Definition of direct sum of Hilbert spaces

Let $$\mathcal{H}=(\mathcal{H},(\cdot, \cdot)_{\mathcal{H}})$$ be a Hilbert space and $$\mathcal{H}_1, \mathcal{H}_2 \subset \mathcal{H}$$ subspaces. When I write $$\mathcal{H}=\mathcal{H}_1\oplus\mathcal{H}_2$$ this means that $$\mathcal{H}_1$$ and $$\mathcal{H}_2$$ are closed subspaces of $$\mathcal{H}$$ such that

$$(1)$$ for every $$z\in\mathcal{H}$$ there exist $$x\in\mathcal{H}_1$$ and $$y\in\mathcal{H}_2$$ such that $$z=x+y$$.

$$(2)$$ The elements $$x\in\mathcal{H}_1$$ and $$y\in\mathcal{H}_2$$ in $$(1)$$ are unique.

Question. The item $$(2)$$ is equivalent to saying that $$(x,y)_{\mathcal{H}}=0$$?

In the second answer of this question it is said that $$(x,y)_{\mathcal{H}}=0$$ implies that $$x$$ an $$y$$ are uniquely determined. So, the converse holds?

1 Answer

The converse doesn't hold.

As an example, take $$H = \Bbb R^2$$ equipped with the Euclidean norm, i.e. $$|(x, y)|^2 = x^2 + y^2$$.

Take $$H_1 = \Bbb R\cdot (1, 0)$$ and $$H_2 = \Bbb R\cdot (1, 1)$$.

It is clear that they satisfy the conditions (1) and (2) in your question. These conditions in fact requires that $$H = H_1 \oplus H_2$$ just as vector spaces.

However this is not a direct sum of Hilbert spaces, as $$v = (1, 0) \in H_1$$ and $$w = (1, 1)\in H_2$$ are not orthogonal.

• What is the definition of direct sum of Hilbert spaces? As in the second answer of this question? Mar 24, 2021 at 0:54
• It is given in the answer that you linked to. An equivalent definition would be: $H$ is a direct sum of $H_1$ and $H_2$ as vector spaces and $H_1$ is orthogonal to $H_2$. Mar 24, 2021 at 0:56
• Given two subspaces $H_1$ and $H_2$ finitely generated using Gram–Schmidt process can I always write $H_1 \perp H_2$? Mar 24, 2021 at 1:18
• Gram-Schmidt is not quite relevant here, as it deals with basis elements. Mar 24, 2021 at 1:22
• Could you explain it better, please? Mar 24, 2021 at 1:37