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A finite-dimensional commutative real algebra with a Euclidean multiplicative norm has dimension at most two. Automatic real bilinearity from a multiplicative norm supplies bilinearity if it was not part of the initial description. On its unit sphere, squaring has derivative $v\mapsto 2xv$, injective because there are no zero divisors. The only equal squares are antipodal pairs, so the map induces an injective immersion ${\mathbb{RP}}^{n-1}\to S^{n-1}$. Compactness and the inverse function theorem make it a diffeomorphism when $n\ge 2$. For $n\ge 3$, the fundamental groups $\mathbb{Z}/2$ and zero contradict this. The real and complex number algebras attain dimensions one and two.

If $r\ge \sqrt{2}$ and a convex polytope satisfies $r^{-1}{B}^n\subseteq P\subseteq B^n$, then it has at least $(1-r^{-2}{)}^{-n/2}\ge e^{n/(2r^2)}$ facets. Its outward facet normals define spherical caps covering the unit sphere; the spherical cap area upper bound supplies the estimate.

The interior consists exactly of strictly copositive matrices. If $x^{T}Ax$ has positive minimum $\delta$ on the compact nonnegative unit sphere, perturbations of operator norm less than $\delta$ preserve this positive lower bound there. Conversely, if a unit nonnegative vector has zero quadratic form, $A-\epsilon I$ leaves the copositive cone for every $\epsilon >0$. Thus such a matrix is not interior. The identity matrix is a convenient interior point.

The multiplicative norm first rules out zero divisors: if $xy=0$ and $x\ne 0$, then $0=|x||y|$ forces $y=0$. If the dimension is zero the claimed bound already holds, so suppose it is positive.

It is worth checking that the printed description of a ring with an underlying real vector space really permits differentiation of multiplication. This follows from the norm identity even if compatibility with real scalars is not separately stated. For fixed $x$, the map $L_x(y)=xy$ is additive by distributivity, andIt is therefore continuous. Additivity gives rational linearity. Approximating a real number by rationals and using continuity proves $L_x(ty)=t{L}_x(y)$ for every real $t$. The same argument holds in the other variable. This is automatic real bilinearity from a multiplicative norm, so multiplication is a bilinear map between finite-dimensional vector spaces, and hence is smooth.

Let $S=\{x\in A:|x|=1\}$ be the unit sphere of the given positive-definite bilinear form. It is a copy of $S^{n-1}$. The squaring mapis well defined because $|x^2|=|x{|}^2=1$, and is smooth. Commutativity gives its derivativeFor $x\in S$ and $v\ne 0$, the norm identity gives $|D{Q}_x(v)|=2|v|$, so the derivative is injective, including on the tangent space $T_{x}S=x^{\perp }$. It takes tangent vectors into $T_{x^2}S$ since $Q$ maps the sphere to itself. Alternatively, polarization of $|xy{|}^2=|x{|}^2|y{|}^2$ gives $⟨xy,xz⟩=|x{|}^2⟨y,z⟩$, so $⟨x^2,2xv⟩=2⟨x,v⟩=0$ directly.

The only identifications under squaring are antipodal:The factorization uses commutativity, and the last implication uses the absence of zero divisors. Conversely $Q(-x)=Q(x)$. It follows that $Q$ induces an injective smooth mapThe antipodal quotient map is a local diffeomorphism, so the injectivity of $DQ$ implies injectivity of $Dq$. Since Real projective space is compact, the stated compact injective immersion is an embedding theorem makes $q$ a closed embedding.

For $n\ge 2$ its source and target have the same dimension $n-1$. The inverse function theorem makes its image open as well as closed in the connected sphere $S^{n-1}$. It is nonempty, so the image is the whole sphere; consequently $q$ is a diffeomorphism. But for $n\ge 3$ this is impossible:Indeed, the antipodal two-sheeted covering map $S^{n-1}\to {\mathbb{RP}}^{n-1}$ has simply connected total space in this range, and is its universal cover, with deck group $\mathbb{Z}/2$. A sphere of dimension at least two is simply connected. These fundamental groups cannot be those of homeomorphic spaces.

We have obtained the dimension bound for a commutative Euclidean normed algebra:The real numbers and complex numbers with their usual products and Euclidean norms attain dimensions one and two. In dimension two, the induced map ${\mathbb{RP}}^1\to S^1$ is entirely compatible with the argument: the circle squaring map identifies exactly the two antipodal preimages. We neither assume a multiplicative identity nor use a classification of real division algebras; the squaring argument proves the bound from the supplied hypotheses.

