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Let $H_1, H_2$ be Hilbert spaces and $T\in B(H_1,H_2)$. Show that the following conditions are equivalent:

A. $T$ is isometry.

B. $T^*T=I_{H_1}$.

C. $T$ preserves the inner product.

D. $T$ "copies/sends" every orthonormal set in $H_1$ to orthonormal set in $H_2$.

E. $T$ is injective and there exist a basis to $H_1$ that $T$ "copies" to an orthonormal set in $H_2$.

A $\Rightarrow$ B is already shown here Show that an isometric linear operator $T:H\to H$ satisfies $T^* T=I$, where $I$ is the identity operator on $H$.

B $\Rightarrow$ C: Assume that $T^*T=I$ then, by definition $\forall x,y\in H_1$ $\langle Tx,Ty\rangle = \langle T^*Tx,y\rangle = \langle x,y\rangle$.

C $\Rightarrow$ D: We assume that $T$ preserves inner product. Let $\{u_a\}_{a\in A} \in H_1$ be an orthonormal set. For every $a\in A$ , $\|Tu_a\|=\|u_a\|$ (by assume). And for every $a\neq b \in A$ $\langle Tu_a,Tu_b\rangle =\langle u_a,u_b\rangle =0$.

So $\{T(u_a)\}_{a\in A}$ is an orthonormal set in $H_2$.

How to do D $\Rightarrow$ E and E $\Rightarrow$ A ? And is that enough for showing that the claims are equivalent?

Mat999
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    Do you mean "E$\Rightarrow$A" rather than "E$\Rightarrow$F", since I can see no statement labelled "F" in your question? If so, then yes, showing that D$\Rightarrow$E and E$\Rightarrow$A would complete the proof that all the statements are equivalent. – lonza leggiera May 03 '21 at 14:48
  • Yeah i meant that! @lonza leggiera – Mat999 May 03 '21 at 18:06

2 Answers2

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D$\Rightarrow$E:

Every Hilbert space has an orthonormal basis, so let $\ \big\{u_a\big\}_{a\in A}\ $ be such a basis of $\ H_1\ $. If $\ \text{D}\ $ holds, then $\ \big\{Tu_a\big\}_{a\in A}\ $ must be an orthonormal set in $\ H_2\ $. Now let $\ x=\sum_\limits{a\in A}x_au_a\ $, $\ y=\sum_\limits{a\in A}y_au_a\ $ be arbitrary members of $\ H_1\ $ with $\ Tx=Ty\ $. Then for any $\ b\in A\ $, \begin{align} \ \big\langle Tx,Tu_b\big\rangle&=\big\langle T\sum_\limits{a\in A}x_au_a,Tu_b\big\rangle\\ &=\big\langle \sum_\limits{a\in A}x_aTu_a,Tu_b\big\rangle\\ &=x_b \end{align} by the orthonormality of $\ \big\{Tu_a\big\}_{a\in A}\ $. Likewise, $\ \big\langle Ty,Tu_b\big\rangle=y_b\ $. Thus, since $\ Tx=Ty\ $, $\ x_b=y_b\ $ for all $\ b\in A\ $ and hence $\ x=y\ $. Therefore $\ T\ $ is injective.

E$\Rightarrow$A:

Suppose $\ \text{E}\ $ holds. Then there exists a basis $\ \big\{u_a\big\}_{a\in A}\ $ of $\ H_1\ $ such that $\ \big\{Tu_a\big\}_{a\in A}\ $ is an orthonormal set in $\ H_2\ $. If $\ x=\sum_\limits{a\in A}x_au_a\ $ is an arbitrary member of $\ H_1\ $, then $\ \|x\|^2=\sum_\limits{a\in A}|x_a|^2\ $ and \begin{align} \|Tx\|^2&=\langle Tx,Tx\rangle\\ &=\Big\langle\sum_\limits{a\in A}x_aTu_a,\sum_\limits{a\in A}x_aTu_a\Big\rangle\\ &=\sum_\limits{a\in A}|x_a|^2\\ &=\|x\|^2\ . \end{align} Therefore $\ T\ $ is an isometry.

