Lecture Notes in Functional Analysis

by William L. Paschke

edition 0.9

Contents

Frontmatter

I. Hilbert Space

II. Bounded Operators

III. Compact Operators

IV. The Spectral Theorem

Index and References

IV. The Spectral Theorem

We begin with two elementary matters that could have been discussed much earlier, namely: the partial ordering of self-adjoint operators in $$$\mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$; and projections, i.e. self-adjoint idempotents. Filling in the details below is recommended as an exercise.

Write $$${ \mathcal{L}\mathopen{}\left( H\right)\mathclose{} }_{\rm SA}= \mathopen{}\left\{\, A\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}\,\middle\vert\, A= A^{*}\,\right\}\mathclose{}$$$. For $$$A$$$ and $$$B$$$ in $$${ \mathcal{L}\mathopen{}\left( H\right)\mathclose{} }_{\rm SA}$$$, say $$$A\leq B$$$ provided $$$B-A\geq 0$$$, i.e. $$$\mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}\leq \mathopen{}\left\langle{}B\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}$$$ for all $$$x\in H$$$. Notice for $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$, $$$A\leq B$$$ implies $$$T^{*}AT\leq T^{*}BT$$$ because $$$\mathopen{}\left\langle{} T^{*}\mathopen{}\left( A\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}, x\right\rangle\mathclose{}= \mathopen{}\left\langle{}A\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, T\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}\leq \mathopen{}\left\langle{}B\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x\right\rangle\mathclose{}$$$ for all $$$x\in H$$$. For $$$A\in { \mathcal{L}\mathopen{}\left( H\right)\mathclose{} }_{\rm SA}$$$, recall from functional calculus $$$f \to f\mathopen{}\left( A\right)\mathclose{}$$$ from $$$\mathrm{C}\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$ to $$$\mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$, $$$f\geq 0$$$ on $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$ implies $$$f\mathopen{}\left( A\right)\mathclose{}\geq 0$$$ because $$$f\mathopen{}\left( A\right)\mathclose{}= {\mathopen{}\left({f}^{\frac{1}{2}}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}}^{2}= \mathopen{}\left({f}^{\frac{1}{2}}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{} ^{*}{f}^{\frac{1}{2}}\mathopen{}\left( A\right)\mathclose{}$$$ so $$$f$$$ and $$$g$$$ in $$$\mathrm{C}\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}, \mathbb{R}\right)\mathclose{}$$$ with $$$f\leq g$$$ implies $$$f\mathopen{}\left( A\right)\mathclose{}\leq g\mathopen{}\left( A\right)\mathclose{}$$$.

Definition IV.1

$$$P\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ is a projection if $$$P= P^{*}= {P}^{2}$$$.

For a closed subspace $$$E\subseteq H$$$, the orthogonal projection of $$$H$$$ onto $$$E$$$ is a projection in this sense. If $$$P$$$ is a projection, then $$$P\mathopen{}\left( H\right)\mathclose{}$$$ is closed and $$$P : H \to P\mathopen{}\left( H\right)\mathclose{}$$$ is the orthogonal projection of $$$H$$$ on $$$P\mathopen{}\left( H\right)\mathclose{}$$$. For projections $$$P$$$ and $$$Q$$$ we have $$$P\leq Q$$$ if and only if $$$P\mathopen{}\left( H\right)\mathclose{}\subseteq Q\mathopen{}\left( H\right)\mathclose{}$$$ if and only if $$$QP= P$$$. In this situation, $$$Q-P$$$ is projection on $$$Q\mathopen{}\left( H\right)\mathclose{}\cap { \mathopen{}\left(P\mathopen{}\left( H\right)\mathclose{}\right)\mathclose{} }^{\perp}$$$ (i.e., $$$Q\mathopen{}\left( H\right)\mathclose{}\cap \mathopen{}\left(1-P\right)\mathclose{}\mathopen{}\left( H\right)\mathclose{}$$$), often written $$$Q\mathopen{}\left( H\right)\mathclose{}\ominus P\mathopen{}\left( H\right)\mathclose{}$$$.

Fix $$$A\in { \mathcal{L}\mathopen{}\left( H\right)\mathclose{} }_{\rm SA}$$$. For $$$λ\in \mathbb{R}$$$, define $$${f}_{λ}$$$ on $$$\mathbb{R}$$$ by $$${f}_{λ}\mathopen{}\left( t\right)\mathclose{}= t-λ+\mathopen{}\left\lvert{}t-λ\right\rvert\mathclose{}$$$. Regard $$${f}_{λ}\in \mathrm{C}\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$. Let $$${P}_{λ}$$$ be projection on $$$\operatorname{Ker}\mathopen{}\left( {f}_{λ}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$.

Lemma IV.2

1. $$$λ\leq μ$$$ implies $$${P}_{λ}\leq {P}_{μ}$$$.
2. $$$λ\lt \min{}\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$ implies $$${P}_{λ}= 0$$$.
3. $$$λ\gt \max{}\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$ implies $$${P}_{λ}= 1$$$.

Proof.

