No. Non-entireness rules out a compactly supported solution, but the Malgrange–Ehrenpreis theorem gives a distributional fundamental solution of a linear differential operator $E$ with $P(D)E=\delta$. Since $v$ has compact support, the convolution $u=E\ast v$ is defined and satisfiesOne may choose a tempered fundamental solution for a constant-coefficient operator, so $u\in {\mathcal{S}}^{\prime }(\mathbb{R})$.

If $u\in {\mathcal{E}}^{\prime }(\mathbb{R})$ solves $P(D)u=v$, Fourier transformation givesThe Paley–Wiener–Schwartz theorem makes $\hat{u}$ entire, so $\hat{v}/P$ is entire.

Conversely, suppose $F(z)=\hat{v}(z)/P(z)$ is entire. Polynomial division estimates away from the finitely many zeros of $P$, together with the maximum principle on fixed disks around those zeros, show that $F$ retains a Paley--Wiener--Schwartz bound, with only the polynomial exponent changed. The converse theorem therefore gives $u\in {\mathcal{E}}^{\prime }(\mathbb{R})$ with $\hat{u}=F$. Then $P(D)u=v$. Thus

Suppose output qubit $j$ is measured. Its probability of outcome zero isBecause $C$ is a Clifford operation, the Pauli group is preserved under conjugation. Propagating $Z_j$ backward through the $N=\mathrm{poly}(n)$ gates therefore produces, including its sign, a tensor-product PauliThe input is a product state, so the expectation factors:Each one-qubit factor follows directly from the classical description of $|{\alpha }_k⟩$, and conjugating a Pauli through each Clifford gate takes constant classical work. Hence $p_0$, and $p_1=1-p_0$, are computable in polynomial time. This is Heisenberg propagation of a Pauli observable through a Clifford circuit.

The process admits a strong classical simulation of a quantum circuit if a classical algorithm, given its circuit and input descriptions, computes either output probability to the requested polynomial precision in time polynomial in the input size and precision parameter. This asks for the probabilities themselves, rather than only efficient samples from their distribution.

Prepare $|\xi ⟩=Q^{-1/2}{\sum }_b|b⟩$ in the second register by part (ii), while preserving the first register $|a⟩$. Applying the diagonal unitary $V$ from part (iii) givesThe preparation can be made coherent using amplitude amplification in place of postselection when this map is needed as a subroutine.

Write $b={\sum }_{j=0}^{m-1}{2}^j{b}_j$. Applying the phase gateto qubit $b_j$ contributes $\exp (2\pi i{2}^j{b}_j/Q)$. The product of these $m$ gates is thereforeSimilarly, for $a={\sum }_j{a}_j{2}^j$ and $b={\sum }_k{b}_k{2}^k$, apply a controlled phase gatebetween every pair $(a_j,b_k)$. The accumulated phase isThe $m^2$ controlled phase gates implement

Apply Hadamard gates to $|0^m⟩$ to prepareReversibly compute the efficiently decidable predicate $[x<Q]$ into an ancillary qubit and measure it. The success probability is $Q/2^m>1/2$, and conditioned on success the first register is exactlyRestart after a failed measurement. After $L=\lceil {\log }_2(1/\delta )\rceil$ attempts, the probability that all attempts fail is below $\delta$, so the state is prepared with probability at least $1-\delta$ using a number of gates polynomial in $m$ and $\log (1/\delta )$.

With $\omega =e^{2\pi i/Q}$, the quantum Fourier transform over the additive group ${\mathbb{Z}}_Q$ is

Use $|1⟩$ as the target register for quantum phase estimation of $U_a$. By part (iii), it is the equal superposition $r^{-1/2}{\sum }_s|{\psi }_s⟩$ of eigenvectors with phases $s/r$. Controlled modular multiplications implement the required powers $U_a^{2^j}$ efficiently. The phase-estimation circuit produceswhere the first register contains an $m$-bit approximation with constant success probability. A quantum measurement in the computational basis therefore outputs an approximation to $s/r$, with $s$ uniformly distributed over $\{0,\dots ,r-1\}$.

