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clarification
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By Example 7.14.13 in Volume 2 of Bogachev's Measure Theory, every Radon measure on $H$ is separable, so that $L^2(H,\gamma)$ is also separable. It is not necessary that $H$ is a Hilbert space, just that every compact subset of $H$ be metrizable.

By Example 7.14.13 in Volume 2 of Bogachev's Measure Theory, every Radon measure is separable, so that $L^2(H,\gamma)$ is also separable. It is not necessary that $H$ is a Hilbert space, just that every compact subset of $H$ be metrizable.

By Example 7.14.13 in Volume 2 of Bogachev's Measure Theory, every Radon measure on $H$ is separable, so that $L^2(H,\gamma)$ is also separable. It is not necessary that $H$ is a Hilbert space, just that every compact subset of $H$ be metrizable.

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user6096
user6096

By Example 7.14.13 in Volume 2 of Bogachev's Measure Theory, every Radon measure is separable, so that $L^2(H,\gamma)$ is also separable. It is not necessary that $H$ is a Hilbert space, just that every compact subset of $H$ be metrizable.