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For nonzero compact operator $A$, step $0<\tau <2/\Vert A{\Vert }^2$, and data admitting a minimum-norm least-squares solution $x^{†}$, zero-initialized Landweber iteration satisfiesThe Picard criterion makes the majorant summable, so the dominated convergence theorem proves this limit. No mesh-independent or singular-value-independent strict contraction factor is required. At $\tau =2/\Vert A{\Vert }^2$, the top singular mode has multiplier $-1$ and may not converge. Without solvable projected data, the iterates need not have a limit in the solution Hilbert space, even when their residuals approach the least-squares infimum.

For a proper lower-semicontinuous convex functional on a real Hilbert space, $p\in \partial J(u)$ means $⟨v,p⟩-J(v)\le ⟨u,p⟩-J(u)$ for all $v$. Taking the supremum gives equality in the Fenchel–Young inequality: $J(u)+J^{\ast }(p)=⟨u,p⟩$. Apply the same argument to $J^{\ast }$ and use the Fenchel-Moreau theorem $J^{\ast \ast }=J$ to obtain the reciprocal subgradient condition.

For positive-semidefinite kernels $k_j$ with Reproducing-kernel Hilbert spaces ${\mathcal{H}}_j$, the Reproducing-kernel Hilbert space of $k={\sum }_j{k}_j$ consists of sums $f={\sum }_j{f}_j$, withThe minimum is attained uniquely. Indeed, in the Hilbert space direct sum $E={⨁}_j{\mathcal{H}}_j$, the tuples summing to zero form a closed subspace $N$: they are the intersection, over inputs $x$, of the kernels of the continuous maps $(f_j)\mapsto {\sum }_j{f}_j(x)$. Every affine fibre has a unique representative in $N^{\perp }$ by orthogonal decomposition by a closed subspace. The vector $(k_j(\cdot ,x){)}_j$ lies in $N^{\perp }$ and represents evaluation in the quotient; its evaluation at $y$ is ${\sum }_j{k}_j(y,x)$, identifying the positive-semidefinite kernel with $k$.

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.

For a bounded operator on a real Hilbert space, $⟨Lv,v⟩=⟨Sv,v⟩$ with $S=(L+L^{\ast })/2$. Consequently the Euler-Lagrange equation of $⟨Lv,v⟩-2⟨f,v⟩$ is $Sv=f$. It is $Lv=f$ only when $L$ is self-adjoint or when the relevant solution also annihilates the skew part. For example $L=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$ has positive quadratic form $\Vert v{\Vert }^2$, but that form contains no information about its skew part.

For a bounded linear operator between Hilbert spaces, the spectral regularization method for quadratic Tikhonov regularization has scalar coefficient $s/(s^2+\alpha )$. Since $(s-\sqrt{\alpha }{)}^2\ge 0$, this coefficient is at most $1/(2\sqrt{\alpha })$. The spectral theorem for normal operators transfers this scalar bound to the operator norm, providing the noise term in the noise-bias decomposition for linear regularization.

For bounded linear operators between Hilbert spaces with $\Vert Bu\Vert \ge \beta \Vert u\Vert$ and $\beta >0$, the operator $K^{\ast }K+\alpha B^{\ast }B$ is coercive and invertible for $\alpha >0$. The reconstruction uniquely minimizes $\Vert Ku-f{\Vert }^2/2+\alpha \Vert Bu{\Vert }^2/2$. Testing its normal equation for coercive Tikhonov regularization by $u$ gives $\Vert R_{\alpha }\Vert \le 1/(2\beta \sqrt{\alpha })$. If $K$ is injective, exact-data consistency and $\alpha \to 0$, $\delta /\sqrt{\alpha }\to 0$ give norm convergence to $K^{†}f$.