Summing the eigenvectors and reversing the finite sums givesThe root-of-unity filter makes the inner sum equal to $r$ for $k=0$ and zero otherwise. Since $a^0=1$,

The multiplicative order $r$ makes the states $|a^k\mathrm{mod}N⟩$, $0\le k<r$, distinct and cyclic under $U_a$. ThereforeThus each $|{\psi }_s⟩$ is an eigenvector withThese are the Fourier eigenvectors of the cyclic modular-multiplication orbit.

Since $\gcd (x,N)=1$, multiplication by $x$ modulo $N$ is a bijection of ${\mathbb{Z}}_N$, with inverse multiplication by the modular inverse $x^{-1}$. Hence $U_x$ permutes the standard orthonormal basis. It preserves all inner products and satisfiesso $U_x$ is a unitary operator.

Put $H={\mathrm{Stab}}_F(x)$. For $g,h\in G$,Thus $f_x$ is constant on every left coset of $H$ and takes different values on different cosets. Its oracle is therefore an oracle for the hidden subgroup problem, and the hidden subgroup is precisely the stabilizer subgroup $H$.

Let $H={\mathrm{Stab}}_F(x)$. The identity belongs to $H$ because $F(e,x)=x$. If $g,h\in H$, the group action law givesso $gh\in H$. Finally, if $g\in H$, thenso $g^{-1}\in H$. The subgroup criterion therefore proves that ${\mathrm{Stab}}_F(x)$ is the stabilizer subgroup of $x$.

Extend the quantum conditional entropy from normalized states to positive operators bySince $\mathrm{Tr}{X}_{AB}=\mathrm{Tr}{X}_B$, the two terms involving $\log t$ cancel under $X\mapsto tX$, so $F(tX)=tF(X)$. Thus $F$ is positively homogeneous, and part (b) extends its concavity from states to the positive cone.

Apply part (c) with $X={\sigma }_{AB}$ and $Y={\rho }_{AB}$. Differentiating the matrix logarithm under the trace givesThe inequality from part (c), after moving $F(\rho )$ to the left, becomesThis is the data-processing inequality for quantum relative entropy under partial trace. Tensoring each output with the appropriate maximally mixed state does not change either side, so it also proves data processing under normalized partial traces. Singular $\sigma$ follows by approximation on its support.

For $t>0$, positive homogeneity and concavity giveAfter subtracting $f(X)$, dividing by $t$, and taking the one-sided directional derivative at zero,

Because the two flags $E_C$ and $F_C$ are orthogonal projections, the flagged state is block diagonal. If $h(\lambda )$ denotes the binary entropy, thenandIts unflagged marginals are ${\rho }_{AB}=\lambda {\rho }_{AB}^{\prime }+(1-\lambda ){\rho }_{AB}^{\prime \prime }$ and ${\rho }_B=\lambda {\rho }_B^{\prime }+(1-\lambda ){\rho }_B^{\prime \prime }$. Substitution in Strong subadditivity of Von Neumann entropy cancels the two binary-entropy terms and givesThis is exactly the concavity of quantum conditional entropy.

A real function $f$ on $I=[0,1]$ is an operator convex function when, for all Hermitian operators $X,Y$ whose spectra lie in $I$ and every $0\le \lambda \le 1$,where $⪯$ is the Loewner order. Reversing the inequality defines an operator concave function, equivalently $-f$ is operator convex.

For $m=4,n=2$, the equation is the one-dimensional porous medium equation $S_{\tau }=(S^3{)}_{xx}$. Use a similarity solution $S={\tau }^{-1/4}f(\xi )$ and $\xi =x{\tau }^{-1/4}$. ThenThe no-mass-flux condition at $x=0$ sets the integration constant to zero:With $f(0)=1$,ConsequentlyIts edge is $R=12r_0{\tau }^{1/2}\propto t^{1/2}$. Moreover,is time-independent. This agrees with part (b), since $2-m+2n=2$.

Set $r=r_0{x}^2$, $\Sigma ={\Sigma }_0{x}^{-3}S$, and $\bar{\nu }={\bar{\nu }}_0{x}^{2m-3n}{S}^n$. Then ${\partial }_r=(2r_{0}x{)}^{-1}{\partial }_x$. Substitution into the viscous evolution of an accretion disk equation, with $\tau =3{\bar{\nu }}_{0}t/(4r_0^2)$, cancels the common dimensional factors and gives