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Let $U$ be a bounded operator on a Hilbert space. Show that the following are equivalent:

I. $U$ is surjective and $\|Uv\|=\|v\|$ for all $v\in H$;

II. $U$ is surjective and $\langle Uv,Uw\rangle=\langle v,w\rangle$ for all $v,w\in H$;

III. $UU^*=U^*U=I$, where $I$ is the identity operator.

An operator satisfying the conditions above is called a unitary operator Examples include translation operators on $L^2(X)$ that come from measure-preserving bijections of $X$.

Let $H$ be a Hilbert space with a fixed orthonormal basis $\{e_n\}_{n\in \mathbb{N}}$, and let $T$ be a compact self-adjoint operator on $H$. Show that there exists a unitary $U$ such that the matrix of $UTU^*$ is diagonal with real entries.

Hint: define $U$ by stipulating that it take a basis of eigenvectors for $T$ (which exists by the spectral theorem) to the given basis.

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The Arveson form of the Polarization Identity for sesquilinear forms is $$ b(x,y) = \frac{1}{4}\sum_{n=0}^{3}b(x+i^ny,x+i^n y). $$ Therefore, if $b$ and $c$ are sesquilinear forms, then $b(x,x)=c(x,x)$ for all $x$ iff $b(x,y)=c(x,y)$ for all $x,y$. Examples of sesquilinear forms include $\langle x,y\rangle$, $\langle Ux,Uy\rangle$ and $\langle U^{\star}Ux,y\rangle$. Therefore $\langle x,y\rangle=\langle Ux,Uy\rangle$ for all $x,y$ iff $\|x\|=\|Ux\|$ for all $x$. And $I=U^{\star}U$ iff $$ \langle x,x\rangle = \langle U^{\star}Ux,x\rangle,\;\;\; x\in X. $$ Equivalently, $\|x\|=\|Ux\|$ for all $x\in X$. If $U^{\star}U=I$, then applying $U$ to the left of both sides gives $UU^{\star}Ux=Ux$ for all $x$, or $UU^{\star}y=y$ for all $y\in\mathcal{R}(U)$. Hence, $UU^{\star}=I$ if $U^{\star}U=I$ and if $U$ is surjective.

For the last part, let $\{e_n \}$ be an orthonormal basis of $H$, and let $\{ f_n\}$ be an orthonormal basis of eigenvectors of $T$. Define $$ U\sum_{n}\alpha_n e_n = \sum_{n}\alpha_n f_n $$ By Parseval's equality, $U$ is an isometry, because $$ \|\sum_n \alpha_n e_n\|^2=\sum_n|\alpha_n|^2 = \|\sum_n \alpha_n f_n\|^2 = \|U\sum_n\alpha_n e_n\|^2. $$ And $U$ is surjective because $\{ f_n\}$ is a basis. Therefore, $$ U^{\star}\sum_n\alpha_n f_n =U^{-1}\sum_n\alpha_n f_n = \sum_n \alpha_n e_n \\ UTU^{\star}\sum_n \alpha_n e_n = UT\sum_n\alpha_n f_n = U\sum_n\lambda_n\alpha_n f_n = \sum_n\lambda_n\alpha_n e_n $$

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Better reduce your theorem..

Theorem (Isometry)
Given Hilbert spaces $\mathcal{H}$ and $\mathcal{K}$.
Let R be a bounded operator from $\mathcal{H}$ into $\mathcal{K}$. TFAE:

  1. $R$ has a left inverse by its adjoint, i.e. $R^*R=1$
  2. $R$ is an isometry, i.e. $\|R\varphi\|=\|\varphi\|$ for all $\varphi\in\mathcal{H}$.
  3. $R$ preserves scalar products, i.e. $\langle R\varphi,R\psi\rangle=\langle\varphi,\psi\rangle$ for all $\varphi,\psi\in\mathcal{H}$.

Proof (Polarization)

Proving by circular chain:

(1=>2): Scalar product induces norm: $$\|R\varphi\|^2=\langle R\varphi,R\varphi\rangle=\langle R^*R\varphi,\varphi\rangle=\langle\varphi,\varphi\rangle=\|\varphi\|^2$$ (2=>3): Norm generates scalar product: $$\langle R\varphi,R\psi\rangle=\frac{1}{4}\sum_{\alpha=0\ldots3}i^\alpha\|(R\varphi)+i^\alpha (R\psi)\|^2=\frac{1}{4}\sum_{\alpha=0\ldots3}i^\alpha\|(\varphi)+i^\alpha (\psi)\|^2=\langle\varphi,\psi\rangle$$

(3=>1): Scalar product defines adjoint: $$\langle R^*R\varphi,\chi\rangle=\langle R\varphi,R\psi\rangle=\langle\varphi,\psi\rangle=\langle 1\varphi,\psi\rangle$$ (3=>1): Scalar product is non-degenerate: $$\langle R^*R\varphi,\psi\rangle\equiv\langle 1\varphi,\psi\rangle\implies R^*R=1$$

Concluding the first lemma.

Lemma (Inverses)
Given plain spaces $X$ and $Y$.
Let $F$ be a function from $X$ into $Y$. TFAE:

  • $F$ is injective resp. surjective.
  • $F$ has a left resp. right inverse, i.e. $LF=1_X$ resp. $FR=1_Y$.

Remark (Uniqueness)
Left resp. right inverses are not necessarily unique!

Lemma (Left vs. Right)
Given plain spaces $X$ and $Y$.
Let $F$ be a function from $X$ into $Y$. Then: $$LF=1_X\quad FR=1_Y\implies L=R$$ That is left and right inverse become unique and agree.

Proof (Identity)

Identity function is a unit: $$L=L1_Y=L(FR)=(LF)R=1_XR=R$$

Concluding the third lemma.

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