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. $\endgroup$ Jan 15, 2018 at 12:00

2 Answers 2


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?


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

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

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 $y_{n} \rightarrow y$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .