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Put $G=X^{T}X$ and $H=X{G}^{-1}{X}^T$. The rank of a matrix assumption makes $G$ invertible, and $H$ is the orthogonal projection matrix onto the column space of the design matrix. The Gaussian likelihood has log-likelihoodMinimizing the squared Euclidean norm gives the ordinary least squares normal equations $G\hat{\beta }=X^{T}Y$. Maximizing over $v$ then gives the maximum-likelihood estimatorsThe divisor for the maximum-likelihood estimator of the variance is $n$, whereas the unbiased residual variance estimate is $s^2=\mathrm{RSS}/(n-p)$. The exceptional event $\mathrm{RSS}=0$ has probability zero under the Gaussian linear model with ${\sigma }^2>0$; on that event the likelihood is unbounded as $v\downarrow 0$.

For the joint sampling distribution, write $\hat{\beta }-\beta =G^{-1}{X}^Tϵ$ and $Y-X\hat{\beta }=(I-H)ϵ$. Their cross-covariance matrix vanishes because $X^T(I-H)=0$. These two vectors are jointly multivariate normal, so they are independent random variables. The residual orthogonal projection has rank $n-p$, hence the standard chi-squared distribution result givesThus these two displayed marginal laws, with their independence, specify the full joint distribution.

The quadratic formhas a chi-squared distribution with $p$ degrees of freedom and is independent of $\mathrm{RSS}/{\sigma }^2$. Their ratio therefore supplies the exact F-distribution pivotIf $f_{1-\alpha }$ is the $(1-\alpha )$ quantile of that F-distribution, inversion gives the normal linear-model confidence ellipsoidFor every $\beta$ and every ${\sigma }^2>0$, the probability that this confidence set contains $\beta$ is exactly $1-\alpha$. To test the null hypothesis, reject precisely when ${\beta }_0$ lies outside this confidence ellipsoid, equivalently when $T({\beta }_0)>f_{1-\alpha }$. Under the null hypothesis the rejection probability is exactly $\alpha$, independent of the unknown variance.

For the numerical calculation, the second column of the design matrix sums to zero and its sum of squares is $60$. Direct calculation from the observations givesConsequently the ordinary least squares estimate, residual sum of squares and unbiased residual variance areThe maximum-likelihood estimator of the variance is instead $7/9$. At the proposed null value the F-distribution statistic isTherefore reject the null hypothesis at the 5% level. The relevant critical value has $2$ numerator and $7$ residual degrees of freedom.

Yes: distinct density operators have a projective measurement with different outcome probabilities. Let $\rho ,\sigma$ be the two density operators and let $D=\rho -\sigma$. It is a nonzero Hermitian operator with trace zero. Its nonzero eigenvalues cannot all have the same sign, since their sum is zero. Let $P_{+}$ be the orthogonal projection onto its positive-eigenvalue subspace. Then the two-outcome projective measurement $\{P_{+},I-P_{+}\}$ satisfiesThe outcome therefore carries statistical information about which quantum ensemble was used.

For example, with equal prior probabilities, guess $\rho$ after $+$ and $\sigma$ after the other outcome. The success probability isSince the positive and negative eigenvalue sums of $D$ have equal absolute value, $\mathrm{Tr}(P_{+}D)=\Vert D{\Vert }_1/2$. ConsequentlyThis is the equal-prior measurement in the Holevo–Helstrom theorem, expressed using the trace distance. The conclusion is some information, not necessarily perfect one-shot discrimination: nonorthogonal density operators generally cannot be distinguished with certainty.

Let $V$ be a finite-dimensional real or complex inner product space, and let $A\subseteq \mathrm{End}(V)$ be a unital subalgebra stable under the adjoint operator. Define its commutant of an operator algebra by $A^{\prime }=\{T:Ta=aT\text{ for all }a\in A\}$. The double commutant theorem for star-algebras asserts $\boxed{A^{\prime \prime }=A}$. The inclusion $A\subseteq A^{\prime \prime }$ follows immediately from the definition; we prove the reverse.