1. Notice $$$λ\leq μ$$$ implies $$${f}_{μ}= g{f}_{λ}$$$ for a continuous function $$$g$$$, so $$${f}_{μ}\mathopen{}\left( A\right)\mathclose{}= g\mathopen{}\left( A\right)\mathclose{}{f}_{λ}\mathopen{}\left( A\right)\mathclose{}$$$. Hence $$${f}_{λ}\mathopen{}\left( A\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= 0$$$ implies $$${f}_{μ}\mathopen{}\left( A\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= 0$$$, that is, $$$\operatorname{Ker}\mathopen{}\left( {f}_{λ}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}\subseteq \operatorname{Ker}\mathopen{}\left( {f}_{μ}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$.
2. $$$λ\lt \min{}\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$ implies $$$A-λ\geq 0$$$ and $$$A-λ$$$ invertible, so $$${f}_{λ}\mathopen{}\left( A\right)\mathclose{}= 2\mathopen{}\left(A-λ\right)\mathclose{}$$$ is invertible, so $$${P}_{λ}= 0$$$.
3. $$$λ\gt \max{}\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$, then $$$A-λ\leq 0$$$ and so $$$\mathopen{}\left\lvert{}A-λ\right\rvert\mathclose{}= λ-A$$$ so $$${f}_{λ}\mathopen{}\left( A\right)\mathclose{}= 0$$$ so $$$\operatorname{Ker}\mathopen{}\left( {f}_{λ}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}= H$$$.

Lemma IV.3

1. $$$A{P}_{λ}= {P}_{λ}A$$$.
2. $$$A{P}_{λ}\leq λ{P}_{λ}$$$.
3. $$$A\mathopen{}\left(1-{P}_{λ}\right)\mathclose{}\geq λ\mathopen{}\left(1-{P}_{λ}\right)\mathclose{}$$$.

Proof.

1. Let $$$x\in \operatorname{Ker}\mathopen{}\left( {f}_{λ}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$. Then $$${f}_{λ}\mathopen{}\left( A\right)\mathclose{}\mathopen{}\left( A\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}= A\mathopen{}\left( {f}_{λ}\mathopen{}\left( A\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}= 0$$$. So $$$A\operatorname{Ker}\mathopen{}\left( {f}_{λ}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}\subseteq \operatorname{Ker}\mathopen{}\left( {f}_{λ}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$, so $$${P}_{λ}A{P}_{λ}= A{P}_{λ}$$$. But also $${P}_{λ}A= {P}_{λ}^{*} A^{*}= \mathopen{}\left({P}_{λ}A{P}_{λ}\right)\mathclose{}^{*}= {P}_{λ}^{*} A^{*} {P}_{λ}^{*}= {P}_{λ}A{P}_{λ}= A{P}_{λ} \text{.}$$
2. For $$$y\in \operatorname{Ker}\mathopen{}\left( {f}_{λ}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$, we have that $$$\mathopen{}\left(A-λ\right)\mathclose{}y= {-}\mathopen{}\left\lvert{}A-λ\right\rvert\mathclose{}y$$$. So for any $$$x\in H$$$, $$\mathopen{}\left\langle{}\mathopen{}\left(A-λ\right)\mathclose{}\mathopen{}\left( {P}_{λ}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x\right\rangle\mathclose{}= \mathopen{}\left\langle{}{P}_{λ}\mathopen{}\left( \mathopen{}\left(A-λ\right)\mathclose{}\mathopen{}\left( {P}_{λ}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left(A-λ\right)\mathclose{}\mathopen{}\left( {P}_{λ}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, {P}_{λ}\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}= {-} \mathopen{}\left\langle{}\mathopen{}\left\lvert{}A-λ\right\rvert\mathclose{}\mathopen{}\left( {P}_{λ}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, {P}_{λ}\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{} \leq 0 \text{,}$$ showing $$$\mathopen{}\left(A-λ\right)\mathclose{}\mathopen{}\left( {P}_{λ}\right)\mathclose{}\leq 0$$$.
3. For any $$$y\in H$$$, $$\mathopen{}\left\langle{}\mathopen{}\left(A-λ\right)\mathclose{}\mathopen{}\left( {f}_{λ}\mathopen{}\left( {A}^{\frac{1}{2}}\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}, {f}_{λ}\mathopen{}\left( {A}^{\frac{1}{2}}\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left(A-λ\right)\mathclose{}\mathopen{}\left( {f}_{λ}\mathopen{}\left( A\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}, y\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left({\mathopen{}\left(A-λ\right)\mathclose{}}^{2}+\mathopen{}\left(A-λ\right)\mathclose{}\mathopen{}\left\lvert{}A-λ\right\rvert\mathclose{}\right)\mathclose{}\mathopen{}\left( y\right)\mathclose{}, y\right\rangle\mathclose{}\geq 0 \text{.}$$ It follows that $$$\mathopen{}\left\langle{}\mathopen{}\left(A-λ\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}\geq 0$$$ for all $$$x\in \overline{ {f}_{λ}\mathopen{}\left( {A}^{\frac{1}{2}}\mathopen{}\left( H\right)\mathclose{}\right)\mathclose{} }$$$. Notice $$\overline{ {f}_{λ}\mathopen{}\left( {A}^{\frac{1}{2}}\mathopen{}\left( H\right)\mathclose{}\right)\mathclose{} }= { \mathopen{}\left(\operatorname{Ker}\mathopen{}\left( \mathopen{}\left({f}_{λ}\mathopen{}\left( {A}^{\frac{1}{2}}\right)\mathclose{}\right)\mathclose{} ^{*}\right)\mathclose{}\right)\mathclose{} }^{\perp}= \operatorname{Ker}\mathopen{}\left( { \mathopen{}\left({f}_{λ}\mathopen{}\left( {A}^{\frac{1}{2}}\right)\mathclose{}\right)\mathclose{} }^{\perp}\right)\mathclose{}= \mathopen{}\left(1-{P}_{λ}\right)\mathclose{}\mathopen{}\left( H\right)\mathclose{} \text{,}$$ so $$0\leq \mathopen{}\left\langle{}\mathopen{}\left(A-λ\right)\mathclose{}\mathopen{}\left( \mathopen{}\left(1-{P}_{λ}\right)\mathclose{}\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}, \mathopen{}\left(1-{P}_{λ}\right)\mathclose{}\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left(A-λ\right)\mathclose{}\mathopen{}\left( \mathopen{}\left(1-{P}_{λ}\right)\mathclose{}\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}, y\right\rangle\mathclose{} \text{.}$$

