Let $(H,\langle \cdot , \cdot \rangle_H)$ be a separable complex Hilbert space. Furthermore let $H^\mathbb{R}$ denote the realification of $H$ (simply ignore the possibility of scalar multiplication with anything other than real scalars). We can equip $ H^\mathbb{R}$ with the inner product $\langle x, y \rangle = Re(\langle x, y \rangle_H)$ and one can show that $( H^\mathbb{R},\langle \cdot, \cdot \rangle)$ is again a Hilbert space.


Is $( H^\mathbb{R},\langle \cdot, \cdot \rangle)$ separable under the new topology induced by $\langle \cdot, \cdot \rangle$? If so, how do I show it?


Note that since for any $x\in H$ we have $\langle x, x\rangle=\overline{\langle x, x\rangle}$ then $$\langle x, x\rangle\in\mathbb{R}$$

Thefore if you look at the metric generated by inner product

$$d(x, y)=\sqrt{\langle x-y, x-y\rangle}$$

it is obvious that they are equal for both inner products. Thus they share all metrical and topological properties.

  • $\begingroup$ D'oh! Thanks, much appreciated. $\endgroup$ – Martin Jan 11 '17 at 20:20

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