Write $N=\dim V$, fix a basis $v_1,\dots ,v_N$, and act diagonally on $V^N$. The linear subspaceis invariant under every diagonal $a$. Since $A$ is closed under adjoints, its orthogonal complement $W^{\perp }$ is invariant too: $⟨aw,z⟩=⟨w,a^{\ast }z⟩=0$ for $w\in W^{\perp }$, $z\in W$. Hence the orthogonal projection $P$ onto $W$ commutes with every diagonal $a$.

Writing $P$ as an $N$ by $N$ block matrix $(P_{ij})$, this says $P_{ij}a=a{P}_{ij}$, so each $P_{ij}\in A^{\prime }$. If $b\in A^{\prime \prime }$, it commutes with every $P_{ij}$, and its diagonal action therefore commutes with $P$ and preserves $W$. Because $1\in A$, $(v_1,\dots ,v_N)\in W$. Therefore there is one $a\in A$ such thatEquality on the basis gives $b=a$, proving the theorem over either scalar field.

The identity hypothesis is necessary. For a possibly nonunital adjoint-stable subalgebra, adjoining the ambient identity does not change its commutant of an operator algebra, so the exact statement is $A^{\prime \prime }=A+\mathbb{F}I$. For example $A=\{0\}$ has $A^{\prime \prime }=\mathbb{F}I$ on a nonzero $V$. Thus the usual exam formulation takes the algebra to contain the identity.

For a closed manifold $M$ with an oriented Riemannian metric, define $\delta$ to be the formal $L^2$ adjoint operator of the exterior derivative, equivalently $\delta =(-1{)}^{n(q+1)+1}\ast d\ast$ on $q$-forms. The Hodge Laplacian is ${\Delta }_q=d\delta +\delta d$. The Hodge decomposition theorem states that its space ${\mathcal{H}}^q=\ker {\Delta }_q$ of harmonic differential forms is finite-dimensional and smooth, and thatThe last isomorphism sends a harmonic differential form to its de Rham cohomology class: every class has exactly one harmonic representative. Out-of-range degree spaces are zero.

We first justify the analytic facts needed in the proof. In local coordinates and a smooth frame for the form bundle, the principal symbol of ${\Delta }_q$ is $g^{ij}{\xi }_i{\xi }_{j}I$. This follows from the exterior/contraction identities for $d$ and $\delta$. Freezing the leading coefficients, the Fourier transform estimate for the constant positive quadratic form gives, for each nonnegative integer $s$,Here is how to pass to variable coefficients. On a sufficiently small coordinate ball the difference of leading coefficients has arbitrarily small supremum norm. After differentiating $s$ times, its top-order contribution is bounded by $\epsilon \Vert u{\Vert }_{H^{s+2}}$; derivatives of the coefficients and first-order terms contribute $C_s\Vert u{\Vert }_{H^{s+1}}$. A partition of unity gives the same types of terms from commutators with the cutoffs. Absorb the epsilon term and use Sobolev interpolation to absorb $\Vert u{\Vert }_{H^{s+1}}$ into a small multiple of $\Vert u{\Vert }_{H^{s+2}}$ plus $C\Vert u{\Vert }_2$. This proves the estimate.

