Show that the cartesian product of Hilbert spaces is a Hilbert space.

Question

Please give me a hint with this exercise.

Let $X$ and $Y$ be vector spaces, $X$ of finite dimension, with inner products $\left<\cdot,\cdot\right>_X$ and $\left<\cdot,\cdot\right>_Y$ respectively. We define on $X\times Y$ $$\left<(x_1,y_1),(x_2,y_2)\right>:=\left<x_1,x_2\right>_X+\left<y_1,y_2\right>_Y.$$ Show that:

1. $\left<\cdot,\cdot\right>$ is a inner product on $X\times Y$.
2. $X\times Y$ is a Hilbert space iff, $Y$ is a Hilbert space.
3. Does the previous result(2) still hold if $X$ is a generic infinite dimensional space?

This is what I did:

1. Done as usual.
2. $[\Rightarrow]$ Let $y_n$ a Cauchy sequence on $Y$, let $x_n$ a Cauchy sequence on $X$, then $(x_n,y_n)$ is a Cauchy sequence on $X\times Y$, due to the fact that $X\times Y$ is Hilbert, then there is $(x,y)\in X\times Y$ such that $$(x_n,y_n)\to (x,y)$$ in particular, $y_n\to y\in Y$. Therefore, $Y$ is Hilbert.

$[\Leftarrow]$ As $X$ is finite dimensional, then is Hilbert, so if $(x_n,y_n)$ is a Cauchy sequence on $X\times Y$ then $x_n$ is Cauchy on $X$ and $y_n$ is Cauchy on $Y$, because $X$ and $Y$ are Hilbert, there are $x\in X$ and $y\in Y$ such that $$x_n\to x,\quad y_n\to y$$ so $$(x_n,y_n)\to (x,y)\in X\times Y$$ so $X\times Y$ is Hilbert.

3. I do not know.

Thank you.

Answer 1

In 3) it is not given that $X$ is complete. If $Y$ is a Hilbert space then $X \times Y$ need not be : Just take $Y=\{0\}$ and $X$ to be incomplete. Let $(x_n)$ be a Cauchy sequence in $X$ which is not convergent. Then $(x_n,0)$ is a Cauchy sequence in $X\times Y$ which is not convergent.