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

Theorem II.22 (Riesz-Fisher Theorem)

Let $$$H$$$ be a Hilbert space and let $$$φ\in {H}^{*}$$$. Then there exists a unique $$$y\in H$$$ such that $$$φ\mathopen{}\left( x\right)\mathclose{}= \mathopen{}\left\langle{}x, y\right\rangle\mathclose{}$$$ for all $$$x\in H$$$. Further, $$$\mathopen{}\left\lVert{}φ\right\rVert\mathclose{}= \mathopen{}\left\lVert{}y\right\rVert\mathclose{}$$$.

Proof. We may assume $$$φ\neq 0$$$. Since $$$φ$$$ is continuous, $$${φ}^{-1}\mathopen{}\left( 0\right)\mathclose{}$$$ is a closed proper subspace of $$$H$$$. So, $$${ {φ}^{-1}\mathopen{}\left( 0\right)\mathclose{} }^{\perp}$$$ is non-trivial (from Proposition I.61) and there is a non-zero vector $$${y}_{0}\in { {φ}^{-1}\mathopen{}\left( 0\right)\mathclose{} }^{\perp}$$$. Thus $$$φ\mathopen{}\left( {y}_{0}\right)\mathclose{}\neq 0$$$ (else $$$\mathopen{}\left\langle{}{y}_{0}, {y}_{0}\right\rangle\mathclose{}= 0$$$). For $$$x\in H$$$ consider $$$x-\frac{φ\mathopen{}\left( x\right)\mathclose{}}{φ\mathopen{}\left( {y}_{0}\right)\mathclose{}}{y}_{0}$$$, which belongs to $$${φ}^{-1}\mathopen{}\left( 0\right)\mathclose{}$$$. Since $$${y}_{0}\perp {φ}^{-1}\mathopen{}\left( 0\right)\mathclose{}$$$, we get that $$\mathopen{}\left\langle{}x-\frac{φ\mathopen{}\left( x\right)\mathclose{}}{φ\mathopen{}\left( {y}_{0}\right)\mathclose{}}{y}_{0}, {y}_{0}\right\rangle\mathclose{}= 0$$ and $$\mathopen{}\left\langle{}x, {y}_{0}\right\rangle\mathclose{}= \frac{φ\mathopen{}\left( x\right)\mathclose{}}{φ\mathopen{}\left( {y}_{0}\right)\mathclose{}}\mathopen{}\left\langle{}{y}_{0}, {y}_{0}\right\rangle\mathclose{} \text{;}$$ that is, $$$φ\mathopen{}\left( x\right)\mathclose{}= \frac{φ\mathopen{}\left( {y}_{0}\right)\mathclose{}}{\mathopen{}\left\langle{}{y}_{0}, {y}_{0}\right\rangle\mathclose{}}\mathopen{}\left\langle{}x, {y}_{0}\right\rangle\mathclose{}$$$. So, $$$y= \frac{\overline{ φ\mathopen{}\left( {y}_{0}\right)\mathclose{} }}{\mathopen{}\left\langle{}{y}_{0}, {y}_{0}\right\rangle\mathclose{}}{y}_{0}$$$ works.

For uniqueness: if also $$$φ\mathopen{}\left( x\right)\mathclose{}= \mathopen{}\left\langle{}x, z\right\rangle\mathclose{}$$$, then $$$\mathopen{}\left\langle{}x, y-z\right\rangle\mathclose{}= 0$$$ for all $$$x\in H$$$; in particular, $$$\mathopen{}\left\langle{}y-z, y-z\right\rangle\mathclose{}= 0$$$.

To show $$$\mathopen{}\left\lVert{}φ\right\rVert\mathclose{}= \mathopen{}\left\lVert{}y\right\rVert\mathclose{}$$$, we have $$$\mathopen{}\left\lvert{}φ\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lvert{}\mathopen{}\left\langle{}x, y\right\rangle\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{}$$$. So, $$$\mathopen{}\left\lVert{}y\right\rVert\mathclose{}\geq \mathopen{}\left\lVert{}φ\right\rVert\mathclose{}$$$. Also, $$$\mathopen{}\left\lvert{}φ\mathopen{}\left( y\right)\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lvert{}\mathopen{}\left\langle{}y, y\right\rangle\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lVert{}y\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{}$$$. So, $$$\mathopen{}\left\lvert{}φ\mathopen{}\left( \frac{y}{\mathopen{}\left\lVert{}y\right\rVert\mathclose{}}\right)\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lVert{}y\right\rVert\mathclose{}$$$. Thus $$$\mathopen{}\left\lVert{}φ\right\rVert\mathclose{}\geq \mathopen{}\left\lVert{}y\right\rVert\mathclose{}$$$.