The first-order energy form $Q(u,u)=\Vert du{\Vert }_2^2+\Vert \delta u{\Vert }_2^2$ satisfies $Q(u,u)+c\Vert u{\Vert }_2^2\ge c^{\prime }\Vert u{\Vert }_{H^1}^2$ for a sufficiently large $c>0$. This follows from the same frozen-symbol argument at first order, or by integrating the principal part and absorbing the lower-order terms. The Lax-Milgram theorem thus constructs a bounded inverse $R=({\Delta }_q+c{)}^{-1}:L^2\to H^1$ in the weak sense. To obtain $H^2$ regularity of this weak solution, use a local difference quotient ${\delta }_{h}u$ and test the weak equation with $-{\delta }_{-h}({\chi }^2{\delta }_{h}u)$. Positivity of $g^{ij}$ bounds $\Vert \chi {\delta }_h\nabla u{\Vert }_2^2$; derivatives of $\chi$ and difference quotients of the smooth coefficients give terms bounded by $C\Vert u{\Vert }_{H^1}\Vert \chi {\delta }_h\nabla u{\Vert }_2$ and $C\Vert u{\Vert }_{H^1}^2$. The forcing contributes at most $C\Vert f{\Vert }_2(\Vert \chi {\delta }_h\nabla u{\Vert }_2+\Vert u{\Vert }_{H^1})$. Absorption gives a bound independent of $h$, so the difference-quotient criterion yields $u\in H^2$. Differentiating the equation and repeating gives the higher estimates and smoothness for smooth forcing. Thus the use of elliptic regularity here has been justified rather than assumed.

The inverse $R$ is compact as an $L^2$ operator by the Rellich-Kondrachov compactness theorem, and self-adjoint by symmetry of the weak energy form. It is positive and injective with dense range. The spectral theorem for compact self-adjoint operators now gives an orthonormal eigenbasis of ${\Delta }_q$, with nonnegative eigenvalues tending to infinity. Each eigenfunction is smooth by the estimates above. In particular ${\mathcal{H}}^q$ is finite-dimensional and there is a positive gap above its zero eigenvalue.

Let $H_q$ be the orthogonal projection onto ${\mathcal{H}}^q$. Define the Green operator of the Hodge Laplacian by multiplying a positive-eigenvalue coefficient by $1/\lambda$ and sending the zero coefficients to zero. The gap makes this bounded on $L^2$, and the elliptic estimate shows it maps $H^s$ to $H^{s+2}$; hence it sends smooth forms to smooth forms. It obeysFor a smooth $q$-form $\alpha$, expand the left side to obtainThese three summands are mutually orthogonal. The energy identityshows that a harmonic differential form has both $dh=0$ and $\delta h=0$, making it orthogonal to the exact and coexact summands by integration by parts. Also $⟨d\beta ,\delta \gamma ⟩=⟨d^2\beta ,\gamma ⟩=0$. This proves the claimed direct sum and its uniqueness.

Finally let $d\alpha =0$ and write $u=G_q\alpha$. We have $d{\Delta }_{q}u={\Delta }_{q+1}du$ because $d^2=0$. Since $d(\alpha -H_q\alpha )=0$, $du$ is harmonic. It is also exact, so its orthogonality to harmonic forms gives $du=0$. The decomposition therefore reduces to $\alpha =H_q\alpha +d\delta u$: every closed differential form is cohomologous to a harmonic one. If a harmonic $h$ is exact, $h=d\beta$, then $\Vert h{\Vert }_2^2=⟨\delta h,\beta ⟩=0$. Hence its class is zero only when $h=0$, establishing the isomorphism with de Rham cohomology.

Assume $n>p\ge 1$, the ordinary nonsaturated Gaussian linear model. The log likelihood is, up to a constant,The normal equations and full column rank give the least-squares estimator $\hat{\beta }=(X^{T}X{)}^{-1}{X}^{T}Y$. Write $H=X(X^{T}X{)}^{-1}{X}^T$ and $Q=I-H$, an orthogonal projection of rank $n-p$. Maximizing in ${\sigma }^2$ yieldsSince $QX=0$, the residual sum of squares is ${\epsilon }^{T}Q\epsilon$. Expanding its expectation uses only $\mathbb{E}{\epsilon }_i{\epsilon }_j={\sigma }^2{\delta }_{ij}$:This does not use Cochran's theorem. It proves bias $-p{\sigma }^2/n$ and givesIn the saturated case $n=p$, residuals vanish identically, the likelihood is unbounded as the positive variance tends to zero, and the unbiased residual estimator is unavailable. Thus the usual variance-estimation request requires $n>p$.

