hilbert because rectangle rule does not hold(||e1+e2||^2+||e1-e2||^2=2(sqrt(2)+1)^2 that is not equal to 2||e1||^2+2||e2||^2=16). – George Giatilis Jul 26 '20 at 04:53
  • I could understand your question (although maybe wrongly?) but your comments are getting to be too much. Please see this Mathjax tutorial on how to type mathematics on this site – Calvin Khor Jul 26 '20 at 04:54
  • In particular your first comment said that the norm was given by the inner product which implies that it satisfies parallelogram law. What is the relationship of $H$ to $N$? the equivalent norm you gave for $\ell^2$ is not induced by any inner product – Calvin Khor Jul 26 '20 at 04:57
  • Ok i understand the general case if i give $H$ the general norm given from inner product just because $H^$ is always complete so $H$ is complete.In my comment I meant $H^$ (not $N^$).What i wanted to say is that $\ell^2$ is isomorphic with $\ell^2$.So i consider $H^$=$\ell^2$ with the ordinary norm and $H$=$\ell^2$ with the || || norm then $H^$ is a Hilbert space(since $\ell^2$ is) but $H$ is not a Hilbert space since rectangle rule does not hold(despite the fact that H is complete). – George Giatilis Jul 26 '20 at 05:33

1 Answers1

3

The answer is No.
A non Hilbert space can be isometric (not just isomorphic) with his Dual. For example $X:=(\ell^2,\lVert $.$\rVert_{\ell^2} +\lVert $.$\rVert_{\infty})$, $X\cong X^*$ ,and obviously is not a hilbert space.

Rick Sanchez C-666
  • 1,050
  • 1
    I think that Since $X*$ is always dual space X will be also dual space(because of isometry).The only problem is that inner product is not transfered through isometry.If we have the norm given from inner product its correct otherwise its not! – George Giatilis Jul 27 '20 at 15:43