Proposition II.23

Let $$$H$$$ and $$$J$$$ be Hilbert spaces, and let $$$T\in \mathcal{L}\mathopen{}\left( H, J\right)\mathclose{}$$$. Then there exists $$$T^{*}\in \mathcal{L}\mathopen{}\left( J, H\right)\mathclose{}$$$ such that $$$\mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}_{J}= \mathopen{}\left\langle{}x, T^{*}\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{}_{H}$$$ for all $$$x\in H$$$, $$$y\in J$$$. Further, $$$T^{*}$$$ is unique, $$$\mathopen{}\left( T^{*}\right)\mathclose{}^{*}= T$$$, and $$$\mathopen{}\left\lVert{} T^{*}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}T\right\rVert\mathclose{}$$$. $$$T^{*}$$$ is called the adjoint of $$$T$$$.

Proof. Fix $$$y\in J$$$ and consider $$$φ : H \to \mathbb{C}$$$ defined by $$$φ\mathopen{}\left( x\right)\mathclose{}= \mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}$$$. We have $$$\mathopen{}\left\lvert{}φ\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}T\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{}= \mathopen{}\left(\mathopen{}\left\lVert{}T\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{}\right)\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}$$$. So, $$$φ\in H^{*}$$$ (with $$$\mathopen{}\left\lVert{}φ\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}T\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{}$$$). Thus we get an implementing vector, call it $$$T^{*}\mathopen{}\left( y\right)\mathclose{}$$$, such that $$$φ\mathopen{}\left( x\right)\mathclose{}= \mathopen{}\left\langle{}x, T^{*}\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{}$$$ for all $$$x$$$. Easily, $$$T^{*} : J \to H$$$ is linear and $$$\mathopen{}\left\lVert{} T^{*}\mathopen{}\left( y\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}φ\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}T\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{}$$$. So, $$$T^{*}\in \mathcal{L}\mathopen{}\left( J, H\right)\mathclose{}$$$ with $$$\mathopen{}\left\lVert{} T^{*}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}T\right\rVert\mathclose{}$$$. At this point it is clear that $$$\mathopen{}\left( T^{*}\right)\mathclose{}^{*}= T$$$, so we have $$$\mathopen{}\left\lVert{}T\right\rVert\mathclose{}= \mathopen{}\left\lVert{} \mathopen{}\left( T^{*}\right)\mathclose{}^{*}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{} T^{*}\right\rVert\mathclose{}$$$. Thus $$$\mathopen{}\left\lVert{} T^{*}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}T\right\rVert\mathclose{}$$$.

Example II.24

The adjoint of a map from $$${\mathbb{C}}^{n}$$$ to $$${\mathbb{C}}^{m}$$$ is obtained by taking the conjugate transpose of the representing matrix: $$$T= \mathopen{}\left( {t}_{ij} \right)\mathclose{}$$$ to $$$T^{*}= \mathopen{}\left( \overline{ {t}_{ji} } \right)\mathclose{}$$$.

Example II.25

If $$$T$$$ is multiplication by a bounded measurable $$$φ$$$ on $$$\mathrm{L}^{\mathrm{2}}\mathopen{}\left( Ω\right)\mathclose{}$$$, then $$$T^{*}$$$ is multiplication by $$$\overline{φ}$$$ because $$\mathopen{}\left\langle{} T^{*}\mathopen{}\left( f\right)\mathclose{}, g\right\rangle\mathclose{}= \mathopen{}\left\langle{}f, T\mathopen{}\left( g\right)\mathclose{}\right\rangle\mathclose{}= \int _{Ω}{}f\overline{T\mathopen{}\left( g\right)\mathclose{}}= \int _{Ω}{}f\overline{φg}= \int _{Ω}{} \mathopen{}\left(\overline{φ}f\right)\mathclose{}\overline{g} = \mathopen{}\left\langle{}\overline{φ}\mathopen{}\left( f\right)\mathclose{}, g\right\rangle\mathclose{} \text{,}$$ i.e. $$$T^{*}\mathopen{}\left( f\right)\mathclose{}= \overline{φ}\mathopen{}\left( f\right)\mathclose{}$$$.

