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For constant wave speed $c>0$, the whole-line Cauchy problemhas the unique classical solution, for $f\in C^2(\mathbb{R})$ and $g\in C^1(\mathbb{R})$,The first term evolves displacement released from rest; the second evolves initial velocity. The factor $1/(2c)$ converts an integral of velocity over distance into displacement. On a finite string the formula applies before the backward characteristic curves reach an endpoint, or globally after an extension that enforces the boundary conditions. ^[ref]

For a completely smooth example, choose a bump function with compact support:where $a>0$ sets the half-width and $A$ sets the initial peak. ThenAt $t=0$ the two copies coincide, recovering peak $A$. At $t=a/(2c)$, $u(0,t)=A{e}^{-1/3}\simeq 0.716531A$. At $t=3a/(2c)$ the pulses are separated, centred at $x=\pm 3a/2$, each of width $2a$ and peak $A/2$. Zero initial velocity forces equal left and right contributions; a purely right-moving initial profile instead requires $g=-c{f}^{\prime }$. Thus “released from rest” and “launched to the right” describe different Cauchy data.

The domain of dependence of $(x_{\ast },t_{\ast })$, for $t_{\ast }\ge 0$, is contained in $[x_{\ast }-c{t}_{\ast },x_{\ast }+c{t}_{\ast }]$ on the initial line. More precisely, displacement contributes only through the two endpoints, while initial velocity contributes throughout the interval. With $c=1$, $x_{\ast }=0.5$ and $t_{\ast }=1.5$, that interval is $[-1,2]$. Changing either initial datum outside it cannot change $u(0.5,1.5)$. If both initial data vanish outside $[-a,a]$, the solution vanishes outside $[-a-ct,a+ct]$: this is finite propagation speed. The expanding interval bounds possible influence; it need not be filled by nonzero displacement, as the separated bumps demonstrate.

Historically, d'Alembert studied the vibrating-string problem in “Recherches sur la courbe que forme une corde tendue mise en vibration”, in the Berlin Academy's volume for the year 1747, pages 214–219. The memoir introduces the differential equation and integrates it through two arbitrary one-variable functions. The original geometric problem makes the formula's enduring appeal clear: arbitrary initial shapes can be handled without first decomposing them into normal modes. ^[ref]

For a brief derivation, use characteristic coordinates $\xi =x-ct$ and $\eta =x+ct$. The chain rule gives $u_{tt}-c^2{u}_{xx}=-4c^2{u}_{\xi \eta }$, so $u=F(\xi )+G(\eta )$. The initial conditions yield $F+G=f$ and $-c{F}^{\prime }+c{G}^{\prime }=g$. Integrating these relations gives the displayed formula; the arbitrary integration constants cancel in their sum.

For the unit ball of $C^p({\mathbb{T}}^n)$ measured in the uniform metric, $c{\epsilon }^{-n/p}\le H(\epsilon )\le C{\epsilon }^{-n/p}\log (1/\epsilon )$. The lower bound uses disjoint bump functions of spatial scale $h$, height proportional to $h^p$ and independently chosen zero/one coefficients: there are $\exp (c{h}^{-n})$ functions separated at scale $h^p$. For the upper bound, Multivariable Jackson approximation gives a trigonometric polynomial of degree $O({\epsilon }^{-1/p})$. Its $O({\epsilon }^{-n/p})$ coefficients are bounded and can be rounded with coefficient accuracy proportional to $\epsilon$ divided by the coefficient count. Counting those rounded coefficient arrays supplies the logarithmic factor. The continuous-function unit ball without derivative bounds has infinite entropy for $\epsilon <1/2$, by an infinite sequence of disjoint height-one bumps.