Use $D_j=-i{\partial }_{x_j}$ and $⟨\xi ⟩=(1+|\xi {|}^2{)}^{1/2}$, so the Fourier transform of $P(D)u$ is $P(\xi )\hat{u}(\xi )$. Write $P_N$ for the homogeneous degree-$N$ part. An elliptic differential operator has $P_N(\xi )\ne 0$ for every real $\xi \ne 0$. By continuity on the unit sphere, $m={\min }_{|\omega |=1}|P_N(\omega )|>0$. The lower-degree terms are bounded by $C|\xi {|}^{N-1}$ for $|\xi |\ge 1$, whencefor sufficiently large $|\xi |$. This proves the high-frequency lower bound for an elliptic polynomial. Degree zero just means a nonzero constant and has no derivative gain.

With the Fourier convention in Question 3, the Sobolev space isThe condition includes that the weighted transform is represented by an $L^2$ function. For open $X$, the Local Sobolev space consists of $u\in {\mathcal{D}}^{\prime }(X)$ for which $\chi u$, extended by zero, belongs to $H^s({\mathbb{R}}^n)$ for every $\chi \in C_c^{\infty }(X)$. We use the elementary Sobolev multiplication by a smooth cutoff fact for every real $s$. It follows from Fourier convolution with the rapidly decreasing $\hat{\chi }$, the weighted inequality $⟨\xi {⟩}^s\lesssim ⟨\xi -\eta {⟩}^{|s|}⟨\eta {⟩}^s$, and the $L^1\ast L^2\to L^2$ convolution bound.

For a compactly supported distribution, continuity on test functions supported in a fixed compact neighborhood gives finite order: for some integer $M$,Insert a compact smooth cutoff function equal to one near the support, times $e^{-ix\cdot \xi }$. The resulting Fourier transform is smooth and satisfies $|\hat{u}(\xi )|\le C^{\prime }⟨\xi {⟩}^M$. The distribution is also tempered by this same finite-order bound. ThusIndeed the square of the weighted bound is integrable exactly when $2(t+M)<-n$. This is negative Sobolev regularity of a compactly supported distribution.

We next prove local regularity by a high-frequency reciprocal parametrix kernel, including its off-diagonal smoothing property. Choose $\psi \in C_c^{\infty }({\mathbb{R}}^n)$ equal to one on a ball containing all real zeros of $P$, and put $b(\xi )=(1-\psi (\xi ))/P(\xi )$, defining it smoothly as zero in the inner ball. Differentiation of the reciprocal and the elliptic lower bound giveThe Fourier multiplier $E=b(D)$ maps $H^s$ to $H^{s+N}$ by its zeroth-order bound, andThe kernel $K={\mathcal{F}}^{-1}b$ is smooth away from the origin: for $x\ne 0$, repeatedly integrate by parts using $e^{ix\cdot \xi }=(i|x{|}^2{)}^{-1}x\cdot {\partial }_{\xi }{e}^{ix\cdot \xi }$. After sufficiently many integrations, the differentiated symbol is integrable. For any desired $x$ derivative, repeat the argument with the additional polynomial ${\xi }^{\beta }$. A large-radius cutoff function justifies each step and its removal uniformly on compact sets away from zero. Thus convolution by $K$ carries a compactly supported distribution to a smooth function at points separated from its support. Also $R$ carries a compactly supported distribution to a Schwartz function, because $\psi \hat{u}$ is smooth with compact support.

For a target compact set in $X$, choose $\chi \in C_c^{\infty }(X)$ equal to one on a neighborhood of it. Then $\chi u$ and $[P(D),\chi ]u$ have compact support, and the commutator is supported where derivatives of $\chi$ occur, away from the target. The parametrix identity givesThe first term is in $H^{s+N}$ since $\chi P(D)u\in H^s$. The other two terms are smooth near the target by the proved kernel property. Since the target was arbitrary,This proof does not discard the cutoff function commutator; it places its support away from the set where regularity is sought.

