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Use the shift derivative of a step response, writing $y_3(t,b)=H(s)K(s)$ with $s=t-b$. Differentiate with respect to the forcing location:The possible delta term vanishes because $K(0)=0$, while $K^{\prime }(s)=g(s)$. Therefore, as a distribution and also as the ordinary piecewise response,Moving the start of a unit step differentiates its forcing into a negative impulse. The zero initial conditions and linearity ensure that the corresponding responses obey the same identity, including the zero-frequency cases.

Use one target ancilla qubit initially in $|0⟩$. Apply a CNOT gate from each data qubit $x_j$ to that same target, for $j=1,\dots ,n$. Each CNOT gate adds its control bit modulo two without changing the control. After all $n$ gates the target contains the parity bit $f(x)=x_1\oplus \cdots \oplus x_n$. Thus the required circuit is the $n$-gate parity fan-in:This is parity computation by CNOT gates. The circuit also satisfies $W|x⟩|y⟩=|x⟩|y\oplus f(x)⟩$ for either target value, and $W^{-1}=W$ because all these shared-target CNOT gates commute and individually square to the identity. The action on general superpositions follows by linearity.

For a linear spatial operator, let $U(t,s)$ denote the formal homogeneous evolution from time $s$ to time $t$, satisfying ${\partial }_{t}U(t,s)h+D(t)U(t,s)h=0$ and $U(s,s)h=h$. This also allows time-dependent coefficients in the spatial operator. The Duhamel principle statesDifferentiation of the integral contributes $g(t)$ at its upper endpoint and $-D(t)$ times the integral, while the initial value is $f_0$. If $D$ is time-independent, one writes $U(t,s)=S(t-s)=e^{-(t-s)D}$. The principle uses linearity; an arbitrary nonlinear differential operator would not justify this superposition formula. No analytic construction of $U$ or boundary conditions is required for this formal statement.

Let $h_0=f_{1,0}-f_{2,0}\ge 0$. The damped free-transport evolution preserves nonnegativity, and a nonnegative kernel makes $\tau$ preserve it as well. Every term in ${\sum }_{n\ge 0}{\tau }^{n}F(h_0)$ is therefore nonnegative. By linearity, the difference $f_1-f_2$ solves the fixed point equation with initial value $h_0$. The assumed uniqueness identifies it with this nonnegative sum on every finite interval. Hence the order preservation for the linear Boltzmann equation givesContinuity upgrades pointwise limits in the series to the stated continuous solution; no pointwise differentiability or minimum principle is needed.

If $f$ and $\tilde{f}$ are two weak solutions with the same initial value and the required local time bound, their difference $w$ satisfies $w=\tau w$. By linearity this implies $w={\tau }^{n}w$ for every $n$. On $[0,T]$ put $A_T={\sup }_{s\le T}\Vert w(s){\Vert }_1<\infty$. The factorial bound for a Volterra iterate now givesFor fixed $T$, the factor tends to zero; its successive-term ratio is $2T/(n+1)$. Therefore $w(t)=0$ as an $L^1$ element at every time on this interval. Since $T$ is arbitrary, the locally time-bounded $L^1$ weak solution is unique globally. This proof uses precisely the additional condition requested, rather than assuming arbitrary pointwise-in-time integrability alone supplies a finite uniform bound.

True. Choose a basis $v_1,\dots ,v_r$ of $V$. By surjective function, choose $u_i\in U$ with $\alpha u_i=v_i$. Define the linear map $\beta$ by $\beta v_i=u_i$ and extend by linearity. Then $\alpha \beta v_i=v_i$ on every basis vector, and thereforeThis constructs a right inverse and does not require $\alpha$ to be injective. If $V$ is the zero vector space, the unique zero map is the required right inverse.

