Extending an operator defined on an orthonormal basis to the full space H, where $H$ is a Hilbert space ans a bounded linear operator

Question

Let $H$ be a separable Hilbert space. Let $\{e_n\}$ be an orthonormal basis for $H$. Let $(y_n)$ be a bounded sequence in $H$. Suppose if I prescribe $T(e_i)=y_i$, can I extend $T$ as a bounded linear operator from $H \to H$. We know span$(e_i)$ is dense in $H$. We can extend linearly to $span(e_i)$. But I am not able to show it is bounded as a linear map from $span(e_i) \to H$. If we an prove this, then clearly it can be extended to a bounded operator to the closure of $span(e_i)$ which is full of $H$. I started thinking of this by the definition of the shift operator which is defined everywhere as follows. Let $H$ be a separable Hilbert space. Let $\{e_n\}$ be an orthonormal basis for $H$. Define $T(e_n)=e_{n+1}$ and extend this to $H$. I am confused how can we extend this to $H$ as a bounded operator.

Answer 1

As David correctly argued in this comment, such a bounded extension is not always possible: choosing $y_n:=e_1$ for all $n$ would mean that $T$ maps $\sum_j\frac{e_j}{j}\in\mathcal H$ to $(\sum_j\frac1j)e_j$, but $\sum_j\frac1j=\infty$, a contradiction. However, cases where this works are

• when $\sum_j\|Te_j\|^2=\sum_j\|y_j\|^2<\infty$ (as pointed out by geetha in this comment). In this case $T$ is even a Hilbert-Schmidt operator, and hence compact
• when the $y_i$ are pairwise orthogonal and bounded:

Proposition. Let any orthonormal basis $(e_i)_{i \in I}$ of $\mathcal{H}$ and a family of pairwise orthogonal vectors $(y_i)_{i \in I} \subseteq \mathcal{H}$ (i.e., $\langle y_i,y_j\rangle=0$ for all $i\neq j$) with $\sup_{i \in I} \| y_i \| < \infty$ be given. Then there exists unique $T \in \mathcal{B}(\mathcal{H})$ with $T e_i = y_i$ for all $i \in I$. In this case $\| T \| = \sup_{i \in I} \| y_i \|$.

Proof. Define $ \mathcal{H}_0 := \operatorname{span} \{ e_i : i \in I \} $ and a linear map $ T_0:\mathcal{H}_0\to\mathcal{H} $ via $ T_0 e_i := y_i$ for all $i \in I$. Now for every $ x \in \mathcal{H}_0 $ there exist $ i_1, \dots, i_m \in I $ such that $ x = \sum_{j=1}^{m} \langle x,e_{i_j} \rangle e_{i_j} $. Using orthogonality, the Pythagorean theorem then implies $ \|x\|^2 = \sum_{j=1}^{m} |\langle x,e_{i_j} \rangle|^2 $ as well as \begin{align*} \|T_0 x\|^2 = \Big\| \sum_{j=1}^{m} \langle x,e_{i_j} \rangle y_{i_j} \Big\|^2 &=\sum_{j=1}^{m} |\langle x,e_{i_j} \rangle|^2 \|y_{i_j}\|^2 \\ &\leq \left( \sup_{i \in I} \|y_i\|^2 \right) \sum_{j=1}^{m} |\langle x,e_{i_j} \rangle|^2 = \left( \sup_{i \in I} \|y_i\|^2 \right) \|x\|^2. \end{align*} This shows $ \|T_0\| \leq \sup_{i \in I} \|y_i\| < \infty $ so by the BLT theorem there exists a bounded extension $T\in\mathcal B(\mathcal H)$ of $T_0$ with the same norm. Finally, uniqueness is evident from continuity. $\square$

This proposition works for the shift operator you mentioned and, more importantly, this shows that a bounded operator is unitary if and only if it maps one (every) orthonormal basis to an orthonormal basis again. Hence for every two orthonormal bases there exists a unitary mapping one to the other.