In the parametric description, $a$ is a point of the straight line and the nonzero vector $b$ gives its direction. The real parameter runs through every point of that line. For the cross product description, take another cross product with $c$ and use the vector triple product identity:Hence every solution has the representationHere $a\cdot b=0$ and $|b|=|c|$, as required. Conversely,Thus the perpendicularity assumption is sufficient, and the solutions form a straight line parallel to $c$. When $d\ne 0$, the vector $d$ is perpendicular both to $c$ and to every position vector on the line; it is the normal to the plane through the origin containing that line. The point $a=c\times d/|c{|}^2$ is the perpendicular foot from the origin. Its distance from the origin is $|d|/|c|$, since $c\cdot d=0$.

For a general parametric straight line, minimize the squared Euclidean distanceThis strictly convex quadratic function has its unique minimum at $\lambda =b\cdot (y-a)/|b{|}^2$, giving the orthogonal projectionFor the cross product description, $a\cdot c=0$, so this becomes

For the two planes, put $s=m\cdot n$ and $c=m\times n$. Nonparallel unit normals give $|s|<1$ and $|c{|}^2=1-s^2$. Seek the perpendicular foot in their span, $a=um+vn$. Its two dot products impose $u+sv=\mu$ and $su+v=\nu$. Solving gives both requested descriptions of the intersection:andIndeed the vector triple product identity gives $x\times (m\times n)=m(x\cdot n)-n(x\cdot m)$. Conversely, since $m,n$ are linearly independent, equality to $\nu m-\mu n$ forces both plane equations. Thus neither description introduces extra points.

At $\alpha =0$, $\Phi$ is the orthogonal projection onto the plane perpendicular to $n$. Its image of a linear map and kernel of a linear map areEvery perpendicular vector is unchanged and every parallel vector is annihilated. The image plane is perpendicular to the kernel axis by the definition of $n^{\perp }$; their dimensions two and one also agree with the rank-nullity theorem.

Fix $P=P_{\theta ,\eta }$. Let $\mathcal{N}$ be the nuisance tangent space: the closed linear span in $L^2(P)$ of score functions of statistical paths that vary only the nuisance parameter $\eta$. By the preceding argument, $\mathcal{N}$ is contained in the mean-zero L2 space $L_0^2(P)$.

Let ${\Pi }_{\mathcal{N}}$ denote orthogonal projection onto this closed subspace of a Hilbert space. The efficient score and scalar efficient information areThe efficient score is the component of the parametric score function that cannot be reproduced by changing the nuisance parameter. The efficient information is its squared L2 norm; it can be zero, so positivity must not be assumed in the definition.

For a pathwise differentiable statistical functional, an influence-function representer is a mean-zero function $\phi \in L^2(P)$ such that every admissible score function $g$ satisfies $D{\psi }_P(g)=P(\phi g)$. The efficient influence function, also called the canonical gradient, is the unique such representer in the statistical tangent space. Equivalently, it is the orthogonal projection of any representer onto that statistical tangent space. The Pythagorean theorem in an inner-product space shows that it has the smallest squared L2 norm among all representers.

Here the statistical tangent space is all of $L_0^2(P_f)$. To verify the closure explicitly, take $h\in L_0^2(P_f)$, truncate it to $h_m=\max (-m,\min (h,m))$, and set $g_m=h_m-P_f{h}_m$. Then $g_m$ is bounded and centered, and $g_m\to h$ in $L^2(P_f)$, by dominated convergence and the Cauchy-Schwarz inequality. Part (c) supplies a representer already in this space. HenceIts variance is $P_f{a}^2-(P_{f}a{)}^2$.

The closure must be taken in the density-weighted space $L^2(P_f)$. An unweighted reading of $L^2[0,1]$ in the printed hint is false. For example, when $f(u)=2u$, the function $h(u)=u^{-3/4}-8/5$ has $P_{f}h=0$ and $P_f{h}^2=36/25$, so bounded centered truncations converge to it in $L^2(P_f)$; nevertheless $h\notin L^2([0,1],du)$. This illustrates density of bounded centered scores and fixes the measure in the closure statement.

