Interpret the normal-operator source condition as the displayed intersection condition: there is $v\in U$ with $w=K^{\ast }Kv\in \partial J(u^{†})$. Fix any $\alpha >0$ and put $\bar{u}=u^{†}+\alpha v$. The subgradient optimality condition for the artificial-data objective isIt holds because its first term equals $-\alpha K^{\ast }Kv=-\alpha w$. By convexity, this is sufficient for global minimality, proving the forward implication. Conversely, if $u^{†}$ minimizes that objective, the same optimality condition givesTaking $v=(\bar{u}-u^{†})/\alpha$ proves the intersection condition. Thus the two conditions are equivalent when the source vector is allowed to be zero.

The printed additional requirement $v\ne 0$ is not equivalent in general. For $K=I$ and $J\equiv 0$, the displayed source condition holds, and choosing $\bar{u}=u^{†}$ makes $u^{†}$ the minimizer, but $\partial J(u^{†})=\{0\}$ forces $v=0$. A nonzero-source version needs an additional hypothesis, such as requiring $w\ne 0$; it cannot be inferred from the printed intersection alone.

Assume the known covariance matrix $\Sigma$ is positive definite. The Gaussian likelihood, up to constants, isIts gradient is $X^T{\Sigma }^{-1}(Y-X\beta )$. The negative Hessian matrix $X^T{\Sigma }^{-1}X$ is positive definite because $X$ has full column rank. Thus the unique maximum-likelihood estimator is the generalized least squares estimatorIt is unbiased, and multiplication of its linear coefficient matrix by $\Sigma$ givesIn fact ${\hat{\beta }}_{\mathrm{GLS}}\sim N_p(\beta ,(X^T{\Sigma }^{-1}X{)}^{-1})$. Equivalently, a whitening transformation premultiplies $Y$ and $X$ by ${\Sigma }^{-1/2}$, after which ordinary least squares applies. There is no extra ${\sigma }^2$ factor when $\Sigma$ is the complete known covariance; such a factor appears if only a shape matrix $\Omega$ is known and $\Sigma ={\sigma }^2\Omega$.

Retain the random-slope linear mixed model of part (e), but replace the within-laboratory independent error covariance by continuous-time autoregressive residual correlation:Errors from different laboratories remain independent and are independent of the random slopes. The resulting marginal covariance within laboratory $j$ isThe exponential decay applies to the conditional errors, not the entire marginal correlation, which also contains the shared random slope.

Use nlme, since lme4::lmer does not fit this general residual correlation structure. With a data frame containing the observed variables, suitable code isHere corCAR1 estimates $\rho =e^{-\kappa }\in (0,1)$, the error correlation one day apart; correlation at separation $\Delta$ is ${\rho }^{|\Delta |}$. This works for noninteger or unequally spaced days. At equally spaced integer days, corAR1(form = ~ days | lab) gives the corresponding discrete-time construction when its parameter is positive. Distinct times within each error series are required: separate tumours with repeated day values should have separate series identifiers or an additional independent-error component, rather than being assigned perfectly correlated errors. The continuous-time autoregressive residual correlation documentation describes the grouping and continuous-time conventions: stat.ethz.ch/R-manual/R-patched/library/nlme/html/corCAR1.html. The fixed days term here follows the fitted output rather than the inconsistent command in part (e).

The Arzelà-Ascoli theorem says that a uniformly bounded, equicontinuous family of continuous functions on a compact metric space is relatively compact in the uniform norm. Thus every sequence in that family has a uniformly convergent subsequence, and its limit is continuous.

Put $r=\Vert X\Vert <1$. By submultiplicativity of the operator norm, $\Vert X^k\Vert \le r^k$, so the Neumann series converges absolutely in the complete finite-dimensional matrix space. Its partial sums satisfyTaking limits gives the inverse and a quantitative remainder bound:To establish two actual Fréchet derivatives, use the resolvent identityFor fixed $X\in U$ and sufficiently small $H$, factor $I-X-H=(I-X)(I-f(X)H)$ and apply the Neumann series to the second factor. This proves that $X+H\in U$, that $f(X+H)$ is locally bounded, and that $f(X+H)-f(X)=O(\Vert H\Vert )$. Subtracting $f(X)Hf(X)$ in the identity leaves $O(\Vert H{\Vert }^2)$. Thus $Df(X)[H]=f(X)Hf(X)$ is the Fréchet derivative.

Near zero, $f(X)=I+X+O(\Vert X{\Vert }^2)$, uniformly in the operator norm. Hence, as a linear operator in $H$,This proves differentiability of $Df$ at zero, givingThe order of multiplication matters: the second Fréchet derivative is the symmetric bilinear map $HK+KH$, not $2HK$.