Lemma IV.4

For $$$λ\leq μ$$$, we have $$$λ\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}\leq A\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}\leq μ\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}$$$.

Proof. Continuing with the proof of the spectral theorem (which we will state at the end), fix $$$a\lt b$$$ with $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\subseteq \mathopen{}\left(a, b\right)\mathclose{}$$$. So $$\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}A\mathopen{}\left(1-{P}_{λ}\right)\mathclose{}\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}\geq λ\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}\mathopen{}\left(1-{P}_{λ}\right)\mathclose{}\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}$$ implies $$A\mathopen{}\left(1-{P}_{λ}\right)\mathclose{}\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}\geq λ\mathopen{}\left(1-{P}_{λ}\right)\mathclose{}\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}= λ\mathopen{}\left({P}_{μ}-{P}_{λ}{P}_{μ}-{P}_{λ}+{P}_{λ}\right)\mathclose{}= λ\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}$$ since $$${P}_{λ}\leq {P}_{μ}$$$. Thus $$$A{P}_{μ}-{P}_{λ}\geq λ\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}$$$. Next, $$$A{P}_{μ}\leq μ{P}_{μ}$$$, so $$\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}A{P}_{μ}\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}\leq μ\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}{P}_{μ}\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}$$ implies $$$A{\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}}^{2}\leq μ{\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}}^{2}$$$ implies $$$A\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}\leq μ\mathopen{}\left({P}_{μ}-{P}_{λ}\right)\mathclose{}$$$.

We now move towards the first version of the spectral theorem: Fix $$$a\lt b$$$ with $$$σ\mathopen{}\left( A\right)\mathclose{}\subseteq \mathopen{}\left(a, b\right)\mathclose{}$$$. Consider a partition of $$$\mathopen{}\left[a, b\right]\mathclose{}$$$: $$$a= {λ}_{0}\lt {λ}_{1}\lt \dotsb\lt {λ}_{n-1}\lt {λ}_{n}= b$$$. Now pick $$${t}_{j}\in \mathopen{}\left[{λ}_{j-1}, {λ}_{j}\right]\mathclose{}$$$ for $$$j\in \mathopen{}\left\{\, 1, 2, \dotsc, n\,\right\}\mathclose{}$$$. Now consider $$$S= \sum_{j=1}^{n}{} {t}_{j}\mathopen{}\left({P}_{{λ}_{j}}-{P}_{{λ}_{j-1}}\right)\mathclose{}$$$.

We claim that $$$\mathopen{}\left\lVert{}A-S\right\rVert\mathclose{}\leq \max_{j}{}\mathopen{}\left({λ}_{j}-{λ}_{j-1}\right)\mathclose{}$$$. Let $$${Q}_{j}= {P}_{{λ}_{j}}-{P}_{{λ}_{j-1}}$$$. Note that $$$\sum_{j=1}^{n}{}{Q}_{j}= 1$$$ and that the $$${Q}_{j}$$$'s are mutually orthogonal: $$${Q}_{j}{Q}_{k}= 0$$$ if $$$j\neq k$$$. Then $$$H= \bigoplus{} {Q}_{j}\mathopen{}\left( H\right)\mathclose{}$$$ so $$$A-S= \sum_{j=1}^{n}{}A{Q}_{j}-\sum_{j=1}^{n}{} {t}_{j}{Q}_{j} = \sum_{j=1}^{n}{} \mathopen{}\left(A-{t}_{j}\right)\mathclose{}{Q}_{j}$$$. Notice that for any operators $$${T}_{1}$$$, $$${T}_{2}$$$, …, $$${T}_{n}$$$ that commute with $$${Q}_{j}$$$, we have that $$$\mathopen{}\left\lVert{}\sum_{j=1}^{n}{} {T}_{j}{Q}_{j} \right\rVert\mathclose{}= \max_{j}{} \mathopen{}\left\lVert{}{T}_{j}{Q}_{j}\right\rVert\mathclose{}$$$ because $$\mathopen{}\left\lVert{}\sum_{j=1}^{n}{} {T}_{j}{Q}_{j} \right\rVert\mathclose{}\geq \mathopen{}\left\lVert{}\sum_{j=1}^{n}{} {T}_{j}{Q}_{j}|{Q}_{k}\mathopen{}\left( H\right)\mathclose{} \right\rVert\mathclose{}= \mathopen{}\left\lVert{}{T}_{k}{Q}_{k}|{Q}_{k}\mathopen{}\left( H\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}{T}_{k}{Q}_{k}\right\rVert\mathclose{} \text{.}$$