Let $V$ denote the coherent exact quantum phase estimation unitary operator from the preceding part, with the data quantum register written first. For every eigenvector of $A$,because $e^{2\pi iA}|u_j⟩=e^{2\pi i{\lambda }_j}|u_j⟩$ and ${\lambda }_j=c_j/2^n$ lies in $[0,1)$, with no phase-aliasing ambiguity. Apply $V$ coherently to the input quantum state and a clean quantum register; do not measure the eigenvalue label. By linearity it produces ${\sum }_j{\beta }_j|u_j⟩|c_j⟩$.

Append a quantum ancilla in $|0⟩$, and use the provided quantum variable rotation with ${\theta }_{c_j}=\arcsin {\lambda }_j$, choosing the nonnegative cosine. The resulting normalized quantum state isThe sum covers both terms; this fixes the potentially ambiguous sum placement in the abbreviated display. The orthonormal basis of eigenvectors and the unit length of each rotated quantum ancilla show that its norm is one. This is quantum spectral filtering by multiplication, not the reciprocal rotation used in the HHL algorithm.

Use a clean quantum ancilla and apply the quantum circuit $C$, then the Pauli Z gate on the ancilla, then the inverse quantum circuit $C^{†}$:This compute-phase-uncompute construction implements $A\otimes I$ on every input with the ancilla initially $|0⟩$, and by linearity on any superposition of those inputs. The action on ancilla-$|1⟩$ inputs need not coincide with $A\otimes I$; the question only requires the clean-ancilla subspace.

Reflection in $y=0$ preserves the prescribed translational velocity and axial angular velocity. A force transforms as a polar vector, whereas a torque transforms as an axial vector; consequently $F_y=G_x=G_z=0$. Reflection in $x=0$ reverses both imposed motions but preserves the normal force component. The linearity of Stokes flow then gives $F_z=0$. Thus only $F_x$ and $G_y$ can be nonzero.

There is a sign inconsistency in the printed question. In right-handed coordinates, $\mathbf{\Omega }=\Omega {\mathbf{e}}_y$ gives a rotational velocity $-\Omega a{\mathbf{e}}_x$ at the sphere's lowest point. The printed shear and force formulas instead use the opposite rotational sense. Below, let $\omega$ be the right-handed component along $+{\mathbf{e}}_y$; the paper's displayed formulas are recovered by setting $\omega =-\Omega$.

In the translating frame, the parabolic gap and leading boundary velocities areThe no-slip boundary condition and lubrication theory giveThe gap is stationary in this frame, so ${\nabla }_{\parallel }\cdot \mathbf{q}=0$. The Reynolds lubrication equation is thereforeSince $h_x=x/a$ and ${\nabla }_{\parallel }^{2}h=2/a$,Differentiating the velocity profile gives the leading shear stressesThe first expression becomes the printed formula after $\omega =-\Omega$.

For the logarithmic lubrication resistance of a sphere near a wall, write $\ell =\log (1/ϵ)$. The logarithmic region is $a\sqrt{ϵ}\ll r\ll a$, where $h\sim r^2/(2a)$. ConsequentlyThese follow from $\int r/hdr=a\ell +O(a)$ and $\int r^3/h^{2}dr=2a^2\ell +O(a^2)$, with a fixed small outer cutoff. The general logarithmic radial numerator is $r^{2n-1}$ over $h^n$; the numerator in the printed integration hint appears to have a typographical error.

The force exerted on the fluid through the upper gap boundary includes pressure acting on its slope:The pressure cancels the $x^2/h^2$ term, givingPressure exerts no torque about the centre of a sphere because its traction is radial. To logarithmic order, the shear has lever arm $-a{\mathbf{e}}_z$, soThe other components vanish by parity: $G_x$ involves an integral of the $xy$ term, and $G_z$ has an integrand odd in $y$. The combined hydrodynamic resistance matrix isIts equal off-diagonal coefficients are required by the Lorentz reciprocal theorem for Stokes flow. It is a positive-definite matrix, as required by viscous dissipation.

