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Let $(e_n)$ be a total orthonormal sequence in a separable Hilbert space $H$ and define the right shift operator to be the linear operator $T \colon H \to H$ such that $T e_n = e_{n+1}$ for $n = 1, 2, 3, \ldots$. Then how to find $Tx$ for an arbitrary element $x \in H$?

By total, we mean that the span of $(e_n)$ is dense in $H$. In this case, every $x \in H$ can be written as $$x = \sum_{n=1}^\infty \langle x, e_n \rangle e_n$$ because the $e_n$ are orthonormal.

What next?

If we could show that $T$ is bounded (and hence continuous), then we can show that $$Tx = \sum_{n=1}^\infty \langle x, e_n \rangle T e_n = \sum_{n=1}^\infty \langle x, e_n \rangle e_{n+1}.$$

But how to show that $T$ is bounded?

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    $\begingroup$ For this definition of $T$ to work, you implicitly assume that $T$ is bounded. Otherwise you can not extend $T$ to $H$, when it is only defined for basis vectors. $\endgroup$ – user42761 Jun 24 '15 at 12:13
  • $\begingroup$ It is a well-known theorem of fucntional analysis that if you have a linear operator $T$ defined on a dense linear subspace of $H$, and $T$ is bounded on this subspace, then there is a unique extension of $T$ onto $H$ such that the extension is linear and bounded with the same operator norm $\endgroup$ – user159517 Jun 24 '15 at 12:14
  • $\begingroup$ @Andre I dont agree. You can start with just this definition and then show that there is a unique bounded operator on $H$ satisfying it. $\endgroup$ – user159517 Jun 24 '15 at 12:20
  • $\begingroup$ @user159517, can you please elaborate on why you disagree with Andre? The extension theorem you've referred to is known to me too. I would appreciate if you could write a detailed answer on how we can start with this definition and then show that there is a unique bounded linear operator satisfying it. $\endgroup$ – Saaqib Mahmood Jun 24 '15 at 12:25
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    $\begingroup$ First you extend $T$ to the linear span of $\{ e_n : n \in \mathbb{N}^+ \}$. Then you note that $T$ is an isometry on that space, $$\left\lVert T\left(\sum_{k = 1}^m c_k \cdot e_{n_k}\right)\right\rVert^2 = \left\lVert \sum_{k = 1}^m c_k e_{n_k + 1}\right\rVert^2 = \sum_{k = 1}^m \lvert c_k\rvert^2 \lVert e_{n_k+1}\rVert^2 = \sum_{k = 1}^m \lvert c_k\rvert^2 = \left\lVert \sum_{k = 1}^m c_k \cdot e_{n_k}\right\rVert^2.$$ Then you extend by continuity to the whole space. $\endgroup$ – Daniel Fischer Jun 24 '15 at 12:30
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Let $D \subset H$ be dense subspace and assume $T: D \to H$ is bounded on $D$ by say $M \in \mathbb R^+$. This means $$ \sup_{ x \in D \setminus \{0\}} \frac{\Vert Tx \Vert} {\Vert x \Vert } = M. $$ We claim that there exists a linear map $\tilde T: H \to H$ which extends $T$ and has norm $M$.

Now, let $y \in H$ be arbitrary. Approximate $y$ by elements of $D$, say $x_n \to y$ in norm. Then $T(x_n)$ converges since $$ \Vert Tx_n - Tx_m \Vert = \Vert {T(x_n-x_m)}\Vert \leq M \Vert x_n-x_m\Vert $$ (Use that $(x_n)$ is Cauchy and that $H$ is complete.) Define $\tilde T(y) := \lim_n T(x_n)$. Note also, that if $x_n' \to y$ then $\lim_n T(x_n) = \lim_n T(x_n')$, which proves that $\tilde T$ is well defined. That $\tilde T$ is linear, you can also check by hand using limit properties.

To see that $\tilde T$ has norm less or equal to $M$, note that if $x_n \to x$ then $$ \Vert \tilde Tx \Vert = \Vert \lim_n T(x_n) \Vert = \lim_n \Vert T(x_n) \Vert \leq M. $$ For equality we know that there exists a sequence $x_n$ in $D$ with $\lim_n \Vert T(x_n) \Vert = M$.

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