# Is the cartesian product of two Hilbert spaces a Hilbert space?

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?

• Yep. It's not difficult to guess a an inner product that would work. Commented Jan 15, 2018 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 you proceed form here?

• The fact that $\|(x_1,x_2)\|_{H_1\times H_2} =\sqrt{\|x_1\|^2_{H_1}+\|x_2\|^2_{H_2}}$ makes possible to show that both projections are continuous, which is subtly used for showing the desired task. Commented May 29 at 0:58

Consider the following inner product on $$H_1\times H_2$$ $$\langle (x_1,x_2), (y_1,y_2)\rangle = \langle x_1,y_1 \rangle + \langle x_2, y_2\rangle$$. It remains to show this is indeed an inner product.

Linearity $$\langle a(x_1,x_2) + b(y_1,y_2), (z_1,z_2) \rangle \\ =\langle (ax_1 + by_1, ax_2 + by_2), (z_1,z_2) \rangle\\ = \langle ax_1 + by_1, z_1\rangle + \langle ax_2 + by_2, z_2\rangle\\ = \langle ax_1 , z_1 \rangle +\langle by_1, z_1\rangle + \langle ax_2 ,z_2 \rangle + \langle by_2, z_2\rangle\\ = a\langle(x_1,x_2), (z_1,z_2) \rangle + b\langle(y_1,y_2), (z_1,z_2) \rangle$$

Symmetry
$$\langle (x_1,x_2), (y_1,y_2)\rangle= \langle x_1, y_1 \rangle + \langle x_2, y_2 \rangle = \langle (y_1,y_2), (x_1,x_2)\rangle$$

Bilinearity
Assume $$\langle(x_1,x_2), (x_1,x_2) \rangle>0$$ then $$\langle(x_1,x_2), (x_1,x_2) \rangle= \langle x_1,x_1\rangle + \langle x_2,x_2 \rangle > 0 \implies$$ either $$\langle x_1, x_1 \rangle > 0$$ or $$\langle x_2, x_2 \rangle > 0$$ so $$(x_1,x_2) \neq (0,0)$$

Fix some Cauchy sequence in $$\{ \left(x_{n}, y_{n}\right)\} \in H_{1} \times H_{2}$$. Then $$x_n$$ and $$y_n$$ are both Cauchy sequences in $$H_1$$ and $$H_2$$ and hence converge to $$x$$ and $$y$$ respectively. Now $$(x,y) \in H_1 \times H_2$$ and we claim this is the limit of $$\{ \left(x_{n}, y_{n}\right)\}$$.

\begin{align*} &\left\|\left(x_{n}, y_{n}\right)-(x, y)\right\|_{H_{1} \times H_{2}} \\ =&\left\langle\left(x_{n}, y_{n}\right)-(x, y),\left(x_{n}, y_{n}\right)-(x, y)\right\rangle^{1 / 2} \\ =&\left\langle\left(x_{n}-x, y_{n}-y\right),\left(x_{n}-x, y_{n}-y\right)\right\rangle^{1 / 2} \\ =& \sqrt{\left\|x_{n}-x\right\|_{H_{1}}^{2}+\left\|y_{n}-y\right\|_{H_{2}}^{2}} \\ & \rightarrow 0 \end{align*}

as $$x_{n} \rightarrow x$$ and $$y_{n} \rightarrow y$$.