In the rotational convention of the printed force formula, set $\Omega =-\omega$ and measure the scalar couple along $-{\mathbf{e}}_y$, so $G=-G_y$. ThenFor sedimentation under the buoyancy-adjusted weight $F$, a uniform sphere is torque-free. Thus $\omega a=U/4$, and the force balance givesThe rotational scalar in the paper's displayed-force convention is $\Omega =-U/(4a)$. The sphere rolls in the right-handed $+{\mathbf{e}}_y$ sense but still slips: its lowest point has laboratory velocity $U-a\omega =3U/4$. The signs and coefficients of the right-handed resistance agree with the rigid-wall terms of Bertin et al., equations (4.4)–(4.5), after reversing the forces and torques there from fluid-on-sphere to sphere-on-fluid.

This is the boundedness criterion for a bilinear form. Since $B$ is continuous at $(0,0)$, there are $r,s>0$ such that $|B(x,y)|\le 1$ whenever $\Vert x\Vert <r$ and $\Vert y\Vert <s$. For nonzero $u,v$, apply this toLinearity in each argument then givesThe same inequality is immediate if either vector is zero, so one may take $M=4/(rs)$.

Integrating the ordinary differential equation four times gives a quartic polynomial. It is useful to use the linearity of the equation and split the solution into a gravity part and a force part:wheresatisfies the clamped and torque-free boundary conditions with $F=0$, whilesatisfies the homogeneous equation and contributes the endpoint force. Direct differentiation verifiesThusthe clamped-free beam under uniform load and endpoint force.

Apply separation of variables to the Laplace equation in polar coordinates by writing $\varphi (r,\theta )=R(r)\Theta (\theta )$. Division by $R\Theta$ givesThe requirement that $\Theta$ be a real $2\pi$-periodic function restricts the separation constants to $n^2$, with angular factors $\cos (n\theta )$ and $\sin (n\theta )$. For $n\ge 1$, the radial ordinary differential equation is an Euler equation with solutions $r^n$ and $r^{-n}$. For the zero mode, $\Theta$ is constant andso $R=a_0+c_0\log r$. By linearity, superposition givesThis is the separated expansion of a harmonic function in a circular region.

For an $n\times n$ matrix $A$, the characteristic polynomial isThe Cayley-Hamilton theorem states that ${\chi }_A(A)=0$.

Over $\mathbb{C}$, choose a basis in which $A$ is upper triangular, with diagonal entries ${\lambda }_1,\dots ,{\lambda }_n$. For the standard invariant flag $V_j=⟨e_1,\dots ,e_j⟩$,The factors $A-{\lambda }_{j}I$ commute, so applying their product in descending order sends $V_n$ successively into $V_{n-1},\dots ,V_0=0$. Hencewhich proves the theorem.

Direct expansion gives the commutator product rule:

Put $C=[B,A]$. Since $C$ commutes with $A$, repeated use of the product rule givesBy linearity, for every polynomial $\phi$,

Let $D(X)=[B,X]$ and suppose $f(A)=0$. For $k=1$,Assume inductively thatBoth $u$ and $C$ are polynomials in, or commute with, $A$, so $u{C}^m=C^{m}u=0$. Apply the derivation $D$ to $u{C}^m=0$ and multiply on the left by $C^m$:Since $D(u)=f^{(k+1)}(A)C$, this saysThe induction is complete. Taking $f={\chi }_A$ and $k=n$ givesThus $[B,A]$ is nilpotent, which is the Jacobson lemma for a commuting commutator.

Let $E_n$ and $E$ be the projection-valued measures of $A_n$ and $A$. Using the projection $P_n:\mathcal{H}\to {\mathcal{H}}_n$, defineThe scalar spectral measures converge weakly whenfor every bounded continuous function $f$ and every $v,w\in \mathcal{H}$. By the spectral theorem for normal operators on a separable Hilbert space, this is equivalent to

The assumed moment identities say precisely that this convergence holds for every monomial $f(x)=x^m$. It follows by linearity for every polynomial. For $v=w$, the $m=2$ identity givesso Markov inequality makes the positive measures $({\mu }_v^{(n)})$ tight. Higher even moments similarly control the tails of any fixed polynomial.