Example II.26

Let $$$T$$$ be the Volterra operator on $$$\mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$: $$$\mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= \int _{0}^{x}{}f\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}t$$$. Then $$\mathopen{}\left\langle{} T^{*}\mathopen{}\left( f\right)\mathclose{}, g\right\rangle\mathclose{}= \mathopen{}\left\langle{}f, T\mathopen{}\left( g\right)\mathclose{}\right\rangle\mathclose{}= \int _{0}^{1}{} f\mathopen{}\left( x\right)\mathclose{}\overline{T\mathopen{}\left( g\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}} \,\mathrm{d}x= \int _{0}^{1}{} f\mathopen{}\left( x\right)\mathclose{}\mathopen{}\left(\int _{0}^{x}{}\overline{g\mathopen{}\left( t\right)\mathclose{}}\,\mathrm{d}t\right)\mathclose{} \,\mathrm{d}x= \int _{0}^{1}{} \int _{0}^{x}{} f\mathopen{}\left( x\right)\mathclose{}\overline{g\mathopen{}\left( t\right)\mathclose{}} \,\mathrm{d}t \,\mathrm{d}x= \int _{0}^{1}{} \int _{t}^{1}{} f\mathopen{}\left( x\right)\mathclose{}\overline{g\mathopen{}\left( t\right)\mathclose{}} \,\mathrm{d}x \,\mathrm{d}t= \int _{0}^{1}{} \int _{0}^{1}{} k\mathopen{}\left( x, t\right)\mathclose{}f\mathopen{}\left( x\right)\mathclose{}\overline{g\mathopen{}\left( t\right)\mathclose{}} \,\mathrm{d}x \,\mathrm{d}t \text{,}$$ where $$k\mathopen{}\left( x, t\right)\mathclose{}= \begin{cases}1, & t\leq x; \\ 0, & \text{otherwise.}\end{cases}$$ By Fubini's theorem, we get $$\int _{0}^{1}{} \int _{0}^{1}{} k\mathopen{}\left( x, t\right)\mathclose{}f\mathopen{}\left( x\right)\mathclose{}\overline{g\mathopen{}\left( t\right)\mathclose{}} \,\mathrm{d}x \,\mathrm{d}t= \int _{0}^{1}{} \mathopen{}\left(\int _{t}^{1}{}f\mathopen{}\left( x\right)\mathclose{}\,\mathrm{d}x\right)\mathclose{}\overline{g\mathopen{}\left( t\right)\mathclose{}} \,\mathrm{d}t= \mathopen{}\left\langle{} T^{*}\mathopen{}\left( f\right)\mathclose{}, g\right\rangle\mathclose{} \text{,}$$ so that $$$\mathopen{}\left( T^{*}\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= \int _{x}^{1}{}f\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}t$$$. Notice that for this operator $$$\mathopen{}\left(T+ T^{*}\right)\mathclose{}\mathopen{}\left( f\right)\mathclose{}$$$ is a constant function. That is, $$$\mathopen{}\left(T+ T^{*}\right)\mathclose{}\mathopen{}\left( f\right)\mathclose{}= \mathopen{}\left\langle{}f, \mathbb{1}\right\rangle\mathclose{}\mathbb{1}$$$, which is the projection of $$$f$$$ on $$$\mathbb{C}\mathbb{1}$$$.