For the unforced, constant-$c$ one-dimensional wave equation, finite propagation speed says that changing Cauchy data outside $[x_{\ast }-c{t}_{\ast },x_{\ast }+c{t}_{\ast }]$ cannot change the solution at $(x_{\ast },t_{\ast })$, where $c>0$ and $t_{\ast }\ge 0$. This interval is the initial-data domain of dependence. Its endpoints are found by following the two characteristic lines for the one-dimensional wave equation backwards from the observation. D'Alembert's formula makes the dependence sharper: initial displacement is sampled at the endpoints, while initial velocity is integrated over the entire interval.^[ref]

For the synthetic choice $c=1$, $(x_{\ast },t_{\ast })=(0.5,1.5)$ has initial-data domain of dependence $[-1,2]$. A change confined near $x=3$ is irrelevant to that observation, however large it is. If both Cauchy data vanish outside $[-a,a]$, their compact support can influence only $[-a-ct,a+ct]$ at $t\ge 0$. Thus an observer at $x>a$ sees exactly zero displacement for $0\le t<(x-a)/c$. There is no nonzero precursor beyond this causal cone for the constant-speed wave equation in the ideal model. This is an exact statement about the partial differential equation, stronger than an experimental failure to detect a small signal.

Support widening for the wave equation is a bound on possible influence, not a claim that every point inside the expanding interval moves. A bump function released with zero initial velocity splits into two separated travelling waves; the middle can become empty. An initial kick can instead leave an interior tail, as in the D'Alembert formula with initial velocity. Consequently finite propagation speed does not imply the Strong Huygens principle. A finite string needs boundary conditions, and the whole-line formula applies locally only until its backward causal cone for the constant-speed wave equation encounters an endpoint.

Compactly supported Hamiltonian vector fields can attain any prescribed tangent vector at a point of a symplectic manifold: prescribe the derivative of a local Hamiltonian and multiply by a bump function. Composing flows for a tangent basis gives an open orbit by the inverse function theorem. Every orbit is open, so on a connected manifold the compactly supported Hamiltonian transformations act transitively.

For a nonnegative smooth bump function supported in an open set, multiplying a Riemannian metric by $1+t\eta$ increases volume there for $t>0$ while leaving the metric elsewhere fixed. For distinct regular compact domains a perturbation supported in the interior of one outside the other breaks equal volume, and therefore breaks any isometry between the domains. The perturbations tend to the original metric in the smooth topology.

For a bounded Fourier multiplier operator $T$ on $L^2$ and a bump function $\varphi$, the displayed convolution identity follows by multiplying Fourier transforms. The convergence holds in $L^2$ and almost everywhere by the approximate identity theorem, after accounting for the integral of the kernel. The limit is $T(f)$ when $\int \varphi =1$. Smoothness and compact support alone do not imply this normalization.

On the whole line, the homogeneous one-dimensional wave equation with constant wave speed $c>0$ isHere $u$ is displacement in the string interpretation; $f$ is initial displacement and $g$ is initial velocity. Its two directions of travel giveThis is D'Alembert's formula, including both Cauchy data. The countertravelling decomposition of the one-dimensional wave equation identifies $F(x-ct)$ as the right-moving travelling wave and $G(x+ct)$ as the left-moving travelling wave. Their sum is permitted by the superposition principle. For $f\in C^2$ and $g\in C^1$, it is a classical solution. More generally, travelling waves with profiles $F$ and $G$ that are locally integrable functions solve the equation as distributions, with derivatives interpreted distributionally.^[ref]

The synthetic diagram uses the smooth bump function $b$ defined in D'Alembert's formula, with $f(x)=b(x)$ and $c=1$. Releasing this displacement from rest, $g=0$, gives $F=G=f/2$. Each emerging travelling wave has half the original peak. Launching the identical initial displacement solely to the right instead requires $g=-c{f}^{\prime }$ and gives $F=f$, $G=0$. At $t=1.5$, the released profiles have peaks $1/2$ at $x=\pm 1.5$, while the launched profile has peak $1$ at $x=1.5$. These are calculated model values. A purely left-moving launch requires $g=c{f}^{\prime }$. The initial velocity therefore selects the balance of directions; the initial shape alone does not.