For the final polynomial, select a multi-index ${\alpha }_0$ with $|{\alpha }_0|=N$ and ${\partial }^{{\alpha }_0}Q$ a nonzero constant. The derivative-ratio hypothesis immediately gives $|Q(\xi )|\ge c|\xi {|}^{\delta N}$ at large frequency. In particular $Q$ has no real zero there. For $N>0$, its degree bound also forces $\delta \le 1$. Differentiating $1/Q$ gives products of ratios ${\partial }^{\beta }Q/Q$, whose total derivative order is $|\alpha |$. Hence the corresponding high-frequency reciprocal satisfiesIts multiplier maps $H^s$ to $H^{s+\delta N}$. Its kernel is still smooth off zero: each frequency integration by parts now lowers the order by $\delta$, so more iterations may be needed, but $\delta >0$ supplies arbitrarily much decay. The same separated-support commutator identity applies without a change. The derivative-ratio Sobolev gain for a polynomial operator is thereforeSimply estimating the commutator by its differential order would lose this sharper gain; the off-diagonal kernel argument uses the full derivative-ratio hypothesis.

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.

An oscillatory integral defines a distribution by cancellation, even when its oscillatory integral amplitude is not integrable in frequency. We use symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)=S_{1,0}^N(X\times {\mathbb{R}}^k)$, where $X\subset {\mathbb{R}}^d$ is open and $N\in \mathbb{R}$. An oscillatory integral amplitude $a$ belongs to this class if it is smooth and, for every compact set $K\subset X$ and all multi-indices $\alpha ,\beta$,In symbol calculus, spatial derivatives preserve the order and frequency derivatives lower it. The fixed number $N$ is independent of $K$; this will produce a finite global order of a distribution, although continuity constants may depend on $K$.

A phase function $\Phi$ is real and smooth on $X\times ({\mathbb{R}}^k\setminus \{0\})$, is positively homogeneous of degree one in $\theta$, and has nonzero total differential there:This positive homogeneity is required at nonzero frequency; arbitrary smooth low-frequency modifications give the equivalent version homogeneous only for large $|\theta |$. In particular, smoothness at zero is not an additional requirement on a general homogeneous phase function. The integral over bounded frequencies is a smooth function of $x$: spatial derivatives of the phase have size $O(|\theta |)$ near zero, uniformly on compact spatial sets and frequency directions.

Choose a cutoff function $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero. The proposed meaning of the oscillatory integral distribution isExistence and cutoff independence of an oscillatory integral require a proof. Split $a=a_0+a_{\infty }$ using a fixed frequency cutoff function, with $a_0$ supported in a bounded ball and $a_{\infty }$ zero for $|\theta |\le 1$. The low-frequency part is already absolutely integrable. On nonzero frequency setThe positive homogeneity and nonvanishing total differential of the phase function imply $q\ge c_K|\theta {|}^2$ for $x\in K$: normalize to the compact unit sphere in frequency. Direct differentiation gives $L{e}^{i\Phi }=e^{i\Phi }$.

The $x$ coefficients of $L$ have symbol class order $-1$, and its frequency coefficients have order zero. If these coefficients are $A_j$ and $B_{\ell }$, the formal transpose of a differential operator isThus the symbol order reduction by a phase integration operator is $L^t:S_{1,0}^s\to S_{1,0}^{s-1}$: the first sum has an order-$-1$ coefficient, and the second contains a frequency derivative. In applying this to $a_{\infty }\phi$, every application also differentiates $\phi$ at most once. Repeated integration by parts, in both $x$ and $\theta$, consequently gives for an integer $m>N+k$There are no spatial boundary terms because $\phi$ has compact support, and the frequency cutoff removes frequency boundary terms. Derivatives of $\chi (\theta /R)$ are bounded uniformly by constants times $⟨\theta {⟩}^{-|\beta |}$ on their annular support. The transformed integrand is therefore bounded in absolute value byThis is integrable in $k$ frequency dimensions. Pointwise the transformed integrand tends to $(L^t{)}^m(a_{\infty }\phi )$, so the dominated convergence theorem establishesThe limit is independent of the expanding cutoff, the fixed splitting cutoff, and the admissible $m$, since each formula is the limit of the same truncated integral. Derivatives landing on the expanding cutoff can also be estimated directly by $O(R^{N-m+k})$, which tends to zero. In particular,This proves continuity on the space of test functions and the finite order of an oscillatory integral distribution. The same $m$ works for all compact sets. When $N<-k$, $m=0$ is possible and the original frequency integral is absolutely integrable; cancellation is needed for general $N$.

