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The symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)$ consists of smooth amplitudes with estimatesThere is no loss of symbol order under an $x$ derivative. A phase function is real and smooth for $\theta \ne 0$, positively homogeneous of degree one in $\theta$, with its full differential $(d_x\Phi ,d_{\theta }\Phi )$ nowhere zero there, at least on the amplitude's conic support. Homogeneous phases need not be smooth at $\theta =0$; that bounded-frequency region is handled separately or the amplitude is cut off near it.

Choose $\rho \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero. The oscillatory integral is defined by regularization in distributions:This is an ordinary integral when decay makes it absolutely convergent. For general order, the definition is independent of the regularizer. To see the mechanism, at large $|\theta |$ setHomogeneity and the nonvanishing full differential give $A\gtrsim |\theta {|}^2$ on compact $x$ sets, and $L{e}^{i\Phi }=e^{i\Phi }$. Its $x$-derivative coefficients have order $-1$ and its $\theta$-derivative coefficients order zero, so each formal transpose $L^t$ lowers the amplitude order by one. Iterating more than $N+k$ times makes the test-function pairing integrable, and the same estimates control regularizer derivatives. This justifies the cutoff function limit and its independence. A fixed bounded-frequency integral is smooth in $x$.

The singular support of a distribution is the complement of the largest open set on which it is represented by a smooth function. For the support-sensitive bound one must use conic support of an oscillatory amplitude: a closed set of limiting high-frequency directions, locally in the base variable. One sufficient precise choice iswith closure taken in this product. If $x_0$ is outside the projection of $C_a\cap \{d_{\theta }\Phi =0\}$, compactness of the unit sphere gives a neighborhood of $x_0$ and a positive uniform lower bound on $|d_{\theta }\Phi |$ wherever the large-frequency amplitude is supported. On that region useIts coefficients have order zero and its transpose lowers order by one. To prove $C^j$ regularity, first differentiate $j$ times in $x$, raising the amplitude order by at most $j$, and then integrate by parts more than $N+j+k$ times. The resulting integrals and their derivatives converge uniformly on compact neighborhoods. Since $j$ is arbitrary, the stationary-direction bound for singular support follows:This is the standard conic interpretation of the support restriction in the question; directions at infinity, rather than finite-frequency stationary points, determine possible singularities.

If $\mathrm{supp}a(x,\cdot )$ literally means the support of the restricted function, the printed inclusion can fail. There is a slice-support obstruction for oscillatory singularities. In one dimension, take $\Phi =x\theta$ and $a=x\eta (\theta )/|\theta |$, where $\eta$ is smooth and even, zero for $|\theta |\le 1$, and one for $|\theta |\ge 2$. This is a symbol of order $-1$, and the full phase differential is nonzero for $\theta \ne 0$. Its regularized integral near $x=0$ isIndeed substitute $u=|x|\theta$ in the tail, split at $u=1$, and write $\cos u=1+(\cos u-1)$ below one. The first term gives $-\log |x|$, while the remainder has an even convergent power series in $x$ plus a constant; the bounded-frequency correction is smooth. Thus $0$ is in the singular support, but $a(0,\cdot )=0$ has empty slice support. The joint closed conic support retains this limiting base point and makes the proved theorem valid. No nonstationarity conclusion is inferred just from the amplitude vanishing at one base point.

For the wave equation, Fourier transformation in $x$ reduces the initial-value problem to ${\partial }_t^2\hat{E}+c^2|\xi {|}^2\hat{E}=0$, with $\hat{E}(\xi ,0)=0$ and ${\partial }_t\hat{E}(\xi ,0)=1$. For $c>0$ its solution iswith value $t$ at zero. Choose $\chi \in C_c^{\infty }({\mathbb{R}}^n)$ equal to one near zero. The low-frequency decomposition of the wave propagator iswhereThe low-frequency term is an ordinary smooth function: the sine quotient extends smoothly in $\xi$ at zero and all derivatives are integrable on the fixed compact frequency set. The high-frequency amplitudes belong to $\mathrm{Sym}({\mathbb{R}}^{n+1},{\mathbb{R}}^n;-1)$, vanish near zero, and the phases have nonzero full differential since $d_x{\Phi }_{\pm }=\xi \ne 0$. Their stationary-direction equations are $x\pm ct\xi /|\xi |=0$. Such a direction can occur only if $|x|=c|t|$. The smooth low-frequency term contributes no singular support, soThis is a statement about singular support: some dimensions have a smooth nonzero tail inside the cone, so it does not claim that the distribution is concentrated only on the cone.