Example II.27

Recall the sets $$$E$$$ and $$$V$$$ and maps $$$i\text{,}t : E \to V$$$ with $$\sup_{v\in V}{} \mathopen{}\left({\#}\mathopen{}\left({i}^{-1}\mathopen{}\left( v\right)\mathclose{}\right)\mathclose{}+{\#}\mathopen{}\left({t}^{-1}\mathopen{}\left( v\right)\mathclose{}\right)\mathclose{}\right)\mathclose{} \lt \infty \text{.}$$ Get $$$T : \mathrm{l}^{0}\mathopen{}\left( V\right)\mathclose{} \to \mathrm{l}^{0}\mathopen{}\left( E\right)\mathclose{}$$$ defined by $$$\mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( e\right)\mathclose{}= f\mathopen{}\left( i\mathopen{}\left( e\right)\mathclose{}\right)\mathclose{}-f\mathopen{}\left( t\mathopen{}\left( e\right)\mathclose{}\right)\mathclose{}$$$. What is the adjoint $$$T^{*} : \mathrm{l}^{0}\mathopen{}\left( E\right)\mathclose{} \to \mathrm{l}^{0}\mathopen{}\left( V\right)\mathclose{}$$$? For $$$γ\in \mathrm{l}^{0}\mathopen{}\left( E\right)\mathclose{}$$$, $$$f\in \mathrm{l}^{0}\mathopen{}\left( V\right)\mathclose{}$$$, $$\mathopen{}\left\langle{} T^{*}\mathopen{}\left( γ\right)\mathclose{}, f\right\rangle\mathclose{}= \mathopen{}\left\langle{}γ, T\mathopen{}\left( f\right)\mathclose{}\right\rangle\mathclose{}= \sum_{e\in E}{} γ\mathopen{}\left( e\right)\mathclose{}\overline{ \mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( e\right)\mathclose{} } = \sum_{e\in E}{} γ\mathopen{}\left( e\right)\mathclose{}\mathopen{}\left(\overline{f\mathopen{}\left( i\mathopen{}\left( e\right)\mathclose{}\right)\mathclose{}}-\overline{f\mathopen{}\left( t\mathopen{}\left( e\right)\mathclose{}\right)\mathclose{}}\right)\mathclose{} = \sum_{v\in V}{}\overline{f\mathopen{}\left( v\right)\mathclose{}}\sum_{\mathopen{}\left\{\, e\,\middle\vert\, i\mathopen{}\left( e\right)\mathclose{}= v\,\right\}\mathclose{}}{}γ\mathopen{}\left( e\right)\mathclose{}-\sum_{w\in V}{}\overline{f\mathopen{}\left( w\right)\mathclose{}}\sum_{\mathopen{}\left\{\, \overline{e}\,\middle\vert\, t\mathopen{}\left( \overline{e}\right)\mathclose{}= w\,\right\}\mathclose{}}{}γ\mathopen{}\left( \overline{e}\right)\mathclose{}= \sum_{v\in V}{}\overline{f\mathopen{}\left( v\right)\mathclose{}}\mathopen{}\left(\sum_{\mathopen{}\left\{\, e\,\middle\vert\, i\mathopen{}\left( e\right)\mathclose{}= v\,\right\}\mathclose{}}{}γ\mathopen{}\left( e\right)\mathclose{}-\sum_{\mathopen{}\left\{\, \overline{e}\,\middle\vert\, t\mathopen{}\left( \overline{e}\right)\mathclose{}= v\,\right\}\mathclose{}}{}γ\mathopen{}\left( \overline{e}\right)\mathclose{}\right)\mathclose{} \text{,}$$ whence $$$T^{*}\mathopen{}\left( γ\right)\mathclose{}\mathopen{}\left( v\right)\mathclose{}= \sum_{\mathopen{}\left\{\, e\,\middle\vert\, i\mathopen{}\left( e\right)\mathclose{}= v\,\right\}\mathclose{}}{}γ\mathopen{}\left( e\right)\mathclose{}-\sum_{\mathopen{}\left\{\, \overline{e}\,\middle\vert\, t\mathopen{}\left( \overline{e}\right)\mathclose{}= w\,\right\}\mathclose{}}{}γ\mathopen{}\left( \overline{e}\right)\mathclose{}$$$ (see Figure II.A for an example).

