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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?

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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.

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  • $\begingroup$ What is the definition of direct sum of Hilbert spaces? As in the second answer of this question? $\endgroup$
    – Guilherme
    Mar 24, 2021 at 0:54
  • $\begingroup$ 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 $H_2$. $\endgroup$
    – WhatsUp
    Mar 24, 2021 at 0:56
  • $\begingroup$ Given two subspaces $H_1$ and $H_2$ finitely generated using Gram–Schmidt process can I always write $H_1 \perp H_2$? $\endgroup$
    – Guilherme
    Mar 24, 2021 at 1:18
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    $\begingroup$ Gram-Schmidt is not quite relevant here, as it deals with basis elements. $\endgroup$
    – WhatsUp
    Mar 24, 2021 at 1:22
  • $\begingroup$ Could you explain it better, please? $\endgroup$
    – Guilherme
    Mar 24, 2021 at 1:37

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