Given a bounded continuous $f$ and $ϵ>0$, choose $R$ so that the measure tails are uniformly small. The Weierstrass approximation theorem supplies a polynomial $p$ withMoment convergence handles $p$; tightness and a sufficiently high even moment handle the two tails. Hence $\int fd{\mu }_v^{(n)}\to \int fd{\mu }_v$. The polarization identity then gives the same conclusion for ${\mu }_{v,w}^{(n)}$. This proves weak convergence of scalar spectral measures.

The assertion fails if only $m=1$ is assumed. Let $\mathcal{H}={\ell }^2(\mathbb{N})$, let ${\mathcal{H}}_n=\mathrm{span}\{e_1,\dots ,e_n\}$, take $A=0$, and letThe reversal matrices are self-adjoint unitaries. For fixed $v,w\in {\ell }^2$,because the finite head of one vector is paired with the vanishing tail of the other. Thus the $m=1$ condition holds. However, $A_n^2=I_{{\mathcal{H}}_n}$, soin general. Taking $f(x)=x^2$ shows that the spectral measures do not converge weakly.

Both $Z\otimes Z$ and $X\otimes I$ are unitary and Hermitian, so their expectations are real. Run the real-part Hadamard test twice, once with controlled $Z\otimes Z$ and once with controlled $X\otimes I$. If the respective ancilla outcome-one probabilities are $p_{ZZ}$ and $p_{XI}$, thenThe identity in part (b) implements each controlled tensor product by gates on its two target wires with the common control. By linearity,

Let $1<p<\infty$ and $q=p/(p-1)$. Every $y\in {\ell }^q$ definesHölder's inequality gives $\Vert f_y\Vert \le \Vert y{\Vert }_q$, and equality follows by using a normalized sequence proportional to $\mathrm{sgn}(y_n)|y_n{|}^{q-1}$.

Conversely, for $f\in ({\ell }^p{)}^{\ast }$ put $y_n=f(e_n)$. On the first $N$ coordinates, choose the same norming vector to obtainLetting $N\to \infty$ shows $y\in {\ell }^q$. On finitely supported sequences, linearity gives $f=f_y$; density of those sequences in ${\ell }^p$ extends this identity to all $x$. Thereforeisometrically.

The dual $X^{\ast }$ is the vector space of bounded linear maps $f:X\to \mathbb{F}$, with normIf $(f_n)$ is Cauchy in this norm, then $(f_n(x))$ is Cauchy in the complete scalar field for every $x$. Define $f(x)={\lim }_n{f}_n(x)$. Pointwise passage to the limit preserves linearity. Given $\epsilon >0$, choose $N$ such that $\Vert f_n-f_m\Vert <\epsilon$ for $m,n\ge N$; sending $m\to \infty$ givesThus $f$ is bounded and $\Vert f_n-f\Vert \le \epsilon$. Hence $X^{\ast }$ is Banach, even if $X$ was only normed.

For fixed $y$, Riesz gives a unique $Sy$ with $\theta (x,y)=⟨x,Sy⟩$. Conjugate linearity in $y$ and the inner-product convention make $S$ linear. Moreover $\Vert Sy\Vert ={\sup }_{\Vert x\Vert \le 1}|\theta (x,y)|\le \Vert \theta \Vert \Vert y\Vert$, and taking both suprema proves $\Vert S\Vert =\Vert \theta \Vert$.

A right inverse of a function $f:X\to Y$ is a function $g:Y\to X$ satisfying $f\circ g={\mathrm{id}}_Y$. It forces $f$ to be a surjective function. A surjective linear map onto a finite-dimensional vector space has a linear right inverse: lift a basis of the target and extend by linearity.

A multilinear operation on vector fields is tensorial when it is linear over smooth functions in every argument. Its value at a point then depends only on the argument vectors at that point. To see this, use a smooth cutoff function to reduce to local fields, expand them in a local frame, and apply linearity over smooth functions to their coefficients. A smoothly valued tensorial operation therefore defines a tensor field. The curvature of an affine connection is tensorial, whereas a covariant derivative differentiates a scalar coefficient in its second argument and is not tensorial there.