Set $d=k-1$ and let $e_m(t_{i+1},\dots ,t_{i+d})$ be the elementary symmetric polynomial of degree $m$ in the interior knots, with $e_0=1$. Expand the Marsden identity in powers of $x$. The coefficient of $x^{d-m}$ on the left is $(-1{)}^m(\genfrac{}{}{0ex}{}{d}{m})t^m$; that coefficient in ${\omega }_i(x)$ is $(-1{)}^m{e}_m(t_{i+1},\dots ,t_{i+d})$. Comparing and cancelling the sign proves the monomial B-spline coefficients:For $m=0$ this gives partition of unity on the basic interval. For $m=1$ it gives the Greville abscissae, the arithmetic means of the $k-1$ interior knots. For the top degree it gives their product. When $k=1$, only $m=0$ occurs and the empty symmetric polynomial is one.

Write $g_t(y)=(y-t{)}_{+}^{k-1}$. Begin with distinct increasing knots so ordinary polynomial interpolation is unambiguous. The two interpolants agree at the $k-1$ common knots $t_{i+1},\dots ,t_{i+k-1}$. Their difference is consequently a scalar multiple of the monic polynomial ${\omega }_i$ of degree $k-1$.

The leading coefficient of an interpolant is its highest divided difference. Thus this scalar isThe equality is precisely the divided-difference recurrence. This proves the Lee interpolation identity:For $k=1$ the common-knot product is empty; the same argument is a difference of constants. Interpret $(y-t{)}_{+}^0$ as $1_{y>t}$ to obtain the usual half-open order-one B-splines. Repeated knots use the standard confluent or limiting interpretation, where the interpolation data are defined.

Sum over $i=1,\dots ,n$. The right side telescopes to ${\ell }_{n+1}(x,t)-{\ell }_1(x,t)$. When $t_k<t<t_{n+1}$, the first interpolation nodes all lie below $t$, so ${\ell }_1=0$. The last interpolation nodes all lie above $t$, so ${\ell }_{n+1}$ interpolates the polynomial $(x-t{)}^{k-1}$ of degree $k-1$ and equals it identically. Hence the Marsden identity follows:

The half-open intervals $[-\pi +2\pi k,\pi +2\pi k)$ tile the line, so exactly one term contributes to ${\sum }_k|f(t+2\pi k){|}^2$. It equals one almost everywhere.

Choose the Shannon scaling mask, a $2\pi$-periodic low-pass filter of a multiresolution analysis which equals one on $[-\pi /2,\pi /2)$ and zero on the rest of $[-\pi ,\pi )$. On the support of $f$ its product with $f(t)$ equals $f(2t)$; outside that support both sides vanish. Its Fourier coefficients giveso it also has the required symbol representation. The inverse Fourier transform gives the Shannon scaling functionThis sinc function has $L^2$ norm one. It illustrates why the Fourier transform convention in part B must allow $L^2$ transforms: the Shannon scaling function is not absolutely integrable on the line.

Use the unnormalized Fourier transform in the question and the Plancherel theorem, with inverse factor $1/(2\pi )$. For a general $L^2$ scaling function, the transform is interpreted in the $L^2$ sense; the printed integral need not be absolutely convergent. All frequency identities are almost-everywhere statements.

Changing variables in the Fourier transform givesThus taking the transform of the scaling refinement equation gives $f(\xi )=m(\xi /2)f(\xi /2)$, orConversely the same calculation and injectivity of the Fourier transform recover refinement, with convergence understood in $L^2$.

To justify both the orthogonality assertion and this convergence precisely, put $P(t)={\sum }_{k\in \mathbb{Z}}|f(t+2\pi k){|}^2$. It belongs to $L^1[-\pi ,\pi ]$ by monotone integration. The Plancherel theorem givesHence orthonormality of all integer translates is equivalent, by uniqueness of Fourier coefficients in L1, to the periodized-energy conditionWhen this holds, the squared norm of ${\sum }_n{a}_n\varphi (2\cdot -n)$ is $\frac{1}{2}{\sum }_n|a_n{|}^2$, so square-summable coefficient series converge in $L^2$. In the converse direction the refinement identity and $P=1$ give, by splitting the periodization of $|f(2t){|}^2$ into even and odd translates,Thus $m$ is bounded and its Fourier coefficients $a_n/2$ are square summable. If $m_N$ are its Fourier partial sum, thenThis verifies the transformed refinement series converges to $f$, completing the equivalence of the two pairs of conditions without imposing an unnecessary $L^1$ assumption on $\varphi$.