What does $$$T^{*}\mathopen{}\left( T\right)\mathclose{} : \mathrm{l}^{0}\mathopen{}\left( V\right)\mathclose{} \to \mathrm{l}^{0}\mathopen{}\left( V\right)\mathclose{}$$$ do? $$\mathopen{}\left( T^{*}\mathopen{}\left( T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( v\right)\mathclose{}= \sum_{\mathopen{}\left\{\, e\,\middle\vert\, i\mathopen{}\left( e\right)\mathclose{}= v\,\right\}\mathclose{}}{}\mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( e\right)\mathclose{}-\sum_{\mathopen{}\left\{\, e\,\middle\vert\, t\mathopen{}\left( e\right)\mathclose{}= v\,\right\}\mathclose{}}{}\mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( e\right)\mathclose{}= \sum_{\mathopen{}\left\{\, e\,\middle\vert\, i\mathopen{}\left( e\right)\mathclose{}= v\,\right\}\mathclose{}}{} \mathopen{}\left(f\mathopen{}\left( v\right)\mathclose{}-f\mathopen{}\left( t\mathopen{}\left( e\right)\mathclose{}\right)\mathclose{}\right)\mathclose{} -\sum_{\mathopen{}\left\{\, e\,\middle\vert\, t\mathopen{}\left( e\right)\mathclose{}= v\,\right\}\mathclose{}}{} \mathopen{}\left(f\mathopen{}\left( i\mathopen{}\left( e\right)\mathclose{}\right)\mathclose{}-f\mathopen{}\left( v\right)\mathclose{}\right)\mathclose{} = f\mathopen{}\left( v\right)\mathclose{}\mathopen{}\left(\underbrace{{\#}\mathopen{}\left({i}^{-1}\mathopen{}\left( v\right)\mathclose{}\right)\mathclose{}+{\#}\mathopen{}\left({t}^{-1}\mathopen{}\left( v\right)\mathclose{}\right)\mathclose{}}_{\mathop{\text{deg}}\mathopen{}\left( v\right)\mathclose{}}\right)\mathclose{}-\sum_{\text{Neighbors w of v}}{}f\mathopen{}\left( w\right)\mathclose{} \text{.}$$ Thus $$$T^{*}\mathopen{}\left( T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}= D\mathopen{}\left( f\right)\mathclose{}-A\mathopen{}\left( f\right)\mathclose{}$$$ where $$$D$$$ is multiplication by the degree function and $$$A$$$ is the adjacency operator.

Figure II.A. To illustrate Example II.27, we have $$$T^{*}\mathopen{}\left( γ\right)\mathclose{}\mathopen{}\left( v\right)\mathclose{}= γ\mathopen{}\left( {e}_{1}\right)\mathclose{}-\mathopen{}\left(γ\mathopen{}\left( {e}_{2}\right)\mathclose{}+γ\mathopen{}\left( {e}_{3}\right)\mathclose{}\right)\mathclose{}$$$
Example II.28

Consider $$$S$$$, the unilateral forward shift on $$$\mathrm{l}^{0}\mathopen{}\left( {\mathbb{Z}}^{+}\right)\mathclose{}$$$ defined as $$$S\mathopen{}\left( {x}_{1}, {x}_{2}, \dotsc\right)\mathclose{}= \mathopen{}\left(0, {x}_{1}, {x}_{2}, \dotsc\right)\mathclose{}$$$. Note that $$$\mathopen{}\left\lVert{}S\mathopen{}\left( \mathbf{x}\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\mathbf{x}\right\rVert\mathclose{}$$$, so $$$S$$$ is bounded. $$\mathopen{}\left\langle{} S^{*}\mathopen{}\left( \mathbf{x}\right)\mathclose{}, \mathbf{y}\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathbf{x}, S^{*}\mathopen{}\left( \mathbf{y}\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left({x}_{1}, {x}_{2}, \dotsc\right)\mathclose{}, \mathopen{}\left(0, {y}_{1}, {y}_{2}, \dotsc\right)\mathclose{}\right\rangle\mathclose{}= \sum_{n=1}^{\infty}{} \overline{{y}_{n}}{x}_{n+1} = \mathopen{}\left\langle{}\mathopen{}\left({x}_{2}, {x}_{3}, \dotsc\right)\mathclose{}, \mathopen{}\left({y}_{1}, {y}_{2}, \dotsc\right)\mathclose{}\right\rangle\mathclose{} \text{.}$$ Thus $$$S^{*}\mathopen{}\left( \mathbf{x}\right)\mathclose{}= \mathopen{}\left({x}_{2}, {x}_{3}, \dotsc\right)\mathclose{}$$$ (backwards shift). $$$S^{*}\mathopen{}\left( S\right)\mathclose{}= I$$$, but $$S\mathopen{}\left( S^{*}\mathopen{}\left( \mathbf{x}\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left(0, {x}_{2}, \dotsc\right)\mathclose{}= I-{P}_{1}\neq I \text{,}$$ where $$${P}_{1}$$$ is the projection on the first entry ($$${P}_{1}\mathopen{}\left( \mathbf{x}\right)\mathclose{}= \mathopen{}\left({x}_{1}, 0, \dotsc\right)\mathclose{}$$$).