4
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Yep. It's not difficult to guess a an inner product that would work. $\endgroup$ Jan 15, 2018 at 12:00

2 Answers 2

7
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $y_{n} \rightarrow y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.