Vary only the probability density function of $X$, using ${\eta }_t=\eta (1+tb)$ with bounded $b$ and $E_{\eta }b(X)=0$. The nuisance score function is $b(X)$, giving the statistical tangent setIts closed linear span is the nuisance tangent space of all centered $L^2(\eta )$ functions of $X$, by density of bounded centered scores.

The conditional expectation of the parametric score function given $X$ is zero:It is therefore orthogonal to this nuisance tangent space. Its orthogonal projection onto that space vanishes, so the efficient score is unchanged. By independence and $E{\epsilon }^2={\sigma }^2$,These equal the parametric score function and Fisher information when $\eta$ is known. There is no loss of information from the unknown covariate density. This is adaptivity to an unknown covariate distribution; it follows from score orthogonality, without having to estimate the nuisance parameter.

First use the additional admitted form of the efficient score. Write $\tilde{\ell }=e\zeta (X)$ and let ${\tau }^2=E_f{\epsilon }^2>0$. The difference $\dot{\ell }-\tilde{\ell }$ is the orthogonal projection onto the nuisance tangent space. Part (c) makes every $e\psi (X)$ orthogonal to that space. Therefore, for all $\psi \in L^2(v)$,The function in braces belongs to L2 space; choosing it as $\psi$ shows it vanishes $v$-almost everywhere. Hence the required conditional deduction isUnder the regular tail condition $sf(s)\to 0$ at both infinities, integration by parts gives $\int s{f}^{\prime }(s)ds=-1$, and the formula becomes $e{h}_{\theta }(x)/{\tau }^2$. This also confirms the sign.

The admitted product form is an extra restriction; it does not follow for every independent-error regression model. To locate the restriction precisely, suppose the error-density nuisance statistical paths preserve both normalization and mean to first order. Their closed mean-preserving error tangent space is $\{\gamma :E_f\gamma =E_f(\epsilon \gamma )=0\}$. The full nuisance tangent space is the orthogonal sum of this space and the centered functions of $X$.

For completeness, bounded functions satisfying the two constraints are dense in the error space. Truncate an arbitrary element, subtract its expected value, and then subtract a multiple of a fixed bounded centered function $q$ with $E_f(\epsilon q)\ne 0$. Such a $q$ exists by truncating $\epsilon$, since ${\tau }^2>0$. The correction coefficients tend to zero by the Cauchy-Schwarz inequality, so the corrected truncations converge in L2 space.

Let $m=E_v{h}_{\theta }(X)$. With $E_f\rho =0$ and $E_f(\epsilon \rho )=1$, the orthogonal projection of $h_{\theta }(X)\rho (\epsilon )$ onto the error nuisance space is $m\{\rho (\epsilon )-\epsilon /{\tau }^2\}$; its projection onto the covariate nuisance space is zero. Thus the general efficient score in independent-error regression isThe orthogonality of the two terms gives the displayed efficient information. For a normal distribution of the error, $\rho (e)=e/{\tau }^2$, so this reduces to the admitted formula. It also does so when $h_{\theta }$ is constant.

A concrete counterexample to the generality of the admission is $g_{\theta }(x)=\theta x$, with $X$ uniform on $[-1,1]$ and independent standard logistic distribution error. Here $m=0$ and $\rho (e)=\tanh (e/2)$, so the actual efficient score is $x\tanh (e/2)$. It cannot equal $e\zeta (x)$ because $\tanh (e/2)/e$ is not constant. This score function is already orthogonal to every nuisance score function: its factor $X$ is centered against error-only directions, and its factor $\rho (\epsilon )$ is centered against covariate-only directions. The requested formula is valid under its stated additional admission, with the general independent-error formula above explaining its limits.