lonza leggiera
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  • Hi @lonza leggiera thank you! How did you get that the inner peoducr equals $x_b$? And when calculating $||Tx||^2$ the inner product of the two summands is just the union under one sum right? – Mat999 May 04 '21 at 07:43
  • The first time I met Hilbert spaces was in a theoretical physics course many years ago. I have no idea if it's a common convention among theoretical physicists, but the inner products of the course's instructors were linear in their second argument and conjugate linear in their first. I often forget that the much more normal convention is the reverse of this. So until I reversed the inner products in my answer a little while ago, I should have had them evaluating to $\ \overline{x}_b\ $ and $\ \overline{y}_b\ $ rather than $\ x_b\ $ and $\ y_b\ $. – lonza leggiera May 04 '21 at 10:10
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    The evaluations of the inner products relies on the orthonormality of $\ \big{Tu_a\big}{a\in A}\ $: $\ \big\langle Tu_a,Tu_b\big\rangle=\delta{ab}\ $, so \begin{align} \Big\langle\sum_\limits{a\in A}x_aTu_a,\sum_\limits{b\in A}y_bTu_b\Big\rangle&=\sum_\limits{a\in A}\sum_\limits{b\in A}x_a\overline{y}b\big\langle Tu_a,Tu_b\big\rangle\ &=\sum\limits{a\in A}\sum_\limits{b\in A}x_a\overline{y}b\delta{ab}\ &=\sum_\limits{a\in A}x_a\overline{y}a\ , \end{align} and, in particular, when $\ x=y\ $, $\langle Tx,Tx\rangle=\sum\limits{a\in A}|x_a|^2\ $ – lonza leggiera May 04 '21 at 10:27
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You have $$\begin{matrix} \quad\text{D} & \Rightarrow\text{ E }\Rightarrow & \text{A}\quad\\[.5ex] \checkmark\,\big\Uparrow && \big\Downarrow\,\checkmark \\[-1ex] \quad\text{C}& \stackrel{\;\Large\checkmark}{\Longleftarrow} & \text{B}\quad \end{matrix}$$ and proving the upper row shall close the circle of implications, thus settling the equivalence of the claims.

D $\Rightarrow$ E
asks to show that $T$ is injective, or equivalently $\ker T=\{0\}\,$:
If $0\neq x\in H_1\,$ then let the unit vector $\frac x{\|x\|}$ be a member of some ONS. Assuming statement D, one then has $Tx\neq 0$ which means that $\ker T$ is trivial.
Next consider an ONB in $H_1$ which in particular is an ONS. By assumption, $T$ will map it to an ONS in $H_2$.

E $\Rightarrow$ A
asks to show that $\,\|Tx\|_{H_2}=\|x\|_{H_1}\,$ for all $\,x\in H_1$.
Parseval's identity $\,\sum_{i\in I} |\langle x, b_{i}\rangle |^2 = \|x\|^2\,$, stating that the Fourier coefficients $\langle x, b_{i}\rangle$ of $\,x\,$ w.r.to an ONB $\{b_i\mid i\in I\}$ are square-summable and yield the squared norm of $x$, will be helpful:
We have $\,Tx = T\big(\sum_{i\in I}\langle x, b_{i}\rangle\, b_i\big) = \sum_{i\in I}\langle x, b_{i}\rangle\, Tb_i\,$ in general. By assumption, the $\{Tb_i\}\subset H_2$ are an ONS (and may be extended to an ONB, but the additional Fourier coefficients will be zero), and furthermore T is injective which yields $$\|Tx\|_{H_2}^2 \:=\: \sum_{i\in I} |\langle x, b_{i}\rangle_{H_1} |^2 \:=\: \|x\|^2_{H_1}\,,$$ completing the proof.

Hanno
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