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

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 $$E$$.

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.