The autocovariance of an MA(1) process comes directly from shared noise terms:Work with the zero-mean white noise convention and ${\sigma }^2>0$. Put ${\mathcal{H}}_{t-1}=\mathrm{span}(X_1,\dots ,X_{t-1})$. Suppose the preceding finite-past innovations $U_1,\dots ,U_{t-1}$ are mutually orthogonal. Their triangular relation to the observations means that they span the same finite past. For $s\le t-2$, $U_s$ is a linear combination of $X_1,\dots ,X_s$, so $\mathrm{Cov}(X_t,U_s)=0$. Moreover,Projection onto this orthogonal basis therefore has only its last coefficient nonzero:The residual is orthogonal to the entire preceding span, proving the induction. Expanding its variance givesThe initialization is $U_1=X_1$, ${\phi }_1={\gamma }_0$, ${\lambda }_1=0$ and $U_0=0$. Each finite covariance matrix is a positive-definite matrix: the last observation in any nonzero finite linear combination contains a noise term absent from earlier observations. Thus all the projection denominators are positive. These are the finite-sample innovations of an MA(1) process, requiring no Gaussian assumption.

Substituting the covariances into the recursion givesThe denominator has a positive limit when ${\lambda }_{t-1}\to \lambda \in [-1,1]$. Taking limits gives $(\lambda -\theta )(\theta \lambda -1)=0$. The permitted interval selectsFor $\theta =0$, every coefficient is zero; at $\theta =\pm 1$, the roots coincide. This is the limiting MA(1) innovations coefficient. The limiting innovation variance is correspondingly ${\sigma }^2\max (1,{\theta }^2)$.

For the observed numerical case, ${\gamma }_0=2$ and ${\gamma }_1=-1$. The exact innovation calculations areTherefore the predictor of $X_3$ based on the two observations and its error variance areFor $X_4$, the available information is still only ${\mathcal{H}}_2$. Since its covariances with both $X_1$ and $X_2$ are zero, its orthogonal projection onto that span is zero:This is two-step prediction for an MA(1) process. Applying the next one-step recursion would instead use an observed $X_3$, which is not available here.

The orthogonal projection has range $\mathcal{S}$, not all of $C[0,1]$: the latter is an infinite-dimensional vector space whereas $\mathcal{S}$ is a finite-dimensional vector space. For a concrete counterexample, $e^x$ cannot belong to any finite-degree piecewise polynomial space, since all its derivatives are nonzero on every open interval. For $k\ge 2$, the B-splines are continuous, so the orthogonal projection is a map $C[0,1]\to \mathcal{S}\subset C[0,1]$. For order one it instead has a generally discontinuous piecewise constant range. For example, a single interval indicator on $[1/4,3/4)$ is the projection of the constant function one onto its span, and is not continuous.

Write $s^{\ast }={\sum }_j{a}_j{N}_j$. Since $M_i$ is a positive scalar multiple of $N_i$, the operator normal equations $⟨f-s^{\ast },N_i⟩=0$ are equivalent toThe mixed-normalization spline Gram matrix is invertible: $G=DH$ where $D_{ii}=k/(t_{i+k}-t_i)>0$ and $H_{ij}=⟨N_i,N_j⟩$ is the ordinary positive-definite matrix of inner products of a linearly independent basis. The positivity and unit-integral normalization of a B-spline give $|b_i|\le \Vert f{\Vert }_{\infty }$. The subpartition of unity for B-splines givesHere the operator norm of a matrix on ${\ell }^{\infty }$ is its maximum absolute row sum. Consequently,This proves the maximum-norm bound for spline projection.

An orthonormal multiresolution analysis consists of closed subspaces of a Hilbert space $(V_j{)}_{j\in \mathbb{Z}}$ of the Hilbert space $L^2(\mathbb{R})$ such thatThere is a scaling function $\varphi$ whose integer function translations form an orthonormal basis of $V_0$. Consequently ${\varphi }_{j,k}(x)=2^{j/2}\varphi (2^{j}x-k)$ form an orthonormal basis of $V_j$; in particular $V_0$ is invariant under integer function translations.

