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A spin-$S$ Hilbert space has dimension $2S+1$. Its Holstein–Primakoff transformation therefore uses only bosonic occupation numbers $0,1,\dots ,2S$. The square root annihilates the upper endpoint. A truncated linear spin-wave approximation formally enlarges this space; it is self-consistent only when boson depletion is small compared with $S$.

For a linear map $T:V\to W$ between finite-dimensional inner product spaces,The first equality follows directly from the defining identity for the adjoint operator; the second follows by taking orthogonal complements. In an infinite-dimensional Hilbert space, the second equality generally requires closure of the image.

An indefinite Hermitian form takes both positive and negative values on $⟨v,v⟩$ for nonzero vectors. For example, $-|v_0{|}^2+|v_1{|}^2+|v_2{|}^2+|v_3{|}^2$ is indefinite. It is not a positive inner product defining a Hilbert space, and a nonzero vector can have zero norm without being orthogonal to all vectors. A positive semidefinite restricted form becomes a positive inner product only after quotienting its radical of a Hermitian form.

An inner-product space is a vector space equipped with an inner product. The inner product induces the norm $\Vert x\Vert =\sqrt{⟨x,x⟩}$ and hence a metric.

An isonormal Gaussian process over a real Hilbert space $H$ is a centered Gaussian process $(W(h){)}_{h\in H}$ satisfyingIt turns the geometry of $H$ into the covariance structure of a Gaussian family.

The space ${\ell }^2$ consists of scalar sequences $x=(x_n)$ satisfying ${\sum }_n|x_n{|}^2<\infty$. With inner product $⟨x,y⟩={\sum }_n{x}_n\overline{y_n}$, it is a Hilbert space.

The space $L^2(X,\mathcal{F},\nu )$ consists of almost-everywhere equivalence classes of measurable functions with $\int |f{|}^{2}d\nu <\infty$. The inner productinduces its norm, and the Riesz-Fischer theorem makes it complete; hence it is a Hilbert space.

A bounded bilinear form $a$ on a real Hilbert space $H$ that is a coercive bilinear form represents every bounded linear functional uniquely: for each $\ell \in H^{\prime }$, there is a unique $u\in H$ with $a(u,v)=\ell (v)$ for every $v\in H$.

For a proof, the Riesz representation theorem gives a bounded linear operator $T$ and a vector $g$ with $a(u,v)=⟨Tu,v⟩$ and $\ell (v)=⟨g,v⟩$. Coercivity and the Cauchy-Schwarz inequality give $\Vert Tu\Vert \ge \alpha \Vert u\Vert$, so $T$ is injective and its image is closed: convergence of $T{u}_n$ makes $u_n$ a Cauchy sequence, whose limit supplies the image limit. If $v$ belongs to the orthogonal complement of the image, then $a(u,v)=0$ for every $u$; setting $u=v$ and using coercivity forces $v=0$. Thus the image is also dense and equals $H$. There is consequently a unique $u$ with $Tu=g$, and $\Vert u\Vert \le \Vert \ell \Vert /\alpha$. The complex version uses a bounded sesquilinear form with $\mathrm{Re}a(u,u)\ge \alpha \Vert u{\Vert }^2$.

For a bounded linear operator $K:U\to V$ between Hilbert spaces, a least-squares solution for $f$ exists exactly when $P_{\overline{\mathcal{R}(K)}}f\in \mathcal{R}(K)$, equivalently $f\in \mathcal{R}(K)\oplus \mathcal{R}(K{)}^{\perp }$. Indeed the orthogonal projection onto the closure uniquely minimizes distance to that closed space, and it can be realized by $Ku$ exactly under this condition. Thus all data admit least-squares solutions exactly when the range is closed.

A least-squares solution minimizes $\Vert Ku-f\Vert$ over the source Hilbert space. It exists exactly when $f\in \mathcal{R}(K)\oplus \mathcal{R}(K{)}^{\perp }$, and then the solutions form a closed affine subspace with direction $\mathcal{N}(K)$.

