The claim is false because vanishing of the amplitude at one spatial point need not control its derivatives nearby. Take $X=\mathbb{R}$, $k=1$,Then $0\in Z(a)$, but distributionallyup to the Fourier-transform sign convention. Thus $0$ lies in the singular support of $I_{\Phi }(a)$ despite belonging to $Z(a)$.

Suppose $x_0$ has a neighborhood on which ${\nabla }_{\theta }\Phi (x,\theta )\ne 0$ for every $\theta \ne 0$. There one may integrate by parts repeatedly withEach adjoint application lowers the effective symbol order. After enough repetitions, the integral and all its $x$-derivatives converge absolutely and define a smooth function near $x_0$. Therefore

Differentiating a function positively homogeneous of degree $M$ in $\theta$ makes $D_{\theta }^{\beta }b$ homogeneous of degree $M-|\beta |$, while $x$-derivatives preserve the degree. On the unit sphere these derivatives are bounded uniformly over compact subsets of $X$. Scaling then givesfor large $\theta$; smoothness controls the remaining compact frequency region. Thus

The Leibniz rule writes every derivative of $a_1{a}_2$ as a finite sum of products of derivatives of the factors. Multiplying their symbol bounds adds the orders, so

Applying $D_x^{\alpha }$ changes only the constants in the defining estimates, while each $D_{\theta }$ lowers the power of $⟨\theta ⟩$ by one. Hence

True. Fourier transformation turns $D^{\alpha }$ into multiplication by ${\xi }^{\alpha }$, andHence $D^{\alpha }:H^s\to H^{s-|\alpha |}$ is bounded.

True. The Fourier transform of a compactly supported distribution is a smooth function of at most polynomial growth. A sufficiently negative Sobolev weight makes its square integrable, so every $u\in {\mathcal{E}}^{\prime }({\mathbb{R}}^n)$ belongs to $H^t({\mathbb{R}}^n)$ for some $t$.

False. The constant distribution $u=1$ is tempered, but its Fourier transform is a multiple of the Dirac delta function, which is not an $L^2$ function after multiplication by any Sobolev weight. Thus $1\notin H^t({\mathbb{R}}^n)$ for every $t$.

True. The Sobolev embedding theorem gives $H^s({\mathbb{R}}^n)\hookrightarrow C^0({\mathbb{R}}^n)$ when $s>n/2$.

True. Repeated use of the Leibniz rule and the polynomial Taylor formula gives

False. Ellipticity concerns the principal part and large frequencies; lower-order terms may create nonzero real roots. For example, $P(\xi )={\xi }^2-1$ is elliptic in one dimension but vanishes at $\xi =\pm 1$.

True. If $s>t$, then $⟨\xi {⟩}^{2t}\le ⟨\xi {⟩}^{2s}$, so $H^s({\mathbb{R}}^n)\subset H^t({\mathbb{R}}^n)$ continuously.

True. Multiplication by a Schwartz function is a bounded map $H^s({\mathbb{R}}^n)\to H^s({\mathbb{R}}^n)$ for every real $s$.

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 $\log (1/\delta )$.