Let $W_j=V_{j+1}\ominus V_j$ be the orthogonal complement of $V_j$ in $V_{j+1}$. Nesting, density and the trivial intersection imply the orthogonal direct sumFor example, telescoping $V_{J+1}=V_L\oplus {⨁}_{j=L}^J{W}_j$ and letting $L\to -\infty$, $J\to \infty$ gives this decomposition; the corresponding orthogonal projections converge strongly to zero and the identity by the intersection and density properties.

To construct the orthonormal wavelet, use $\varphi \in V_1$ to write the scaling refinement equation $\varphi ={\sum }_n{a}_n\varphi (2\cdot -n)$, with $a\in {\ell }^2$ and ${\sum }_n|a_n{|}^2=2$. Put $m(t)=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$. The orthonormal translates and Fourier periodization criterion proved in part (b), applied to the scaling refinement equation, yieldsThis is the quadrature mirror filter identity. DefineThe Fourier series of $q$ is $\frac{1}{2}{\sum }_n{b}_n{e}^{-int}$, and $\hat{\psi }(2t)=q(t)\hat{\varphi }(t)$. The two rows $(m(t),m(t+\pi ))$ and $(q(t),q(t+\pi ))$ form a unitary $2\times 2$ matrix almost everywhere: both have squared norm one and their inner product is zero. This wavelet completion of a multiresolution filter proves not just orthogonality but completeness. Splitting a finer-scale coefficient Fourier series into its values at $t$ and $t+\pi$, this invertible two-channel matrix resolves it into a coarse-scale channel and its complementary channel. Thus the integer function translations of $\psi$ form an orthonormal basis of $W_0$.

The dilated functions ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$ therefore form an orthonormal basis of each $W_j$. The orthogonal direct sum proves the conclusion: every orthonormal multiresolution analysis produces an orthonormal wavelet basis $\{{\psi }_{j,k}:j,k\in \mathbb{Z}\}$ of $L^2(\mathbb{R})$.

We first prove the needed Fredholm alternative for a compact operator without assuming self-adjointness. If $C$ is compact, the finite-rank approximation theorem for compact operators on a Hilbert space follows directly by covering the compact closure of $C$'s unit-ball image by finitely many balls of radius $\epsilon$, spanning their centers by a finite-dimensional space $E$, and using the orthogonal projection $P_E$. Then $\Vert C-P_{E}C\Vert \le \epsilon$.

Choose a finite-rank $F$ with $\Vert C-F\Vert <1$, put $E_0=C-F$, and factorThe first factor is invertible by the Neumann series, and $G$ has finite rank. With $V=\mathrm{ran}G$ and $H=V\oplus V^{\perp }$, the second factor has the triangular formIt is injective precisely when the finite-dimensional block $I_V-G{|}_V$ is injective, and onto precisely when that block is onto. These properties are equivalent in finite dimension. This proves that $I-C$ is injective if and only if it is onto; in that case its inverse is bounded.

For $\lambda \ne 0$, apply this result to $C=K/\lambda$. If $\lambda$ lies in the spectrum of a bounded operator $K$, then $K-\lambda I$ is not invertible, hence is not injective. Therefore every nonzero spectral point of $K$ is an eigenvalue. On its eigenspace $E_{\lambda }$, the operator $K$ equals $\lambda I$. Compactness of $K$ would make the closed unit ball of $E_{\lambda }$ compact after dividing by $\lambda$; part (a) forces $\dim E_{\lambda }<\infty$.

It remains to rule out infinitely many distinct eigenvalues away from zero. Suppose $|{\lambda }_n|\ge \epsilon >0$ are distinct, with respective eigenvectors $v_n$. Distinct eigenvalues give linearly independent eigenvectors. Let $V_n=\mathrm{span}(v_1,\dots ,v_n)$ and choose a unit vector $h_n\in V_n\cap V_{n-1}^{\perp }$. Since $V_n$ is invariant and $(K-{\lambda }_{n}I)V_n\subset V_{n-1}$,For $m<n$, $K{h}_m\in V_{n-1}$, and orthogonality gives $\Vert K{h}_n-K{h}_m\Vert \ge |{\lambda }_n|\ge \epsilon$. This contradicts compactness. Thus for each $\epsilon >0$ there are only finitely many distinct spectral points with modulus at least $\epsilon$.