Use the Fourier transform convention $\hat{h}(\xi )=\int h(x)e^{-2\pi ix\cdot \xi }dx$. The deleted-ball kernel need not be integrable at infinity, so first work with finite annuli. WriteWe will prove a uniform bound, convergence as $R\to \infty$, and then convergence as $\epsilon \downarrow 0$. This makes all Fourier-transform assertions precise for the annularly cancelling singular convolution kernel.

Fix $\xi \ne 0$ and put $y=\xi /(2|\xi {|}^2)$, so $e^{-2\pi iy\cdot \xi }=-1$. For $R>r\ge 4|y|$, let $\Omega =\{r<|x|<R\}$ and defineThe Hörmander integral kernel condition givesOn changing variables $u=x-y$, the phase factor shows $J_{r,R}=-{\int }_{\Omega -y}K(u)e^{-2\pi iu\cdot \xi }du$. The symmetric difference between $\Omega$ and $\Omega -y$ lies in the two shells whose radii are within $|y|$ of $r$ and $R$. On each such shell, the kernel's size bound gives, by polar integration,Thereforeand adding the two comparisons proves the oscillatory tail estimate for a singular kernel:The integral on the right is at most $B$. Moreover, for fixed $\xi$, it tends to zero as $r\to \infty$ because the translated difference is absolutely integrable outside $2|y|$. The shell term also tends to zero. Thus the far Fourier integral is Cauchy and converges as its outer radius tends to infinity.

Set $r_0=4|y|=2/|\xi |$. If $\epsilon <r_0$, annular cancellation allows subtraction of the constant phase on the near annulus:The same estimate applies to a shorter near annulus when $R<r_0$. The far annulus is bounded by $B/2+C_{d}A$, using the tail estimate; if $\epsilon \ge r_0$, apply that estimate with $r=\epsilon$. At $\xi =0$ the annular integral vanishes by cancellation. We have provedFor each fixed $\epsilon >0$, the size bound makes $K_{\epsilon }\in L^2$, and $K{1}_{\{\epsilon <|x|<R\}}\to K_{\epsilon }$ in $L^2$ as $R\to \infty$. By the Plancherel theorem, their Fourier transforms converge in $L^2$ to ${\hat{K}}_{\epsilon }$. The pointwise improper limits just established identify this transform, up to a null set, withFor example identification follows by selecting an almost-everywhere convergent subsequence from the $L^2$ convergence. HenceThe near-annulus integral with the constant phase subtracted converges absolutely at zero, since its bound is proportional to $|x{|}^{1-d}$. The far integral has already converged. Therefore $m_{\epsilon }(\xi )\to m(\xi )$ for every $\xi \ne 0$, with the same uniform bound; the value at zero may be set arbitrarily.

For $f\in L^2$, the convolution $K_{\epsilon }\ast f$ is absolutely convergent for each $x$ by Cauchy-Schwarz inequality. Its Fourier transform is $m_{\epsilon }\hat{f}$: first prove this for Schwartz functions and then approximate any $L^2$ input. The convolutions converge uniformly under this approximation by the same Cauchy-Schwarz bound, and the multiplier outputs converge in $L^2$ by their uniform operator bound, so these two definitions agree. Define the bounded Fourier multiplier operator $T$ by $\widehat{Tf}=m\hat{f}$. Dominated convergence theorem and the Plancherel theorem now give

For the final request, use a normalized bump function, $\int \varphi =1$, so that its dilates form an approximate identity. Multiplying Fourier transforms givesThese products are integrable by Cauchy-Schwarz inequality, since $T{\varphi }_{\epsilon },f\in L^2$, and also define $L^2$ functions because $m$ and $\hat{\varphi }$ are bounded. Thus the convolution identity is justified even though $T{\varphi }_{\epsilon }$ need not be in $L^1$:The allowed standard approximate identity results, applied to $Tf\in L^2$, prove convergence to $Tf$ in norm and almost everywhere. This is the mass normalization in mollified singular integrals; it does not require proving almost-everywhere convergence of the sharp truncations $T_{\epsilon }f$.