On the Hilbert space ${\ell }^2(\mathbb{N})$, the left shift deletes the first coordinate. It is the adjoint operator of the unilateral shift operator $S$. The identities $LS=I$ and $SLf=f-f_1{e}_1$ show that it is also the Moore–Penrose inverse of an operator for $S$; both maps have operator norm one.

A measurable function belongs to $L_{\mathrm{loc}}^2(U)$ when its squared absolute value has finite integral on every compact subset of $U$. Functions are identified if they agree almost everywhere. This is a local integrability condition, not a specified global Hilbert space norm. For example, $x$ belongs to $L_{\mathrm{loc}}^2(\mathbb{R})$ but its mean-square averages on expanding intervals diverge.

The maximally mixed density operator on an $N$-dimensional Hilbert space assigns equal weights to every vector of any orthonormal basis. It is invariant under every unitary operator. Its purity is $\mathrm{Tr}({\rho }_{\ast }^2)=1/N$, the minimum possible because the nonnegative eigenvalues sum to one and obey the Cauchy-Schwarz inequality. For $N>1$ it is a mixed state; for $N=1$ it is the sole pure state. A rank-$k$ projector has probability $k/N$ in this state.

On a $d$-dimensional Hilbert space, the Von Neumann entropy is maximized uniquely by $I/d$. Indeed, nonnegativity of quantum relative entropy giveswith equality exactly when $\rho =I/d$.

For a bounded operator $A:X\to Y$ between Hilbert spaces, $A^{†}$ has domain $\mathcal{R}(A)\oplus \mathcal{R}(A{)}^{\perp }$, maps into $\mathcal{N}(A{)}^{\perp }$, and inverts the restriction of $A$ to $\mathcal{N}(A{)}^{\perp }$. It is bounded exactly when $\mathcal{R}(A)$ is closed.

A Hermitian operator on a $2^n$-dimensional Hilbert space, with $2^n$ distinct eigenvalues all belonging to $\{c/2^n:0\le c<2^n\}$, must occupy this entire grid. This follows from the pigeonhole principle: the set of possible values and the set of distinct eigenvalues have the same cardinality. Its minimum eigenvalue is zero, so the operator is necessarily singular. Distinctness therefore does not imply invertibility; it forces a one-dimensional kernel in this specific complete-grid setting.

For bounded linear operators $K:U\to V$ and $B:U\to W$ between real Hilbert spaces, $\alpha >0$ and $\Vert Bu\Vert \ge \beta \Vert u\Vert$ with $\beta >0$, the unique minimizer of $\Vert Ku-f{\Vert }^2/2+\alpha \Vert Bu{\Vert }^2/2$ satisfies $(K^{\ast }K+\alpha B^{\ast }B)u_{\alpha }=K^{\ast }f$. Differentiate the objective in every direction and use the adjoint operator to obtain this equation. The coercive quadratic penalty makes its left-hand operator boundedly invertible. Setting $B=I$ gives the Tikhonov normal equation; omitting the penalty instead gives the normal equation for a linear inverse problem.

The closed unit ball of a Hilbert space is compact in norm exactly when the space is finite-dimensional. Finite-dimensional compactness follows from the Heine-Borel theorem. In infinite dimension, an orthonormal sequence has pairwise distance $\sqrt{2}$ and no convergent subsequence. Restricting a compact operator that equals a nonzero multiple of the identity on a subspace therefore forces that subspace to be finite-dimensional.

For a bounded linear operator on a complex Hilbert space, the numerical radius is the supremum of the absolute values in its numerical range of an operator. It is a norm equivalent to the operator norm, with $w(T)\le \Vert T\Vert \le 2w(T)$. For a self-adjoint operator the two norms agree.

This Hilbert space has norm $\Vert u{\Vert }_X^2=\Vert u{\Vert }_2^2+\Vert u^{\prime }{\Vert }_2^2+\Vert xu{\Vert }_2^2$. Completeness follows from completeness of H1 space and closedness of the multiplication operator $u\mapsto xu$. Smooth cutoffs followed by convolution with a mollifier show that smooth compactly supported functions are dense. The space is the natural energy domain of the dimensionless quantum harmonic oscillator, whereas the operator domain additionally requires $u^{\prime \prime }$ and $x^{2}u$ to lie in L2 space.