A Schwartz function is a smooth function for which every seminormis finite. These seminorms define the Fréchet topology of the Schwartz space $\mathcal{S}({\mathbb{R}}^n)$. The tempered distributions form its continuous linear dual ${\mathcal{S}}^{\prime }$. We use bilinear distribution pairing, with no complex conjugation on the test function.

Differentiation under the Fourier integral and integration by parts giveTherefore each Schwartz seminorm of $\hat{\phi }$ is bounded by a finite sum of $L^1$ norms of polynomially weighted derivatives of $\phi$. Insert the integrable weight $⟨x{⟩}^{-n-1}$ to bound those norms by finitely many Schwartz seminorms. This proves that $\mathcal{F}:\mathcal{S}\to \mathcal{S}$ is continuous.

To justify inversion without a merely formal exchange of integrals, insert $e^{-\epsilon |\xi {|}^2/2}$ in the inverse integral. The Gaussian Fourier integral givesThe normalized Gaussian is an approximate identity, so the right side tends to $\phi (x)$; the left side converges by dominated convergence theorem since $\hat{\phi }$ is integrable. It follows thatReflection is continuous in Schwartz seminorms, so ${\mathcal{F}}^{-1}=(2\pi {)}^{-n}\mathcal{R}\mathcal{F}$ is continuous as well. This proves the Fourier transform isomorphism of the Schwartz space.

Define the distributional Fourier transform byIt agrees with the ordinary integral transform whenever Fubini applies. Transposing the Schwartz inverse gives an inverse on ${\mathcal{S}}^{\prime }$, and the same squared-transform identity holds. For the usual strong dual topology, a seminorm is $p_B(u)={\sup }_{\phi \in B}|⟨u,\phi ⟩|$ for a bounded set $B\subset \mathcal{S}$. Since a continuous linear Schwartz map takes bounded sets to bounded sets, $p_B(\hat{u})=p_{\mathcal{F}B}(u)$ proves continuity, and similarly for the inverse. Continuity also holds in the weak dual topology. Thus the transform is a continuous isomorphism on tempered distributions, with the stated normalization.

For a real symmetric positive-definite matrix $G$, there is $c>0$ with $g(x)=x^{T}Gx\ge c|x{|}^2$. The reciprocal is locally integrable in dimension three: the radial factor near zero is $r^2{r}^{-2}dr$. At infinity, rapid decay of a Schwartz test function makes it integrable. More quantitatively,This single seminorm bound proves a tempered distribution. No principal-value extension is needed at the origin.

First calculate the isotropic transform. For $f_{\epsilon }(x)=e^{-\epsilon |x|}/|x{|}^2$ and $k=|\xi |>0$, spherical integration and the supplied sine-integral identity giveThe original functions converge in ${\mathcal{S}}^{\prime }$ to $|x{|}^{-2}$ by dominated convergence theorem against tests. The transforms are bounded by $2{\pi }^2/|\xi |$, which is locally integrable in three dimensions and integrable against Schwartz functions at infinity. HenceLet $y=G^{1/2}x$. The Jacobian is $(\det G{)}^{-1/2}$ and the dual vector is $G^{-1/2}\xi$. The Fourier transform of a reciprocal positive quadratic form is consequentlyThe frequency-origin value is understood distributionally; there is no additional delta term.

For the complex extension use a symmetric matrix $A=G+iB$ with real symmetric $B$ and positive real part $G$. Symmetry is natural for a quadratic form; a skew-symmetric part contributes nothing. The bound $|x^{T}Ax|\ge x^{T}Gx$ again gives a regular tempered reciprocal. Both sides of the prospective formula depend holomorphically on the matrix while its real part is positive: compact parameter sets give a common $|x{|}^{-2}$ bound for pairing and differentiated integrands. Continue from the real positive matrices along $A_z=G+izB$. In a complex neighborhood of $z=0$, imaginary $z$ gives real positive matrices, so the one-variable identity theorem supplies the equality; connected continuation along $0\le z\le 1$ reaches $A$.

The analytic determinant square root for accretive symmetric matrices must follow that continuation, rather than an arbitrary scalar principal root of $\det A$. Explicitly, put $C=G^{-1/2}B{G}^{-1/2}$ and let its real eigenvalues be $b_j$. Thenwhere each factor has positive real part. This is the determinant square root normalized positively on real positive matrices. AlsoThus the quadratic-form square root has the unambiguous branch with positive real part. The accretive complex quadratic reciprocal Fourier transform isUniform local integrability of the right side justifies its analytic continuation as a tempered distribution, not merely pointwise away from the origin.