For an absolutely one-homogeneous functional $J=\mathrm{TV}$ on a real Hilbert space, a subgradient $q$ at $u$ is characterized by $⟨q,v⟩\le J(v)$ for every $v$ and $⟨q,u⟩=J(u)$. A bounded vector field with $|z|\le 1$ and $q=\mathrm{div}z$ proves the inequality through the dual definition, provided cutoff approximation is valid in the ambient space. Equality is called a calibration. It certifies a global minimizer of total variation denoising, rather than only a minimizer within a chosen ansatz.

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.

A positive definite symmetric operator is uniformly positive definite if $⟨Lv,v⟩\ge \gamma \Vert v{\Vert }^2$ for some $\gamma >0$. For a bounded linear operator defined on a whole Hilbert space, this is a symmetric coercive operator, and the Lax-Milgram theorem supplies a unique solution to $Lu=f$ for every $f$. For a differential operator, the corresponding bounded coercive form is used on its form space.

A linear isometry from a subspace of a finite-dimensional Hilbert space into the same ambient space extends to a unitary operator. Complete an orthonormal basis of the input subspace and its isometric image to full orthonormal bases, then map the first to the second. The equal complement dimensions make this possible. Orthogonality of the specified image columns, not normalization alone, is essential. In an infinite-dimensional space arbitrary isometries need not be surjective; the finite-dimensional same-space hypothesis cannot be dropped.

For a bounded sequence $(u_n)$ in a Hilbert space, put $c_h={\mathrm{limsup}}_{N\to \infty }|N^{-1}{\sum }_{n=0}^{N-h-1}⟨u_{n+h},u_n⟩|$. If $H^{-1}{\sum }_{h=1}^H{c}_h\to 0$, then $N^{-1}{\sum }_{n=0}^{N-1}{u}_n\to 0$ in norm. Since the $c_h$ are bounded, it suffices that $(c_h)$ has convergence in density of a sequence to zero. The proof applies the Cauchy-Schwarz inequality to an average of shifted sums.

A variational problem seeks a stationary point or minimizer of a functional on a specified function space. For a symmetric coercive bilinear form $a$ and bounded linear functional $\ell$ on a real Hilbert space, minimizing $J(v)=a(v,v)/2-\ell (v)$ is equivalent to the weak formulation $a(u,v)=\ell (v)$ for every test vector. The Lax-Milgram theorem gives a unique solution, and $J(u+w)-J(u)=a(w,w)/2$ proves unique minimization. Specifying the admissible space and its boundary constraints is part of the problem.

For a unitary operator $U$ on a Hilbert space, the Cesaro averagesconverge strongly to the orthogonal projection onto the fixed-point subspace $\ker (U-I)$.

If $x_n\rightharpoonup x$ in a Hilbert space, some subsequence $(x_{n_k})$ has Cesaro means converging in norm to $x$. For $x=0$, choose the subsequence so successive vectors have summably small inner products; expansion of the squared norm of its averages then proves convergence.

Every bounded sequence in a Hilbert space has a weakly convergent subsequence. The closed span of a sequence is separable; diagonal convergence of coordinates along an orthonormal basis gives a bounded limiting linear functional, represented by a vector of the Hilbert space. Applied to a bounded family of mollifications, this proves that their strong Lebesgue limit inherits the weak derivative bound.

For $M>0$ with integral one, the Cauchy-Schwarz inequality makes the mass functional bounded on the weighted Hilbert space. The vector $M$ has norm one, so the velocity-reset projection is the orthogonal projection onto its span. Consequently $L=\Pi -I$ is bounded and self-adjoint. This Hilbert geometry differs from the norm-one positive reset projection on phase-space $L^1$.

If $L$ is bounded and self-adjoint on a complex Hilbert space and $K$ is compact and self-adjoint, thenA singular Weyl sequence for $L$ stays singular for $L+K$, because compact operators send weak convergence to norm convergence and therefore $K{f}_n\to 0$. Applying the same argument with $-K$ proves the reverse inclusion. Finite-multiplicity isolated eigenvalues can move under such perturbations; the essential spectrum of a bounded self-adjoint operator is unchanged.