# If $H^*$ is isomorphic with $H$ is H always a Hilbert space?

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

• Yes, because the dual space is complete Jul 26, 2020 at 4:29
• yes the norm is the ordinary given from inner product Jul 26, 2020 at 4:35
• One can ask if a Banach space $X$ isometrically isomorphic to its dual is a Hilbert space. Jul 26, 2020 at 4:46
• The vector space with inner product need not be $H$? Jul 26, 2020 at 4:46
• @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) Jul 26, 2020 at 4:52

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.
• 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! Jul 27, 2020 at 15:43