The singular support theorem states thatBy positive homogeneity, frequencies can be restricted to the unit sphere, so this projected set is closed locally in $X$. More precisely, the stationary-direction bound for singular support restricts the frequency directions to the closed conic support of an oscillatory amplitude; using a closed directional support avoids losing limits occurring at arbitrarily high frequency. The simpler displayed bound is sufficient here.

To prove the bound, take $x_0$ outside the displayed stationary set. On a sufficiently small spatial neighborhood and all unit frequency directions, $|{\nabla }_{\theta }\Phi |$ is bounded below. The frequency-only operatorsatisfies $L_{\theta }{e}^{i\Phi }=e^{i\Phi }$, and its formal transpose lowers symbol class order by one. A spatial derivative of order $|\alpha |$ of $e^{i\Phi }a$ has amplitude order at most $N+|\alpha |$. Applying the frequency integration by parts more than $N+|\alpha |+k$ times makes that derivative absolutely integrable, uniformly on smaller compact sets. Every spatial derivative therefore exists and is continuous there. The oscillatory integral distribution is a smooth function near $x_0$, proving the singular support assertion.

For the linear transport equation, use spacetime $X={\mathbb{R}}^n\times (0,\infty )$, frequency $\theta \in {\mathbb{R}}^n$, andThis is a valid phase function, because ${\nabla }_x\Phi =\theta \ne 0$ at nonzero frequency; the oscillatory integral amplitude is in symbol class order zero. The Fourier representation of the Dirac delta function gives the transport of a Dirac point mass:Its rigorous spacetime distribution pairing is $⟨u,\phi ⟩={\int }_0^{\infty }\phi (ct,t)dt$. Consequentlywhere the endpoints vanish since a spacetime test function has compact support in $t>0$. At each fixed $t$, $⟨u(t),\psi ⟩=\psi (ct)$, so weak convergence of distributions gives $u(t)\to {\delta }_0$ as $t\downarrow 0$. This is the required initial trace.

Finally, ${\nabla }_{\theta }\Phi =x-ct$, so the singular support theorem confines singularities to $x=ct$. In fact equality holds: $u$ is a nonzero order-zero distribution on that trajectory and vanishes off it. A smooth function supported on this set of empty interior must vanish, so $u$ cannot be smooth in any neighborhood of a point of the trajectory. ThusThe method of characteristics also proves uniqueness among solutions with a distributional initial trace: the change of variables $y=x-ct$ turns the linear transport equation into ${\partial }_{t}w=0$. Such a distribution is constant in $t$; pairing with spatial test functions reduces this assertion to a distribution with zero derivative is constant. Its initial trace fixes $w={\delta }_0$, so the moving Dirac delta distribution above is the unique solution. There is transport of the singularity along the characteristic and no smoothing.

The relevant fixed coordinate subspace is $\{(u,0):u\in {\mathbb{R}}^k\}$. Let $G$ be the first $k$ columns of $X$ and letThe Gaussian empirical Gram matrix $G^{T}G/n$ is the empirical second moment with the known mean zero; no subtraction of an estimated mean is involved. For $\theta =(u,0)$, the ratio under consideration is $|u^{T}Du|/\Vert u{\Vert }_2^2$. Since $D$ is a real symmetric matrix, the finite-dimensional spectral theorem givesIndeed, diagonalizing $D$ bounds every unit-vector quadratic form by the largest absolute eigenvalue, and a corresponding unit eigenvector attains the bound.

We first construct a unit sphere net from ball covering. Enlarge the numerical covering constant, if necessary, to $A_0\ge 1$. Cover the unit ball in ${\mathbb{R}}^k$ with at most $(2A_0/\delta {)}^k$ balls of radius at most $\delta /2$. For each such ball meeting the unit sphere, choose a point of the sphere in it and discard the others. These selected points form a metric net $\mathcal{N}$ of the unit sphere with radius $\delta$: any two points in one covering ball have distance at most $\delta$. This argument ensures that the net points have unit length even if the original covering centers did not.

Take $\delta =1/4$ and put $B=8A_0$, so $|\mathcal{N}|\le B^k$. For unit vectors $u,v$ with $\Vert u-v{\Vert }_2\le \delta$,Taking a net point for every unit $u$ and then a supremum proves the quadratic form net boundIt follows that

Fix $v\in \mathcal{N}$. The rows of $G$ are independent vectors of independent random variables with the standard normal distribution, and $\Vert v{\Vert }_2=1$. Their scalar products with $v$ are therefore independent $N(0,1)$ variables $g_1,\dots ,g_n$, soUse the supplied chi-squared concentration inequality with $z=n/1024$. Its threshold isConsequently $P(|v^{T}Dv|>1/4)\le 2e^{-n/1024}$. The union bound gives the stronger fixed-subspace estimateThis is Gaussian Gram matrix concentration on a fixed subspace.