For the proof of the reverse inequality, $${\mathopen{}\left\lVert{}\sum_{j=1}^{n}{} {T}_{j}\mathopen{}\left( {Q}_{j}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{} \right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}\sum_{j=1}^{n}{} {Q}_{j}\mathopen{}\left( {T}_{j}\mathopen{}\left( {Q}_{j}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right)\mathclose{} \right\rVert\mathclose{}}^{2}= \sum_{j=1}^{n}{} {\mathopen{}\left\lVert{}{T}_{j}\mathopen{}\left( {Q}_{j}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right\rVert\mathclose{}}^{2} = \sum_{j=1}^{n}{} {\mathopen{}\left\lVert{}{T}_{j}\mathopen{}\left( {Q}_{j}\mathopen{}\left( {Q}_{j}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}\right\rVert\mathclose{}}^{2} \leq \sum_{j=1}^{n}{} {\mathopen{}\left\lVert{}{T}_{j}\mathopen{}\left( {Q}_{j}\right)\mathclose{}\right\rVert\mathclose{}}^{2}{\mathopen{}\left\lVert{}{Q}_{j}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2} \leq \max_{k}{} {\mathopen{}\left\lVert{}{T}_{k}\mathopen{}\left( {Q}_{k}\right)\mathclose{}\right\rVert\mathclose{}}^{2}\sum_{j=1}^{n}{} {\mathopen{}\left\lVert{}{Q}_{j}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2} = \max_{k}{} {\mathopen{}\left\lVert{}{T}_{k}\mathopen{}\left( {Q}_{k}\right)\mathclose{}\right\rVert\mathclose{}}^{2}{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}^{2} \text{.}$$ So $$$\mathopen{}\left\lVert{}A-S\right\rVert\mathclose{}= \max_{j}{} \mathopen{}\left\lVert{}\mathopen{}\left(A-{t}_{j}\right)\mathclose{}{Q}_{j}\right\rVert\mathclose{}$$$. By Lemma IV.4, $$${λ}_{j-1}{Q}_{j}\leq A{Q}_{j}\leq {λ}_{j}{Q}_{j}$$$. So $$$\mathopen{}\left({λ}_{j-1}-{t}_{j}\right)\mathclose{}{Q}_{j}\leq \mathopen{}\left(A-{t}_{j}\right)\mathclose{}{Q}_{j}\leq \mathopen{}\left({λ}_{j}-{t}_{j}\right)\mathclose{}{Q}_{j}$$$ which makes $$${-} \mathopen{}\left({t}_{j}-{λ}_{j-1}\right)\mathclose{} \leq \mathopen{}\left(A-{t}_{j}\right)\mathclose{}{Q}_{j}\leq \mathopen{}\left({λ}_{j}-{t}_{j}\right)\mathclose{}$$$. We conclude that $$$\mathopen{}\left\lVert{}\mathopen{}\left(A-{t}_{j}\right)\mathclose{}{Q}_{j}\right\rVert\mathclose{}\leq \max{} \mathopen{}\left({t}_{j}-{λ}_{j-1}, {λ}_{j}-{t}_{j}\right)\mathclose{} \leq {λ}_{j}-{λ}_{j-1}$$$. Thus $$$\mathopen{}\left\lVert{}A-S\right\rVert\mathclose{}\leq \max{} \mathopen{}\left({λ}_{j}-{λ}_{j-1}\right)\mathclose{}$$$.

We have now proved the spectral theorem for bounded self-adjoint operators. It remains to be seen what the spectral theorem has to do with the spectrum of such an operator.

Theorem IV.5

Given $$$A= A^{*}\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ and $$$a\lt b$$$ such that $$$σ\mathopen{}\left( a\right)\mathclose{}\subseteq \mathopen{}\left(a, b\right)\mathclose{}$$$ there is an increasing projection-valued function $$$λ\mapsto {P}_{λ}$$$ with $$${P}_{a}= 0$$$ and $$${P}_{b}= 1$$$ and each $$${P}_{λ} A$$$ such that $$$A= \int _{a}^{b}{}λ\,\mathrm{d}{P}_{λ}$$$ in the sense that $$$A$$$ is the norm-limit of sums $$$\sum_{j=1}^{n}{} {t}_{j}\mathopen{}\left({P}_{{λ}_{j}}-{P}_{{λ}_{j-1}}\right)\mathclose{}$$$ over partitions $$$a= {λ}_{0}\lt {λ}_{1}\lt \dotsb\lt {λ}_{n}= b$$$ marked by $$${t}_{j}\in \mathopen{}\left[{λ}_{j-1}, {λ}_{j}\right]\mathclose{}$$$, as the mesh of the partition goes to $$$0$$$.