An orthonormal multiresolution analysis consists of closed subspaces $(V_j{)}_{j\in \mathbb{Z}}$ of the L2 space $L^2(\mathbb{R})$ with $V_j\subset V_{j+1}$, dense union, intersection $\{0\}$, and the dilation rule $f\in V_j⟺f(2\cdot )\in V_{j+1}$. A scaling function $\varphi$ has integer translates forming an orthonormal basis of $V_0$. Equivalently,form an orthonormal basis of $V_j$. Let $W_j=V_{j+1}\ominus V_j$. Density and the trivial intersection imply $L^2(\mathbb{R})={⨁}_{j\in \mathbb{Z}}{W}_j$.

Nesting gives the scaling refinement equation $\varphi (x)={\sum }_n{a}_n\varphi (2x-n)$, with $h_n=a_n/\sqrt{2}$ the coefficients in the normalized finer-scale basis. Orthogonality of the translates gives ${\sum }_n{a}_n\overline{a_{n-2r}}=2{\delta }_{r0}$. Its Fourier form is the quadrature mirror filter identity $|m(t){|}^2+|m(t+\pi ){|}^2=1$, where $m(t)=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$.

Set $b_n=(-1{)}^n\overline{a_{1-n}}$ and define $\psi (x)={\sum }_n{b}_n\varphi (2x-n)$ in $L^2$. Its high-pass symbol is $q(t)=-e^{-it}\overline{m(t+\pi )}$. The matrixis unitary almost everywhere. Under the identification of $V_1$ with its finer-scale coefficient space, this two-channel unitary transform splits that space into the translates of $\varphi$ and the translates of $\psi$. The latter are therefore an orthonormal basis of $W_0$. Dilating gives an orthonormal wavelet basisThis wavelet completion of a multiresolution filter explains how the multiresolution analysis supplies a single orthonormal wavelet. The argument uses coefficient-space completeness, not just pairwise orthogonality.

Use the weak notion of a typical set. For a word $z^n=(z_1,\dots ,z_n)$ write $P_Z^n(z^n)={\prod }_{i=1}^n{P}_Z(z_i)$. The definition isA weakly typical sequence therefore has probability between $2^{-n(H(Z)+\delta )}$ and $2^{-n(H(Z)-\delta )}$. Words containing zero-probability symbols are excluded. These two bounds and the weak law of large numbers provide the three conclusions below.

Write $\mathcal{D}={\bar{N}}^{-1}{\partial }_t$, $\bar{\Pi }=\mathcal{D}\bar{\varphi }$ and $\Delta =a^{-2}{\nabla }^2$. Since the background scalar has no spatial gradient,Also $\delta P=\bar{\Pi }\mathcal{D}\delta \varphi -{\bar{\Pi }}^2\Psi -V_{,\varphi }\delta \varphi$. These are the linearized scalar-field matter projections; gradient energy first contributes quadratically.

There is a source convention defect in the shear formula. For the stated positive shift $N_i=a^2{B}_{,i}$ and spatial metric with $+2E_{,ij}$, direct substitution into the extrinsic curvature gives the positive-shift scalar shear conventionThere is no extra Laplacian in the definition of ${\chi }_{+}$. Equivalently use $\chi =-{\chi }_{+}$: then the trace-free term has the printed minus sign and $\kappa =3(\mathcal{D}\Phi +H\Psi )-\Delta \chi$. This latter convention retains the printed momentum-constraint form. The source mixes these two choices.

At first order $\delta (K^2)=-6H\kappa$ and $\delta (K^i{}_j{K}^j{}_i)=-2H\kappa$. Combining their difference with $\delta {}^{(3)}R=4\Delta \Phi$ givesFor the momentum constraint, differentiating the tensor above givesFor nonzero Fourier wavenumber, and absorbing the homogeneous integration mode into the background, the consistent momentum constraint is

Linearizing the scalar equation, the shift-advection term and lapse-gradient times scalar-gradient term vanish at this order because the background is spatially homogeneous. The remaining equation isInsert $\delta \Pi$, differentiate its lapse term, and use $\mathcal{D}\bar{\Pi }+3H\bar{\Pi }=-V_{,\varphi }$. The two background-acceleration terms combine into $2\Psi V_{,\varphi }+3H\bar{\Pi }\Psi$. Hence the perturbed Klein-Gordon equation with background lapse isThis scalar evolution equation and the Hamiltonian constraint agree with the displayed targets after the shear conventions are repaired.

For the subsequent essential spectrum arguments, the discrete-spectrum definition must use $\mathrm{ran}(L-\lambda I)$ closed, rather than the printed unshifted range. With that correction, $0\notin {\Sigma }_{\mathrm{e}}(L)$ means exactly that $N=\ker L$ is finite-dimensional and $\mathrm{ran}L$ is closed; this includes the case $0$ is in the resolvent set.

A closed-range bound on the kernel complement supplies the useful equivalenceFor the forward implication, $L:N^{\perp }\to \mathrm{ran}L$ is a bounded bijection between Banach spaces, so the bounded inverse theorem applies. For the reverse implication, any Cauchy sequence of image points has a Cauchy sequence of preimages in $N^{\perp }$, and completeness gives a preimage of its limit.

