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All the transformations below fix the origin. A planar rotation preserves lengths and angles and turns every vector through one common oriented angle. Its rotation matrix isAn orthogonal reflection in a line fixes that line and reverses its perpendicular direction; reflection in the horizontal axis has reflection matrix $\mathrm{diag}(1,-1)$. A uniform dilation multiplies every vector by the same positive scale $s$, represented by $sI$; for example $2I$ doubles lengths. A shear mapping fixes one line pointwise and shifts vectors parallel to that line by an amount proportional to their transverse coordinate. A horizontal shear mapping has matrixwith $S_1$ a nontrivial example.

The first given matrix is the clockwise planar rotation through $\pi /4$, namely $A=R_{-\pi /4}$. Direct multiplication givesThus $C$ is a horizontal shear mapping of strength $2$. Because $A^{-1}=R_{\pi /4}$,So $B$ is the composition of that shear mapping followed by an anticlockwise planar rotation through $\pi /4$; order matters. It is not individually one of the four elementary types under these definitions. In particular, its determinant is $1$, but it is not an orthogonal matrix, excluding both a planar rotation and an orthogonal reflection. It is not a scalar matrix, excluding a uniform dilation; its eigenvalues $\sqrt{2}+1$ and $\sqrt{2}-1$ exclude a pure shear mapping, whose eigenvalues are both $1$. As a symmetric positive-definite matrix, $B$ can also be described as stretching two perpendicular principal directions by these reciprocal factors, preserving area.

Completing squares reveals the geometry of the sheared quartic double-well gradient flow:The critical points satisfy $2(x+y)=0$ and $4x^3-2x+2y=0$, so they are $(0,0)$, $(1,-1)$ and $(-1,1)$. The Hessian matrix isAt the origin its determinant is $-8$, so $(0,0)$ is a saddle point, with $V=0$. At either other critical point, the determinant is $16$ and the upper-left entry is $10$, so it is a positive-definite matrix. They are strict local minima; the completed squares show that both are global minima, with $V=-1$.

A contour at level $c$ obeysThere are no contours for $c<-1$, two isolated minimum points for $c=-1$, and two separate closed ovals for $-1<c<0$. At $c=0$ the two lobes meet at the saddle point; near the origin their tangents are $y=(-1\pm \sqrt{2})x$. For $c>0$ one closed contour surrounds both minima.

Along the gradient flow, the gradient-flow dissipation identity isAt the stated initial point, $V(1,-1/2)=-3/4$. Its trajectory stays in the invariant sublevel set $V\le -3/4$. This set is compact because $V$ is a coercive function. It also excludes every point with $x=0$, where $V=y^2\ge 0$; continuity therefore keeps the trajectory in its positive-$x$ component.

To justify its limit rather than merely infer it from the sketch, integrate the dissipation identity: ${\int }_0^{\infty }\Vert \nabla V{\Vert }^{2}dt$ is finite. On this compact set the gradient and its time derivative are bounded, so $\Vert \nabla V{\Vert }^2$ is uniformly continuous. A nonnegative uniformly continuous function with finite integral tends to zero: otherwise separated intervals of a fixed positive height and width would force an infinite integral. Every accumulation point is consequently a critical point. In the positive-$x$ component below this energy level the only one is $(1,-1)$. Compactness then implies convergence, andThis is a compact gradient-flow trapping criterion; the drawn trajectory illustrates, rather than replaces, the convergence proof.

The autocovariance of an MA(1) process comes directly from shared noise terms:Work with the zero-mean white noise convention and ${\sigma }^2>0$. Put ${\mathcal{H}}_{t-1}=\mathrm{span}(X_1,\dots ,X_{t-1})$. Suppose the preceding finite-past innovations $U_1,\dots ,U_{t-1}$ are mutually orthogonal. Their triangular relation to the observations means that they span the same finite past. For $s\le t-2$, $U_s$ is a linear combination of $X_1,\dots ,X_s$, so $\mathrm{Cov}(X_t,U_s)=0$. Moreover,Projection onto this orthogonal basis therefore has only its last coefficient nonzero:The residual is orthogonal to the entire preceding span, proving the induction. Expanding its variance givesThe initialization is $U_1=X_1$, ${\phi }_1={\gamma }_0$, ${\lambda }_1=0$ and $U_0=0$. Each finite covariance matrix is a positive-definite matrix: the last observation in any nonzero finite linear combination contains a noise term absent from earlier observations. Thus all the projection denominators are positive. These are the finite-sample innovations of an MA(1) process, requiring no Gaussian assumption.

Substituting the covariances into the recursion givesThe denominator has a positive limit when ${\lambda }_{t-1}\to \lambda \in [-1,1]$. Taking limits gives $(\lambda -\theta )(\theta \lambda -1)=0$. The permitted interval selectsFor $\theta =0$, every coefficient is zero; at $\theta =\pm 1$, the roots coincide. This is the limiting MA(1) innovations coefficient. The limiting innovation variance is correspondingly ${\sigma }^2\max (1,{\theta }^2)$.

For the observed numerical case, ${\gamma }_0=2$ and ${\gamma }_1=-1$. The exact innovation calculations areTherefore the predictor of $X_3$ based on the two observations and its error variance areFor $X_4$, the available information is still only ${\mathcal{H}}_2$. Since its covariances with both $X_1$ and $X_2$ are zero, its orthogonal projection onto that span is zero:This is two-step prediction for an MA(1) process. Applying the next one-step recursion would instead use an observed $X_3$, which is not available here.

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.