For the particular form,Substitution gives the required normalized expressionFor every nonzero real $\xi$, the radicand has positive real part, so the specified square root exists uniquely. Both sides are regular tempered distributions, despite their locally integrable singularities at zero.

Use the usual strictly increasing spline knot sequence, and initially assume $k\ge 2$. The explicit formula for the order-$k$ divided difference givesEach summand is a truncated power function in $t$, with its junction at $t_j$. It is a piecewise polynomial function of degree at most $k-1$ and has $k-2$ continuous derivatives. Finite summation preserves these properties.

For $t\ge t_{i+k}$ all summands vanish. For $t\le t_i$, all spline knot data come from the polynomial $u\mapsto (u-t{)}^{k-1}$. An order-$k$ divided difference annihilates a polynomial of smaller degree, so $M_i(t)=0$ there too. Thus the support is contained in $[t_i,t_{i+k}]$.

On the last open spline knot interval only the $j=i+k$ term survives; it is a nonzero multiple of $(t_{i+k}-t{)}^{k-1}$. On the first open spline knot interval, compare the truncated spline knot data with the untruncated polynomial data: only the $j=i$ value differs. Since the untruncated divided difference is zero, the result is a nonzero multiple of $(t_i-t{)}^{k-1}$. Consequently both outer pieces are nonzero, giving exact closed support. The Cox-de Boor recurrence used below also proves positivity throughout the open support. ThereforeThis is simple-knot B-spline regularity. For $k=1$ the B-splines are interval indicators; they are piecewise constant and need not be continuous. The notation $C^{k-2}$ should then be read as no continuity requirement, not as a classical negative-order differentiability space.

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.

For the probability generating function $G(t)=\mathbb{E}[t^X]$, differentiation givesThe second derivative is a factorial moment, not the raw second moment.

In the Galton-Watson process, condition on $X_n=m$. The next generation is a sum of $m$ independent offspring counts, and so its conditional generating function is $G(t{)}^m$. Averaging over $X_n$ givesThe conditional mean is $m\mu$ and the conditional variance is $m{\sigma }^2$. Thus iterated expectation gives $\mathbb{E}[X_n]={\mu }^n$ from one ancestor, and the law of total variance gives $v_{n+1}={\mu }^2{v}_n+{\sigma }^2{\mu }^n$, $v_0=0$. Iterating yields the branching-process generation momentsFor the critical case $\mu =1$, the correct limiting value is $v_n=n{\sigma }^2$. If $\mu =0$, nonnegative offspring are zero almost surely, so every later generation has zero mean and variance; handle this directly rather than use an ambiguous zero power. At generation zero the mean is one and variance zero.

The branching-process extinction criterion states that extinction probability is the smallest fixed point of $G$ in $[0,1]$: probabilities of extinction by successive generations are the iterates of $G$ from zero, increasing to that fixed point. Here the equation is $4q^3-7q+3=0$, factoring as $(q-1)(2q-1)(2q+3)=0$. The two admissible roots are $1/2$ and one, so

Expand around the expectation $\mu$:Thus variance as the minimum mean squared error of a constant givesThe finite second moment also guarantees the finite first absolute moment needed below.

For a probability density function $f$, splitting at $a$ givesLet $F$ be the cumulative distribution function. Since a density gives no mass at a singleton, differentiating under this integral yields $H^{\prime }(a)=F(a)-(1-F(a))=2F(a)-1$. This derivative is nondecreasing, so $H$ is convex andThis is the median minimizes expected absolute loss property. The minimizer need not be unique: a gap in the density can give an interval of medians, all with the same loss.

For the normalized second-order equation, an ordinary point is one where both coefficient functions $p,q$ are analytic. A singular point of a second-order linear ODE is a point that is not ordinary. It is a regular singular point if $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ extend analytically to $x_0$; a singular point failing that condition is irregular.

Here $p=0$, $q=1/x$. Thus zero is singular but regular, because $xp=0$ and $x^{2}q=x$ are analytic. Write the analytic solution as $y={\sum }_{n\ge 0}{a}_n{x}^n$. Its constant equation gives $a_0=0$, and for $n\ge 1$,Using $a_1=1$ givesThe factorial denominators show convergence for every finite $x$.

For the other branch, integrating the first correction equation twice and imposing its two conditions gives $y_1^{\prime }=-\log x$ andThen $y_2^{\prime \prime }=\log x-1$, so $y_2^{\prime }=x\log x-2x+2$ andEach correction tends to zero at the origin and its derivative vanishes at one. Although the second solution has a finite limiting value one, it has a logarithmically divergent endpoint derivative. Indeed the differential equation gives $y^{\prime \prime }\sim -1/x$, and integration givesA finite initial derivative cannot be prescribed for this second branch. The logarithm is consistent with the integer separation of the two exponents in the Frobenius method; this is not the repeated-exponent case.

