How to show that the cartesian product of Hilbert spaces is a Hilbert space?

Question

Let $E_1,...,E_n$ be inner product spaces, $\langle [x_1,...,x_n],[y_1,...,y_n]\rangle$ defines an inner product in $E = E_1 \times ... \times E_n$. Show that if $E_1,...,E_n$ are Hilbert spaces (complete inner product spaces), then E is also a Hilbert space.

Since $E$ has an inner product, we only need to show that $E$ is also complete. How do I do that?

The definition of complete space: A normed space $E$ is called complete if every Cauchy sequence in $E$ converges to an element in $E$.

Answer 1

Assuming that you are defining the inner product on $E$ as the sum of inner products on the $E_i$, i.e. that $$ \langle (x_1,\dots, x_n) , (z_1,\dots, z_n) \rangle = \sum_{i=1}^n \langle x_i, z_i \rangle_{E_i}, $$ then the norm induced by this inner product is:

\begin{align*} \| (x_1,\dots, x_n)\|_E &= \sqrt{\langle (x_1,\dots, x_n),(x_1,\dots, x_n) \rangle} \\ &= \sqrt{\sum_{i=1}^n \langle x_i, x_i \rangle_{E_i}} \\ &= \sqrt{\sum_{i=1}^n \| x_i \|^2_{E_i}}. \end{align*}

Now, in order to show complete-ness of $E$, we need to argue that every Cauchy sequence in $E$ converges to a point in $E$. Let $\{y_m\}_{m \ge 1} := \{(x_{1m},\dots, x_{nm})\}_{m\ge 1}$ be a Cauchy sequence in $E$. Then it follows that the $i$-th 'coordinate' sequence, $\{ x_{im} \}_{m \ge 1} \subset E_i$ is Cauchy in $E_i$, and by the completeness of $E_i$, this sequence converges to an element $y^*_i \in E_i$. This holds for each $i=1,\dots,n$. Now, letting $y^* := (y^*_1, \dots, y^*_n) \in E$, we have

\begin{align*} \| y_m - y^*\|_E &=\sqrt{\sum_{i=1}^n \|x_{im}-y^*_{i} \|_{E_i}^2} \to 0 \end{align*} as $m \to \infty$, which shows that any Cauchy sequence in $E$ converges to an element of $E$, and hence $E$ is complete.