Example IV.6

$$$\mathrm{M}_{α}$$$ is multiplication on $$$\mathrm{L}^{\mathrm{2}}\mathopen{}\left( X, μ\right)\mathclose{}$$$ by $$$α\in \mathrm{L}^{\mathrm{∞}}\mathopen{}\left( X, μ\right)\mathclose{}$$$, i.e. $$$\mathopen{}\left(\mathrm{M}_{α}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= α\mathopen{}\left( x\right)\mathclose{}ξ\mathopen{}\left( x\right)\mathclose{}$$$. $$$\mathrm{M}_{α}-λ+\mathopen{}\left\lvert{}\mathrm{M}_{α}-λ\right\rvert\mathclose{}$$$ is multiplication by $$$α-λ+\mathopen{}\left\lvert{}α-λ\right\rvert\mathclose{}$$$. $$$\operatorname{Ker}\mathopen{}\left( {f}_{λ}\mathopen{}\left( \mathrm{M}_{α}\right)\mathclose{}\right)\mathclose{}= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left\{\, x\,\middle\vert\, , α\mathopen{}\left( x\right)\mathclose{}\leq λ, \,\right\}\mathclose{}\right)\mathclose{}$$$ (i.e. $$$\mathrm{L}^{\mathrm{2}}\mathopen{}\left( X\right)\mathclose{}$$$ functions vanishing almost everywhere off $$$\mathopen{}\left\{\, x\,\middle\vert\, , α\mathopen{}\left( x\right)\mathclose{}\leq λ, \,\right\}\mathclose{}$$$) and $$${P}_{λ}$$$ is projection on $$$\operatorname{Ker}\mathopen{}\left( {f}_{λ}\mathopen{}\left( \mathrm{M}_{α}\right)\mathclose{}\right)\mathclose{}$$$.

In the setup of Theorem IV.5, consider $$$g\in \mathbb{C} \mathopen{}\left[a, b\right]\mathclose{}$$$. What is $$$\int _{a}^{b}{}g\mathopen{}\left( λ\right)\mathclose{}\,\mathrm{d}{P}_{λ}$$$? Given $$$ε\gt 0$$$, get $$$δ\gt 0$$$ such that $$$\mathopen{}\left\lvert{}g\mathopen{}\left( s\right)\mathclose{}-g\mathopen{}\left( t\right)\mathclose{}\right\rvert\mathclose{}\lt ε$$$ whenever $$$\mathopen{}\left\lvert{}s-t\right\rvert\mathclose{}\lt δ$$$. Given a partition $$$a= {λ}_{0}\lt {λ}_{1}\lt \dotsb\lt {λ}_{n}= b$$$ with $$${λ}_{j}-{λ}_{j-1}\lt δ$$$ for all $$$j$$$ and any refinement $$$a= {μ}_{0}\lt {μ}_{1}\lt \dotsb\lt {μ}_{r}= b$$$. If we mark the $$$λ$$$-partition with $$${t}_{j}\in \mathopen{}\left[{λ}_{j-1}, {λ}_{j}\right]\mathclose{}$$$ and the $$$μ$$$-partition with $$${s}_{i}\in \mathopen{}\left[{μ}_{i-1}, {μ}_{i}\right]\mathclose{}$$$ and form $$$T= \sum_{j=1}^{n}{} g\mathopen{}\left( {t}_{j}\right)\mathclose{}\mathopen{}\left({P}_{{λ}_{j}}-{P}_{{λ}_{j-1}}\right)\mathclose{}$$$ and $$$S= \sum_{i=1}^{r}{} g\mathopen{}\left( {s}_{i}\right)\mathclose{}\mathopen{}\left({P}_{{μ}_{i}}-{P}_{{μ}_{i-1}}\right)\mathclose{}$$$, then $$$\mathopen{}\left\lVert{}S-T\right\rVert\mathclose{}\lt ε$$$ because everything commutes. Indeed, $$\mathopen{}\left\lVert{}S-T\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\sum_{j=1}^{n}{} \mathopen{}\left(S-T\right)\mathclose{}\mathopen{}\left({P}_{{λ}_{j}}-{P}_{{λ}_{j-1}}\right)\mathclose{} \right\rVert\mathclose{}= \max_{j}{} \mathopen{}\left\lVert{}\mathopen{}\left(S-T\right)\mathclose{}\mathopen{}\left({P}_{{λ}_{j}}-{P}_{{λ}_{j-1}}\right)\mathclose{}\right\rVert\mathclose{}$$ and each of these is less than $$$ε$$$ by the following calculation: Purely for convenience, let's take $$$j= 1$$$. Write the refined partition of $$$\mathopen{}\left[{λ}_{0}, {λ}_{1}\right]\mathclose{}$$$ as $$${λ}_{0}= {μ}_{0}\lt {μ}_{1}\lt \dotsb\lt {μ}_{k}= {λ}_{1}$$$. Then $$\mathopen{}\left\lVert{}\mathopen{}\left(S-T\right)\mathclose{}\mathopen{}\left({P}_{{λ}_{1}}-{P}_{{λ}_{0}}\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\sum_{i=1}^{k}{} \mathopen{}\left(S-T\right)\mathclose{}\mathopen{}\left({P}_{{μ}_{i}}-{P}_{{μ}_{i-1}}\right)\mathclose{} \right\rVert\mathclose{}= \max_{1\leq i\leq k}{} \mathopen{}\left\lVert{}\mathopen{}\left(S-T\right)\mathclose{}\mathopen{}\left({P}_{{μ}_{i}}-{P}_{{μ}_{i-1}}\right)\mathclose{}\right\rVert\mathclose{} = \max_{i}{} \mathopen{}\left\lVert{}\mathopen{}\left(g\mathopen{}\left( {s}_{i}\right)\mathclose{}-g\mathopen{}\left( {t}_{1}\right)\mathclose{}\right)\mathclose{}\mathopen{}\left({P}_{{μ}_{i}}-{P}_{{μ}_{i-1}}\right)\mathclose{}\right\rVert\mathclose{} \leq \max_{i}{} \mathopen{}\left\lvert{}g\mathopen{}\left( {s}_{i}\right)\mathclose{}-g\mathopen{}\left( {t}_{1}\right)\mathclose{}\right\rvert\mathclose{} \lt ε \text{,}$$ because $$$\mathopen{}\left\lvert{}{s}_{i}-{t}_{1}\right\rvert\mathclose{}\lt δ$$$ for these $$$i$$$.

