Relationship between $L^2(\mathbb{R})$ with Lebesgue measure and $L^2(\mathbb{R})$ with Gaussian measure

Question

Suppose $\mu$ is absolutely continuous measure with respect to Lebesgue measure. Let's take $$\mu(A)=\frac{1}{\sqrt{2\pi}}\int_A e^{-\frac{x^2}{2}}dx$$ as the Gaussian measure and the Gaussian function is the Radon-Nikodym derivative. Denote $L^2(\mathbb{R})$ and $L^2(\mathbb{R},\mu)$ as the $L^2$ space with Lebesgue measure and Gaussian measure respectively. Since for each $f\in L^2(\mathbb{R})$, $$\|f\|_{L^2(\mathbb{R},\mu)}^2=\int|f|^2d\mu=\int|f|^2\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\leq\frac{1}{\sqrt{2\pi}}\int|f|^2dx=\frac{1}{\sqrt{2\pi}}\|f\|_{L^2(\mathbb{R})}^2,$$ we know that $$L^2(\mathbb{R})\subset L^2(\mathbb{R},\mu).$$ And notice that for given $\xi\in\mathbb{R}$, $e^{i\xi\cdot}\in L^2(\mathbb{R},\mu)$ but $e^{i\xi\cdot}\notin L^2(\mathbb{R})$. Thus, the inclusion is nontrivial. I am wondering whether the inclusion relationship is true for the general case of measure which is absolutely continuous with respect to Lebesgue measure. Is there a name for the space $L^2(\mathbb{R},\mu)$ when $\mu$ is Gaussian measure defined above? Welcome any discussion and suggestion! Thanks in advance!

Answer 1

Let $d\mu=\dfrac{1}{x}dx$ and consider $f(x)=\chi_{(0,1)}(x)\dfrac{1}{x^{1/4}}\in L^{2}(\mathbb{R})$, but then $\displaystyle\int_{\mathbb{R}}\chi_{(0,1)}(x)\dfrac{1}{x^{1/2}}\dfrac{1}{x}dx=\infty$, so $f\notin L^{2}(\mathbb{R},\mu)$.