Let $(H_1,\langle , \rangle_1)$ and $H_2,\langle , \rangle_2)$ be two complex Hilbert spaces.

Is $H_1\times H_2$ a Hilbert space?

I think that we can define the following inner-product on $H_1\times H_2$: $$\langle X, Y\rangle=\langle x_1, y_1\rangle_1+\langle x_2, y_2\rangle_2,$$ for all $X=(x_1,x_2)\in H_1\times H_2$ and $Y=(y_1,y_2)\in H_1\times H_2$.

Question: why $(H_1\times H_2,\langle \cdot, \cdot\rangle)$ is complete?

  • $\begingroup$ Yep. It's not difficult to guess a an inner product that would work. $\endgroup$ – Theo Bendit Jan 15 '18 at 12:00

Hint the norm on $H_1\times H_2$ is given by

$$\|(x_1,x_2)\|_{H_1\times H_2} =\sqrt{\|x_1\|^2_{H_1}+\|x_2\|^2_{H_2}}$$

and under this norm it is easy to check the completeness of $H_1\times H_2$ since having a cauchy sequence $(x^n_1,x^n_2)_n$ in $H_1\times H_2$ will directly implies that $(x^n_i)_n$ is a Cauchy sequence(Why?) in the Hilbert (Complete space ) space $(H_i, \|\cdot\|_{H_i})$ $i=1,2$

because $$\|x^n_i- x^m_i\|_{H_i}\le \|(x^n_1,x^n_2)-(x^m_1,x^m_2)\|_{H_1\times H_2} \to 0$$ can proceed form here?


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