Remark IV.7

Because $$$\mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ is complete it follows that there exists $$${T}_{g}\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ —written $$${T}_{g}= \int _{a}^{b}{}g\mathopen{}\left( λ\right)\mathclose{}\,\mathrm{d}{P}_{λ}$$$ —such that for all $$$ε\gt 0$$$ there exists a partition $$$F$$$ such that $$$\mathopen{}\left\lVert{}{T}_{g}-S\right\rVert\mathclose{}\lt ε$$$ for any sum $$$S= \sum{} g\mathopen{}\left( {s}_{i}\right)\mathclose{}\mathopen{}\left({P}_{{μ}_{i}}-{P}_{{μ}_{i-1}}\right)\mathclose{}$$$ in which $$$\mathopen{}\left\{\, {μ}_{0}, {μ}_{1}, \dotsc, {μ}_{n}\,\right\}\mathclose{}$$$ refines $$$F$$$.

Some easy properties:

1. $$${T}_{g}$$$ commutes with $$${P}_{λ}$$$.
2. $$${T}_{c{g}_{1}+{g}_{2}}= c{T}_{{g}_{1}}+{T}_{{g}_{2}}$$$.
3. $$${T}_{g}^{*}= {T}_{\overline{g}}$$$.
4. $$${T}_{{g}_{1}{g}_{2}}= {T}_{{g}_{1}}{T}_{{g}_{2}}$$$.
5. $$$\mathopen{}\left\lVert{}T\mathopen{}\left( g\right)\mathclose{}\right\rVert\mathclose{}\leq \max_{\mathopen{}\left[a, b\right]\mathclose{}}{}\mathopen{}\left\lvert{}g\right\rvert\mathclose{}$$$.

The only one of these that offers any resistance is item 4. For this, get a partition $$${λ}_{0}\lt {λ}_{1}\lt \dotsb\lt {λ}_{n}$$$ (by common refinement of a partition that works for $$${g}_{1}$$$ and one that works for $$${g}_{2}$$$) such that $$$\sum{} {g}_{1}\mathopen{}\left( {t}_{j}\right)\mathclose{} \, \Delta {P}_{{λ}_{j}}$$$ and $$$\sum{} {g}_{2}\mathopen{}\left( {t}_{j}\right)\mathclose{} \, \Delta {P}_{{λ}_{j}}$$$ are close to $$${T}_{{g}_{1}}$$$ and $$${T}_{{g}_{2}}$$$ respectively (where $$$\Delta {P}_{{λ}_{i}}= {P}_{{λ}_{i}}-{P}_{{λ}_{i-1}}$$$), and then $$\mathopen{}\left(\sum{} {g}_{1}\mathopen{}\left( {t}_{j}\right)\mathclose{} \, \Delta {P}_{{λ}_{j}} \right)\mathclose{}\mathopen{}\left(\sum{} {g}_{2}\mathopen{}\left( {t}_{j}\right)\mathclose{} \, \Delta {P}_{{λ}_{j}} \right)\mathclose{}= \sum{} {g}_{1}\mathopen{}\left( {t}_{j}\right)\mathclose{}{g}_{2}\mathopen{}\left( {t}_{j}\right)\mathclose{} \, \Delta {P}_{{λ}_{j}} \text{.}$$ since $$${P}_{{λ}_{j}} {P}_{{λ}_{k}} = 0$$$ for $$$j\neq k$$$.

Theorem IV.8

Let $$$A= A^{*}\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ with $$$σ\mathopen{}\left( A\right)\mathclose{}\subseteq \mathopen{}\left(a, b\right)\mathclose{}$$$ and spectral resolution $$$λ\mapsto {P}_{λ}$$$ as in Theorem IV.5. Then for $$$g\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{}$$$ we have $$$\int _{a}^{b}{} g\mathopen{}\left( λ\right)\mathclose{} \,\mathrm{d}{P}_{λ}= \mathopen{}\left(g|σ\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( A\right)\mathclose{}$$$ where the left-hand side is the norm limit of Riemann-Stieltjes sums and the right-hand side comes from functional calculus on $$$\mathrm{C}\mathopen{}\left( σ\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$.

