So I was wondering say you have a Hilbert such as the inner product is defined as $$ <f,g>= \int fge^{-x^2}dx $$ where $ f,g \in M $ where $M$ is the set of functions on which the inner product is defined.
Now say you have a mapping such as $$ \phi :M \rightarrow M' $$ $$ f \rightarrow e^{-x^2/2}f $$
and you have a inner product defined such as $$ <f,g>=\int fgdx $$
Where $ f,g \in M' $. So I was wondering what would the relation be between these two Hilbert spaces then.
Define $T\colon M\to M'$ by $(Tf)(x)=e^{-x^2/2}f(x)$. ClearThen, for amy $f,g\in M$ we have $$ \langle f,g\rangle=\int_{\mathbb R}f(x)\,g(x)\,e^{-x^2}\,dx=\int_{\mathbb R}f(x)\,e^{-x^2/2}\,g(x)\,e^{-x^2/2}\,dx=\langle Tf,Tg\rangle. $$ Since clearly $T$ is bijective, $T$ is an isometry and $M$ and $M'$ are isometric.