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My lecturer said that to make the Hilbert space $(H\oplus H',⟨\cdot,\cdot⟩ )$ we need to (1) make the cartesian product $H \oplus H' = H\times H'$, (2) give it an inner product, and - the confusing part to me - (3) complete the resulting space.

I don't see when this would be required, as the inner product is already continuous, and I can't think of any elements we need to add in say $\mathbb{R}^d$ or $\bigoplus_{n∈ \mathbb{Z}}\operatorname{span}\{e^{2π inx}\} = L^2([0,1]/\sim, \mathbb{C})$ if we don't complete the space.

What product/direct sum of Hilbert spaces $H_i$ would require completion?

Calvin Khor
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I agree that, if you make the direct sum of a finite number of Hilbert spaces, then you do not need to complete anything. Indeed, the finite product of complete metric spaces is also complete. However, things change quite a little bit when you switch to infinite products: indeed, the algebraic direct sum of an infinite family of vector spaces consists not on the whole cartesian product (on which it is unclear how to define an inner product), but rather on the elements of it that are zero except for a finite number of entries. For example, the algebraic direct sum of countable copies of $\mathbb{R}$ is $$ \bigoplus_{\mathbb{N}}^{\text{algebraic}} \mathbb{R}=\left\{ (x_n)_{n\in\mathbb{N}} \ :\ x_n=0\ \text{for all sufficiently big }n\right\}.$$ You make this into a pre-Hilbert space by defining an inner product on it in this way: $$ \left( (x_n), (y_n)\right) = \sum_{n\in \mathbb{N}} x_ny_n, $$ (here you need the fact that sequences vanish eventually: otherwise you would get a possibily-non-convergent infinite series here). Finally, taking the completion you get the space usually known as $\ell^2$.

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Giuseppe Negro
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