Proof. We have shown that $$$g\mapsto \int _{a}^{b}{} g\mathopen{}\left( λ\right)\mathclose{} \,\mathrm{d}{P}_{λ}\equiv {T}_{g}$$$ is a *-homomorphism from $$$\mathrm{C}\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{}$$$ into $$$\mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ with $$$\mathopen{}\left\lVert{}{T}_{g}\right\rVert\mathclose{}\leq \max_{\mathopen{}\left[a, b\right]\mathclose{}}{}\mathopen{}\left\lvert{}g\right\rvert\mathclose{}$$$. We have that $$$\int _{a}^{b}{} 1 \,\mathrm{d}{P}_{λ}= I$$$ and $$$\int _{a}^{b}{} λ \,\mathrm{d}{P}_{λ}= A$$$, so for any polynomial $$$p\mathopen{}\left( x\right)\mathclose{}$$$ we have that $$$\int _{a}^{b}{} p\mathopen{}\left( A\right)\mathclose{} \,\mathrm{d}{P}_{λ}= p\mathopen{}\left( A\right)\mathclose{}$$$. Take $$$g\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{}$$$, and write $$$ĝ= g|σ\mathopen{}\left( A\right)\mathclose{}$$$. Get a sequence $$$\mathopen{}\left( {p}_{n} \right)\mathclose{}$$$ of polynomials with $$${p}_{n} \to g$$$ uniformly on $$$\mathopen{}\left[a, b\right]\mathclose{}$$$. So $$${p}_{n}\mathopen{}\left( A\right)\mathclose{}= \int _{a}^{b}{} {p}_{n}\mathopen{}\left( λ\right)\mathclose{} \,\mathrm{d}{P}_{λ} \to \int _{a}^{b}{} g\mathopen{}\left( λ\right)\mathclose{} \,\mathrm{d}{P}_{λ}$$$ but also $$${p}_{n} \to ĝ$$$ uniformly on $$$σ\mathopen{}\left( A\right)\mathclose{}$$$, so $$${p}_{n}\mathopen{}\left( A\right)\mathclose{} \to ĝ\mathopen{}\left( A\right)\mathclose{}$$$.

Remark IV.9

Theorem IV.8 says that for $$$g\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{}$$$, the integral $$$\int _{a}^{b}{} g\mathopen{}\left( λ\right)\mathclose{} \,\mathrm{d}{P}_{λ}$$$ depends only on the restriction of $$$g$$$ to $$$σ\mathopen{}\left( A\right)\mathclose{}$$$. It follows that an open real interval misses $$$σ\mathopen{}\left( A\right)\mathclose{}$$$ if and only if $$$λ\mapsto {P}_{λ}$$$ is constant on that interval.

Eigenvalues of $$$A$$$ show up in the spectral resolution as jump discontinuities, as we shall now explain.

Lemma IV.10

Let $$$\mathopen{}\left({E}_{n}\right)\mathclose{}$$$ be a decreasing sequence of projections, i.e. $$${E}_{n}\geq {E}_{n+1}$$$ for all $$$n$$$, and let $$$E$$$ be the orthogonal projection on $$$\bigcap_{n}{}{E}_{n}\mathopen{}\left( H\right)\mathclose{}$$$. Then $$${E}_{n}\mathopen{}\left( ξ\right)\mathclose{} \to E\mathopen{}\left( ξ\right)\mathclose{}$$$ for all $$$ξ\in H$$$.

Proof. The nonnegative sequence $$$\mathopen{}\left\lVert{}\mathopen{}\left({E}_{n}-E\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}$$$ is decreasing. Call the limit $$$r$$$. Given $$$ε\gt 0$$$, get $$$N$$$ such that $$${\mathopen{}\left\lVert{}\mathopen{}\left({E}_{n}-E\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}}^{2}\lt {r}^{2}+ε$$$ for $$$n\geq N$$$. If $$$m\gt n\geq N$$$, then vectors $$$\mathopen{}\left({E}_{n}-{E}_{m}\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}$$$ and $$$\mathopen{}\left({E}_{m}-E\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}$$$ are orthogonal, so $${\mathopen{}\left\lVert{}\mathopen{}\left({E}_{n}-E\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}\mathopen{}\left({E}_{n}-{E}_{m}\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}\mathopen{}\left({E}_{m}-E\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}}^{2}$$ and thus $$${\mathopen{}\left\lVert{}\mathopen{}\left({E}_{n}-{E}_{m}\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}}^{2}$$$ is the difference of two numbers in $$$\mathopen{}\left[{r}^{2}, {r}^{2}+ε\right)\mathclose{}$$$, which makes $$${\mathopen{}\left\lVert{}\mathopen{}\left({E}_{n}-{E}_{m}\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}}^{2}\lt ε$$$. The sequence $$$\mathopen{}\left( {E}_{n}\mathopen{}\left( ξ\right)\mathclose{} \right)\mathclose{}$$$ therefore converges, say to $$$η$$$. For any $$$k$$$, we have $$${E}_{k}\mathopen{}\left( η\right)\mathclose{}= \lim_{n}{} {E}_{k}\mathopen{}\left( {E}_{n}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{} = \lim_{n}{} {E}_{n}\mathopen{}\left( ξ\right)\mathclose{} = η$$$. Thus $$$η= E\mathopen{}\left( η\right)\mathclose{}= \lim_{n}{} E\mathopen{}\left( {E}_{n}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{} = E\mathopen{}\left( ξ\right)\mathclose{}$$$.