Retain the derivative marks and coefficients from the printed system. In matrix form it is $M\dot{u}+Ku=b(t)$, withSince $\det M=11$, multiplication by its inverse givesThe linear transformation $u=x+y$, $v=x-y$ diagonalizes this system:Both start at zero. An integrating factor gives $u=v=e^{-t}-e^{-3t}$, soDirect substitution into the two original equations confirms the different forcing coefficients; the exact cancellation in $y$ depends on those coefficients.

Let $\Delta ={\theta }_1-{\theta }_0>0$, with $\alpha ,{\beta }_{\max }>0$. Newton law of cooling gives the warming and cooling profilesIn the stipulated linear model, $\beta (t)={\beta }_{\max }(\theta (t)-{\theta }_0)/\Delta$. The bacterial count obeys $\dot{N}=-\beta (t)N$, so its logarithmic reduction is the cumulative thermal destruction under exponential warming and cooling:Writing $q=e^{-\alpha T}$, the warming integral divided by ${\beta }_{\max }$ is $T-(1-q)/\alpha$ and the cooling contribution is $(1-q{)}^2/\alpha$. Hence the required implicit equation isThe bracket is zero at zero and has derivative $(1-q)(1+2q)>0$ for $T>0$, tending to infinity with $T$. There is therefore one positive solution.

For the hardier species, $T$ is unchanged in this model if $\alpha$ and the achieved ${\beta }_{\max }$ remain the same. Raising the oven temperature changes $\Delta$, but its factor cancels against the changed slope of the linear destruction law. The entire normalized temperature history and hence the destruction integral remain the same; the comparison uses both the warming and equal-duration cooling stages.

For uniform convergence to $k$ on $A$, for every $\eta >0$ there is $N$ such that $|k_n(x)-k(x)|<\eta$ for all $n\ge N$ and all $x\in A$. The bound $|\sin (nx)/\sqrt{n}|\le 1/\sqrt{n}$ proves uniform convergence to zero, whereas its derivative is $\sqrt{n}\cos (nx)$. On an interval containing zero these derivatives diverge at zero, so they cannot converge uniformly. This supplies a counterexample to differentiating a uniformly convergent sequence without additional hypotheses.

For uniform derivative convergence with an anchored value, put $g=\lim k_n^{\prime }$ and $a=\lim k_n(x_0)$. A uniform limit of continuous functions is continuous. The fundamental theorem of calculus givesThus $k_n(x)\to k(x)=a+{\int }_{x_0}^{x}g(t)dt$, with error bounded byThis is uniform on bounded subintervals, and on all of $A$ if it is bounded. The fundamental theorem of calculus now gives $k^{\prime }=g$, so $k$ is continuously differentiable. For an unbounded $A$, no global uniform convergence of $k_n$ follows merely from the displayed hypotheses, but local uniform convergence and the claimed differentiability do.

For the series, use partial sums and differentiate their individual terms:For $|x|\le 1$, their absolute values are at most $2n/(n^3+1)$, a summable bound. The Weierstrass M-test makes the derivative series uniformly convergent, while every partial sum is zero at $x=0$. The anchored-value result therefore proves continuous differentiability on $(-1,1)$, with derivative equal to the displayed termwise derivative series.

For $z\ne 0$, the denominator zeroes are $z_n=1/[\pi (n+\frac{1}{2})]$. Its derivative is $\sin (1/z)/z^2$, nonzero at each such point. ThusThese poles accumulate at zero. Consequently zero is a non-isolated singularity, as described by reciprocal cosine with accumulating poles. It is not an isolated essential singularity: no punctured disc about zero is a domain of holomorphy for this reciprocal function. Although $\cos (1/z)$ itself has an essential singularity at zero, taking its reciprocal introduces this accumulation of poles.

At $z=1$, numerator and denominator both have simple zeroes. Their derivatives give the removable valueAt each positive integer $n\ge 2$, the numerator is nonzero and the denominator has a simple zero, so these are simple poles. At zero the expansion on the principal sheet iswhich contains a fractional power. Thus zero is a branch point with pole-like leading growth, not an ordinary isolated pole. The negative real axis is the chosen branch cut. On local continuations across its nonzero points, negative integers also give simple poles because the continued square root is $\pm i\sqrt{|n|}$, never one. They lie outside the open principal branch domain; noninteger negative points are merely cut points, not intrinsic singularities.