Now decompose a bounded sequence as $f_n=p_n+v_n$ with $p_n\in N$, $v_n\in N^{\perp }$. If $L{f}_n$ converges, the lower bound makes $(v_n)$ a Cauchy sequence. The finite-dimensional vector space $N$ makes the bounded $p_n$ have a convergent subsequence. Their sum has a norm-convergent subsequence.

Conversely, if every bounded sequence with convergent images has a norm-convergent subsequence, the kernel cannot be infinite-dimensional: an orthonormal sequence in it would have zero images and no convergent subsequence. If the range were not closed, the lower-bound equivalence would provide unit vectors $v_n\in N^{\perp }$ with $L{v}_n\to 0$. Any norm limit would lie in both $N$ and $N^{\perp }$, hence be zero, contradicting its unit norm. This proves the required sequential properness for a self-adjoint operator equivalence.

The spectral-shift repair is essential for later parts. On ${\ell }^2$, take $L=\mathrm{diag}(2,1,1/2,1/3,\dots )$. Its range is not closed, but $2$ is an isolated eigenvalue with one-dimensional eigenspace and closed shifted range. The printed definition would incorrectly place $2$ in the essential spectrum, although no singular Weyl sequence exists there: on the complement of that eigenspace, $\Vert (L-2I)v\Vert \ge \Vert v\Vert$.

No. Reuse the diagonal operator on sequence space $L(x_n)=(x_n/n)$ on ${\ell }^2$. Its kernel is zero, hence finite-dimensional. Let $y^{(N)}$ be the truncation of $(1/n)$ to its first $N$ coordinates. Each $y^{(N)}=L(1,\dots ,1,0,\dots )$ lies in the range, while $y^{(N)}\to (1/n)$ in ${\ell }^2$. The limit is not in the range because its formal preimage is not square summable. Thus finite-dimensional kernel does not imply closed range. Equivalently the unit vectors $e_n\in (\ker L{)}^{\perp }$ violate every positive closed-range bound on the kernel complement.

If unit vectors $f_n$ satisfy $(L-\lambda I)f_n\to 0$, a bounded inverse would give $1\le \Vert (L-\lambda I{)}^{-1}\Vert \Vert (L-\lambda I)f_n\Vert \to 0$. Hence $\lambda$ belongs to the spectrum of a bounded operator.

Conversely a spectral point is real by the preceding argument. Put $A=L-\lambda I$, again a self-adjoint operator. If ${\inf }_{\Vert f\Vert =1}\Vert Af\Vert >0$, then $A$ is injective with closed range. Its range is dense by image-kernel orthogonality for an adjoint, so it is bijective with a bounded inverse, a contradiction. Thus this infimum is zero; choose unit $f_n$ with $\Vert A{f}_n\Vert <1/n$. We have provedSuch a sequence is a spectral Weyl sequence. No weak convergence is required here; nonreal $\lambda$ are ruled out by the lower bound in the preceding solution.

For a bounded self-adjoint operator, $g$ is orthogonal to its range exactly when $⟨Lf,g⟩=⟨f,Lg⟩=0$ for every $f$, or equivalently $Lg=0$. ConsequentlyThis is image-kernel orthogonality for an adjoint. Taking the double orthogonal complement gives the closure; it does not assert that the range itself is closed.

The intended self-adjoint operator identity is $⟨Lf,g⟩=⟨f,Lg⟩$, with the inner product linear in its first argument. The printed second $Lf$ must be $Lg$. Taken literally, the printed identity forces $L=0$: put $g=0$ and then vary $g$. Its spectral conclusion is then trivial, but it does not define self-adjointness.

For the corrected identity, $⟨Lv,v⟩$ is real. If $\mathrm{Im}\lambda \ne 0$, the Cauchy-Schwarz inequality givesThus $L-\lambda I$ is injective and bounded below, so its range is closed: a convergent image sequence has a Cauchy sequence of preimages. Its range is dense because its orthogonal complement is $\ker (L-\bar{\lambda }I)$, also zero by the same lower bound. Hence it is surjective with inverse norm at most $1/|\mathrm{Im}\lambda |$. Therefore

No. On the complex l2 sequence space, let $L(x_n)=(x_n/n)$. This is a bounded self-adjoint operator. At $\lambda =0$, $\ker L=\{0\}$, so zero is not an eigenvalue (an eigenvector must be nonzero). But $y=(1/n)\in {\ell }^2$ has no preimage in ${\ell }^2$: its only formal preimage is the constant-one sequence. Thus $L$ is not surjective andAlternatively $\Vert L{e}_n\Vert =1/n\to 0$ excludes a bounded inverse. This example separates the spectrum of a bounded operator from its point spectrum.