Remark IV.11

For $$$λ\in \mathbb{R}$$$, let $$${Q}_{λ}$$$ be the orthogonal projection on $$$\overline{ \bigcup_{θ\lt λ}{} {P}_{θ}\mathopen{}\left( H\right)\mathclose{} }$$$ and $$${R}_{λ}$$$ be the orthogonal projection on $$$\bigcap_{θ\gt λ}{} {P}_{θ}\mathopen{}\left( H\right)\mathclose{}$$$. We have $$${Q}_{λ}\leq {P}_{λ}\leq {R}_{λ}$$$. The lemma above yields $$$\lim_{θ\to{λ}^{\mathrm{+}}}{} {P}_{θ}\mathopen{}\left( ξ\right)\mathclose{} = {R}_{λ}\mathopen{}\left( ξ\right)\mathclose{}$$$ for all $$$ξ$$$, and likewise for the limit from below by taking orthogonal complements.

Proposition IV.12

For all $$$λ$$$, $$${R}_{λ}-{Q}_{λ}$$$ is the orthogonal projection on $$$\operatorname{Ker}\mathopen{}\left( A-λ\right)\mathclose{}$$$. That is, $$$λ$$$ is an eigenvalue of $$$A$$$ if and only if $$${Q}_{λ}\neq {R}_{λ}$$$.

Proof. For any $$$ξ$$$, we have $$$\mathopen{}\left\langle{}\mathopen{}\left\langle{}{\mathopen{}\left(A-λ\right)\mathclose{}}^{2}\mathopen{}\left( ξ\right)\mathclose{}\right\rangle\mathclose{}, ξ\right\rangle\mathclose{}= \int _{a}^{b}{} {\mathopen{}\left(θ-λ\right)\mathclose{}}^{2} \,\mathrm{d} \mathopen{}\left\langle{}{P}_{θ}\mathopen{}\left( ξ\right)\mathclose{}, ξ\right\rangle\mathclose{}$$$. If this scalar Riemann-Stieltjes integral vanishes, then $$$\mathopen{}\left\langle{}{P}_{θ}\mathopen{}\left( ξ\right)\mathclose{}, ξ\right\rangle\mathclose{}$$$ (i.e., $$${\mathopen{}\left\lVert{}{P}_{θ}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}}^{2}$$$) must be constant for $$$θ\gt λ$$$ and for $$$θ\lt λ$$$, i.e. $$${\mathopen{}\left\lVert{}{P}_{θ}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}ξ\right\rVert\mathclose{}}^{2}$$$ (and hence $$${P}_{θ}\mathopen{}\left( ξ\right)\mathclose{}= ξ$$$) for $$$θ\gt λ$$$, and $$${P}_{θ}\mathopen{}\left( ξ\right)\mathclose{}= 0$$$ for $$$θ\lt λ$$$. Thus, $$${R}_{λ}\mathopen{}\left( ξ\right)\mathclose{}= ξ$$$ and $$${Q}_{λ}\mathopen{}\left( ξ\right)\mathclose{}= 0$$$. We have shown that $$$\operatorname{Ker}\mathopen{}\left( A-λ\right)\mathclose{}\subseteq \mathopen{}\left({R}_{λ}-{Q}_{λ}\right)\mathclose{}\mathopen{}\left( H\right)\mathclose{}$$$. Conversely, if $$${R}_{λ}\mathopen{}\left( ξ\right)\mathclose{}= ξ$$$ and $$${Q}_{λ}\mathopen{}\left( ξ\right)\mathclose{}= 0$$$, then the function $$$θ\mapsto \mathopen{}\left\langle{}{P}_{θ}\mathopen{}\left( ξ\right)\mathclose{}, ξ\right\rangle\mathclose{}$$$ jumps from $$$0$$$ to $$${\mathopen{}\left\lVert{}ξ\right\rVert\mathclose{}}^{2}$$$ across $$$λ$$$ and is otherwise flat, so the value of the integral above is $$${\mathopen{}\left\lVert{}ξ\right\rVert\mathclose{}}^{2}$$$ times the value of the integrand at $$$λ$$$. This is of course $$$0$$$, making $$$\mathopen{}\left(A-λ\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}= 0$$$.

We turn now to unbounded (i.e. not necessarily bounded) operators, aiming ultimately at the spectral theorem for the densely defined self-adjoint ones. A natural example is the operator $$$T$$$ with domain $$$\mathopen{}\left\{\, ξ\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}\right)\mathclose{}\,\middle\vert\, , \int _{\mathbb{R}}{} {x}^{2}{\mathopen{}\left\lvert{}ξ\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}}^{2} \,\mathrm{d}x\lt \infty, \,\right\}\mathclose{}$$$ defined by $$$\mathopen{}\left(T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= xξ\mathopen{}\left( x\right)\mathclose{}$$$.

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