Taking $\epsilon =1/j$ shows that the nonzero spectrum is countable; if it is infinite, its distinct elements can be enumerated as a sequence tending to zero. Since the spectrum is closed, zero then also belongs to it. In an infinite-dimensional space zero belongs to the spectrum even if there are only finitely many nonzero spectral points: an invertible compact $K$ would make $I=K^{-1}K$ compact, contradicting part (a). The only possible accumulation point is zero, and every nonzero eigenspace is finite-dimensional. Zero need not be an eigenvalue. The argument uses invariant spans, not an orthogonal eigenbasis, which is unavailable for a general non-self-adjoint compact operator.

We prove compact perturbation invariance of Fredholm operators using a two-sided inverse modulo compact operators. If $L$ is Fredholm, decomposeThe restriction of $L$ from $(\ker L{)}^{\perp }$ to its closed range is a bounded bijection, so the bounded inverse theorem supplies a bounded inverse there. Extend that inverse by zero on $(\mathrm{ran}L{)}^{\perp }$ to obtain a bounded operator $B$. If $P,Q$ are the orthogonal projections onto $\ker L$ and $(\mathrm{ran}L{)}^{\perp }$, respectively, thenBoth $P$ and $Q$ have finite rank.

Set $T=L+K$. Products of a compact operator and a bounded operator are compact: on one side compactness preserves compact images, and on the other the bounded operator maps the unit ball into a bounded ball. Hencewhere $C,D$ are compact.

Here is why these identities force $T$ to be Fredholm. On $\ker T$, $C=-I$, so part (a) and compactness make $\ker T$ finite-dimensional. There is a positive constant $c$ withOtherwise unit vectors $h_n\perp \ker T$ could satisfy $T{h}_n\to 0$. From $h_n=-C{h}_n+BT{h}_n$ and compactness, a subsequence would converge strongly to a unit vector $h\perp \ker T$ with $Th=0$, which is impossible. The lower bound proves closed range: for a convergent sequence $T{h}_n$, first discard the kernel components, then the remaining $h_n$ are Cauchy and their limit maps to the desired range limit.

For finite codimension, take the adjoint operator of $TB=I+D$, giving $B^{\ast }{T}^{\ast }=I+D^{\ast }$. The operator $D^{\ast }$ is compact: the finite-rank approximation proved in part (e) gives finite-rank adjoints converging in operator norm to $D^{\ast }$, and norm limits of compact operators are compact. On $\ker T^{\ast }$, $D^{\ast }=-I$, so this kernel is finite-dimensional. Finally,Because the range is closed, its cokernel is isomorphic to this finite-dimensional orthogonal complement. This verifies all three Fredholm operator conditions for $L+K$.

For the converse, start with $L+K$ and perturb by the compact operator $-K$. Therefore $L$ is Fredholm if and only if $L+K$ is Fredholm, with no self-adjointness assumption.

For an autoregressive moving-average model, the driving-to-output transfer function is $A(z)=\theta (z)/\varphi (z)$, and the inverse transfer function is $A(z{)}^{-1}=\varphi (z)/\theta (z)$. The causality and invertibility root criteria for an ARMA model require these respective rational functions to have power series about zero converging on a disk larger than the unit disk. If the two polynomials have no common factor, the conditions becomeIndeed, outside-disk roots leave a radius of convergence greater than one, so the coefficients decay geometrically and are absolutely summable. Conversely an uncancelled pole inside or on the unit disk prevents the required stable power series. If factors are common, apply the criterion after cancellation, to the noise-driven solution rather than additional homogeneous components. The backshift operator translates these power series into the desired one-sided filters.