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UseThe adjoint operator is consequentlyWith a unitary Fourier normalization, this is simply ${\mathcal{F}}^{\ast }={\mathcal{F}}^{-1}$.

At fixed incident direction and wavenumber, defineTaking the complex conjugate of the kernel givesThe least-squares minimizer of $\Vert TV-f_{\infty }{\Vert }^2$ satisfies the normal equation for a linear inverse problem $T^{\ast }TV=T^{\ast }{f}_{\infty }$. Fourier inversion on the measured transfer-vector set is exactly the corresponding Moore--Penrose reconstruction $T^{†}{f}_{\infty }$; hence the formal solution in part (i) is the minimum-norm least-squares solution when the data are incomplete or inconsistent.

The Riesz representation theorem says that for every bounded linear functional $f$ on a Hilbert space $H$, there is a unique $z\in H$ such thatand $\Vert f\Vert =\Vert z\Vert$.

For $T\in L(H)$ and fixed $y\in H$, the mapis a bounded linear functional, withRiesz therefore gives a unique vector, denoted $T^{\ast }y$, such thatfor every $x$. Uniqueness in the representation theorem shows that $y\mapsto T^{\ast }y$ is linear. Moreover,so the adjoint operator belongs to $L(H)$ and $\Vert T^{\ast }\Vert \le \Vert T\Vert$.

If $(e_j)$ is an orthonormal basis and $A=(a_{ij})$ is the matrix of $T$, then the matrix of $T^{\ast }$ isits entries are $\overline{a_{ji}}$.

For positive observations $y$ and positive predicted intensities $s$, this is the generalized Kullback–Leibler divergence with observation in the first argument. Its scalar derivative in $s$ is $1-y/s$ and its second derivative is $y/s^2>0$. It differs from the divergence of a reconstructed density from a reference density by which argument is varied. A linear prediction $s=Kx$ gives derivative $K^{\ast }(1-y/(Kx))$ through the adjoint operator, on suitable admissible function spaces. This fidelity arises from the negative log likelihood of a Poisson observation model, after removal of terms depending only on observations.

A self-adjoint operator $A$ is positive definite when $⟨Av,v⟩>0$ for every nonzero vector $v$ in its domain. In finite dimension it is invertible and has positive eigenvalues. In infinite dimension, strict positivity of individual quadratic-form values need not give a bounded inverse; the stronger uniform bound $⟨Av,v⟩\ge c\Vert v{\Vert }^2$ with $c>0$ supplies coercivity.

For maps $\beta :U\to V$ and $\gamma :V\to W$ between finite-dimensional inner-product spaces, with $\mathrm{im}\beta =\ker \gamma$, the operator $\beta {\beta }^{\ast }+{\gamma }^{\ast }\gamma$ is positive definite. Its quadratic form is $\Vert {\beta }^{\ast }v{\Vert }^2+\Vert \gamma v{\Vert }^2$, whose zero set is $\mathrm{im}\beta \cap (\mathrm{im}\beta {)}^{\perp }=\{0\}$. This uses the adjoint operator relation $\ker {\beta }^{\ast }=(\mathrm{im}\beta {)}^{\perp }$.

A densely defined operator $T$ on a Hilbert space is self-adjoint when it equals its adjoint operator, including equality of their operator domains:For a bounded linear operator defined on the whole Hilbert space, this is equivalent tofor all vectors $v,w$. For an unbounded self-adjoint operator, the domain condition is essential: the inner-product identity on $D(T)$ alone gives only a symmetric operator.

The continuous dual space of $H^s({\mathbb{R}}^d)$ is $H^{-s}({\mathbb{R}}^d)$ under the extension of the L2 inner product. The Fourier transform definition $\Vert f{\Vert }_{H^s}^2=\int (1+|\xi {|}^2{)}^s|\hat{f}(\xi ){|}^{2}d\xi$ proves boundedness of the pairing by the Cauchy-Schwarz inequality, and the Riesz representation theorem proves that every continuous functional has this form. Consequently a bounded linear operator $K:H^r\to H^{r+a}$ has an adjoint operator, relative to this pairing, from $H^{-r-a}$ to $H^{-r}$. For an operator that smooths by two orders for every $r$, this yields $K^{\ast }:H^2\to H^4$ by taking $r=-4$.

If $Au=f$ and $A^{\ast }\mu \in \partial J(u)$, then $u$ is a J-minimizing solution. For every other exact solution $v$, the subgradient inequality gives $J(v)\ge J(u)+⟨\mu ,Av-Au⟩=J(u)$. This is a certificate of global optimality supplied by the range of the adjoint operator.

This range condition puts the exact solution in the operator range of the adjoint operator. It controls the small-singular value components of the solution and supplies convergence rates for spectral regularization methods. It is the quadratic-penalty case of a source condition in variational regularization.

For a real vector $x$ with support of a vector $S$, this certificate is a vector $h$ whose measurements under the adjoint operator match the active sign function values and lie strictly between minus one and one on the inactive coordinates. If $A_S$ is injective, such a certificate proves that $x$ is the unique basis pursuit minimizer. For a nonzero $v\in \ker A$, injectivity ensures $v_{S^c}\ne 0$, and $0=⟨A^{\ast }h,v⟩$ implies the fixed-sign null space condition. The strict inequality is interpreted coordinatewise if $S^c$ is empty. The condition with the sign function written here is for real variables; complex basis pursuit uses the unit phases of active coordinates instead.

A densely defined operator $T$ on a Hilbert space is symmetric whenEquivalently, $T$ agrees with its adjoint operator on $D(T)$, while $D(T^{\ast })$ may be larger. A self-adjoint operator additionally requires equality of these operator domains.

A densely defined linear operator on a complex Hilbert space is self-adjoint when its adjoint operator has exactly the same domain and agrees there. Symmetry only requires $A\subseteq A^{\ast }$ and does not guarantee self-adjointness. A self-adjoint operator is a closed operator with real spectrum. Domains matter for differential operators and unbounded multiplication operators.