Isometric operator between $\mathcal{H} \oplus \mathcal{H}$ and $\mathcal{H}$

Question

Let $\mathcal{H}$ be an infinitely-dimensional Hilbert space. I've seen the claim that there exists an isometric linear operator $\mathcal{H} \oplus \mathcal{H} \to \mathcal{H}$, however I couldn't find any proof online. I tried to show that the cardinalities of the orthonormal bases of the two spaces are the same, but I got stuck. Any help is greatly appreciated.

Answer 1

Something like this should work for separable $\mathcal H$. Remember that $\mathcal H\oplus\mathcal H$ is naturally endowed with the 2-norm $\|(\psi,\phi)\|_\oplus^2 := \|\psi\|^2 + \|\phi\|^2$, where the norm appearing on the right-hand side is the one induced by the inner product on $\mathcal H$. Now choose a complete orthonormal system $(e_j)_{j=0}^\infty$ of $\mathcal H$. If $(\psi,\phi)\in\mathcal H\oplus\mathcal H$, define $$T(\psi,\phi) := \sum_{j=0}^\infty \psi_{j} e_{2j} + \sum_{j=0}^\infty \phi_{j}e_{2j+1} $$ where $\psi_j$ and $\phi_j$ indicate the coefficient of $\psi$ and $\phi$, respectively, in the direction $e_j$: $$\psi = \sum_{j=0}^\infty \psi_j e_j, \qquad \phi = \sum_{j=0}^\infty \phi_j e_j. $$ Isometricity of $T$ follows from Parseval's identity (applied twice) and from the fact that the closure of the span of $(e_{2j})_{j=0}^\infty$ and the closure of the span of $(e_{2j+1})_{j=0}^\infty$ are mutually orthogonal Hilbert subspaces of $\mathcal H$.

Edit. If $\mathcal H$ is not separable, the same general idea can be applied, with inevitable modifications. First of all, we need to introduce the notion of overcountable sum: if $\Lambda$ is an index set of any cardinality, we define $$\sum_{\lambda \in \Lambda} \alpha(\lambda) := \sup_{I \subset\Lambda\ \text{finite}} \sum_{i \in I} \alpha(i),$$ whenever $\alpha : \Lambda\to\mathbb R$. This reduces to the usual notion of finite sum and countable sum if $\Lambda$ is finite or countably infinite, respectively. One may show that if the right-hand side is finite, then the set $\{\lambda \in \Lambda : \alpha(\lambda)\neq 0\}$ is either finite or countably infinite (see e.g. V. Moretti, Spectral Theory and Quantum Mechanics, 2nd edition, Theorem 3.24(b)).

At this point, we call $\mathfrak b\subset \mathcal H$ a complete orthonormal system if any two elements are orthogonal to each other and $\mathfrak b^\perp = \{0\}$. In this setting, one may prove a generalisation of Bessel's inequality: for all $\psi \in \mathcal H$ and all orthonormal systems $\mathfrak b$ (need not be complete), $$\sum_{e \in \mathfrak b} |\langle \psi,e\rangle|^2 \leq \|\psi\|^2.$$ This readily implies that the left-hand side is finite, and thus only finitely many or countably infinitely many inner products $\langle \psi,e\rangle$ are non-zero. Furthermore, if $\mathfrak b$ is complete, a generalisation of Parseval's identity holds true: for all $\psi\in\mathcal H$, $$\sum_{e\in\mathfrak b} |\langle \psi,e\rangle|^2 = \|\psi\|^2. $$ As before, only finitely many or countably infinitely many elements of $\mathfrak b$ actually contribute to the sum on the left-hand side. This subset of $\mathfrak b$ will vary depending on the particular element $\psi$ one is considering. Enumerate it by a sequence $(e_j^\psi)_{j=0}^\infty$ (append infinitely many $0$'s at the end if only finitely many basis elements were needed). The particular choice of ordering does not matter in the end, because all intervening series converge absolutely.

Now, select a fixed countably infinite orthonormal subset $(f_j)_{j=0}^\infty\subset\mathcal H$, and define $$T(\psi,\phi):=\sum_{j=0}^\infty \langle \psi,e_j^\psi\rangle f_{2j} + \sum_{j=0}^\infty \langle \phi,e_j^\phi\rangle f_{2j+1}.$$ Then $$\begin{split} \|(\psi,\phi)\|_\oplus^2 &= \|\psi\|^2+\|\phi\|^2 \\[0.7em] &= \sum_{e\in\mathfrak b} |\langle \psi,e\rangle|^2 + \sum_{e\in\mathfrak b} |\langle \phi,e\rangle|^2 \\ &= \sum_{j=0}^\infty |\langle \psi, e_j^\psi\rangle|^2 + \sum_{j=0}^\infty |\langle \phi,e_j^\phi\rangle|^2 \\ &= \left\| \sum_{j=0}^\infty \langle \psi,e_j^\psi\rangle f_{2j}\right\|^2 + \left\|\sum_{j=0}^\infty \langle \phi,e_j^\phi\rangle f_{2j+1}\right\|^2 \\[0.7em] &= \|T(\psi,\phi)\|^2. \end{split}$$ Over the next-to-last equal sign, we are using Parseval's identity in $\widetilde{\mathcal H} := \overline{\operatorname{span}_j \{f_j\}}$, a closed subspace of $\mathcal H$ (and thus a Hilbert space in its own right). The last equal sign follows from the Pythagorean theorem, due to the orthogonality of the direct sum $\widetilde{\mathcal{H}} = \mathcal{H}_+\oplus\mathcal{H}_-$, where $$\mathcal{H}_+ := \overline{\operatorname{span}_j\{f_{2j}\}},\qquad \mathcal{H}_- := \overline{\operatorname{span}_j\{f_{2j+1}\}}.$$