Since $1\le k<p$, we have $p\ge 2$. Under $n\ge Ck\log p$,Choose the numerical constant $C\ge 1024(1+\log B/\log 2)$. Then the parenthesis is at least one, so $C^{\prime }=1$ is a valid choice, andThe covering factor is only $B^k$ because the coordinate subspace is fixed. The ambient dimension enters through the assumed sample-size bound.

Let $m$ be the number of facets of the convex polytope $P$. Its facet description can be writtenBecause $r^{-1}{B}^n\subset P$, each supporting closed half-space has $h_j\ge 1/r$. For any $\theta \in S^{n-1}$, follow the ray $t\theta$ to its last point in $P$. Its radius $t_{\theta }$ is at most one because $P\subset B^n$. At least one facet is active there, givingThus the unit sphere is covered by the $m$ spherical caps centred at $u_j$ with angular radius $\rho =\arccos (1/r)$. Since $r\ge \sqrt{2}$, this radius lies in $[\pi /4,\pi /2)$ and the spherical cap area upper bound applies. Subadditivity of surface area givesUsing $\log (1-x)\le -x$ for $0<x<1$,The facet lower bound for ball approximations is exponential in the dimension.

Choose a maximal subset $U\subset S^{n-1}$ whose distinct points have Euclidean distance greater than one. The volumetric bound for Euclidean metric nets makes the construction finite: the open Euclidean balls of radius $1/2$ about the points of $U$ are disjoint and all lie in $(3/2)B^n$, soMaximality means that $U$ is a metric net of radius one on the unit sphere. Thus every $\theta \in S^{n-1}$ has some $u\in U$ with $\Vert u-\theta \Vert \le 1$. Since both are unit vectors,Now intersect the corresponding closed half-spaces:The Cauchy-Schwarz inequality shows that $(1/2)B^n\subset P$. Conversely, write any nonzero $x\in P$ as $x=t\theta$ and choose $u$ as above. Then $t/2\le t(u\cdot \theta )=u\cdot x\le 1/2$, so $t\le 1$. Hence $P\subset B^n$, which also proves boundedness. This convex polytope has at most $|U|\le 3^n$ facets, since redundant inequalities can only reduce their number. It meets the required inclusions with $r=2$.

A pointed cone has no nonzero line through the origin: equivalently $K\cap (-K)=\{0\}$. If $A\in K\cap (-K)$, its quadratic form vanishes on the nonnegative orthant. Testing the coordinate vectors gives $A_{ii}=0$. Testing $e_i+e_j$ then givesSince $A$ is a symmetric matrix, every entry is zero. Thus $\boxed{K\cap (-K)=\{0\}}$.

The identity matrix gives an interior point. For a symmetric perturbation $E$ with operator norm $\Vert E{\Vert }_2<1$, the Cauchy-Schwarz inequality givesThis open norm ball around $I$ lies even in the positive semidefinite cone, hence in $K$. Consequently $\boxed{I\in \mathrm{int}K}$ and the interior is nonempty.

More generally, the interior of the copositive cone consists exactly of strictly copositive matrices. Positivity on the compact nonnegative unit sphere has a positive minimum and persists under small perturbations; a zero there is destroyed by an arbitrarily small negative multiple of $I$.

Write $n=2m-1$. On ${\mathbb{R}}^{2m}$ define the orthogonal linear mapIt satisfies $J^T=-J$ and $J^{T}J=I$. Hence $⟨x,Jx⟩=0$ and $\Vert Jx\Vert =\Vert x\Vert$. Restricting to the unit sphere givesThis continuous unit tangent field is available exactly because the ambient dimension is even.

Suppose such an $f$ exists. Orthogonality and unit lengths makea continuous map into the unit sphere, since its squared norm is one. This is a homotopy from the identity to the antipodal map. Their degrees would coincide, but for even $n\ge 2$ they are respectively $1$ and $(-1{)}^{n+1}=-1$, a contradiction. Thus no such continuous map exists for even $n$. At $n=0$ there is not even a unit vector orthogonal to either point of $S^0\subset \mathbb{R}$, so nonexistence holds there as well. This is a sphere form of the Hairy ball theorem.