If “bump function” means merely smooth with compact support, the stated limit needs the unit-integral qualification. For arbitrary such $\varphi$, the limit is $(\int \varphi )Tf$ instead. This is also the almost-everywhere limit: at each Lebesgue point of $Tf$, subtract $(\int \varphi )Tf(x)$ inside the convolution and bound the remaining local average by the boundedness and compact support of $\varphi$. As a concrete counterexample without normalization, take $K(x)=1/(\pi x)$ in dimension one and $\varphi =2\rho$ for a unit-integral mollifier $\rho$. This kernel has the required size, odd annular cancellation, and translation-difference integral $\log (3)/\pi$ on $\{|x|>2|y|\}$ for every $y\ne 0$. Its limit operator is the nonzero Hilbert transform; by the Hilbert-transform Fourier multiplier, it is an $L^2$ isometry. For any nonzero $f\in L^2$ the mollified limit is $2Tf$, not $Tf$.

Let $Mf$ be the uncentered maximal function of $|f|$, defined by its averages over balls. First establish the strong $L^p$ estimate needed below. The uncentered maximal weak-type inequality gives the weak $L^1$ bound with $C_d=3^d$, and plainly $Mf\le \Vert f{\Vert }_{\infty }$ for bounded inputs. For $f\in L^p$, split it into its parts on $\{|f|>a/2\}$ and $\{|f|\le a/2\}$. The first part is in $L^1$, since $|f|1_{\{|f|>a/2\}}\le (a/2{)}^{1-p}|f{|}^p$, and the second part contributes at most $a/2$ to the maximal function. Sublinearity therefore givesIntegrating this bound with layer cake representation and Tonelli theorem yieldsThis proves the strong Lp bound from weak L1 and L-infinity bounds directly. In particular $Mf$ is finite almost everywhere.

Now split the positive Riesz potential of $|f|$ at a radius $r$. In the near region, decompose into annuli $2^{-j-1}r<|x-y|\le 2^{-j}r$. On the $j$th annulus,The geometric series converges because $\alpha >0$. In the far region, Hölder's inequality, with $p^{\prime }=p/(p-1)$, givesThe radial integral is finite exactly because $p<d/\alpha$. Combining these bounds, including the fixed Riesz normalization constant, givesFor $0<Mf(x)<\infty$, choose $r=(\Vert f{\Vert }_p/Mf(x){)}^{p/d}$. If $Mf(x)=0$, every ball average vanishes and $f=0$ almost everywhere, so the assertion is trivial. The zero-norm case is likewise trivial. The resulting Hedberg inequality isThis proves absolute convergence of the defining integral for almost every $x$, also for signed or complex $f$. Put $\theta =\alpha p/d\in (0,1)$. The proposed exponent satisfies $q=p/(1-\theta )$, so $(1-\theta )q=p$. Raising the pointwise estimate to the $q$th power and using the proved maximal bound givesThus the Hardy–Littlewood–Sobolev inequality follows from the estimates just derived, rather than being quoted in place of its proof.

For uniqueness of the target exponent, suppose such a uniform estimate holds with index $r$ instead. Choose a nonzero nonnegative bump function $f$ and let $f_t(x)=f(tx)$ for $t>0$. Homogeneity of the Riesz potential givesThe assumed estimate at $t=1$ makes the nonzero output's $L^r$ norm finite. For all $t>0$ it must therefore satisfywith $d/r=0$ when $r=\infty$. Letting $t$ tend to zero or infinity forces the power of $t$ to vanish:Hence no other target index is possible.

For the Sobolev embedding, take $d>1$ and $1<p<d$, so the preceding result is available with $\alpha =1$. For $u\in C_c^{\infty }({\mathbb{R}}^d)$, the vector field $(x-y)/|x-y{|}^d$ has zero divergence in $y$ away from $x$. Integrating by parts outside the ball $|x-y|\le \epsilon$, its inner boundary contributes $|{\mathbb{S}}^{d-1}|$ times the spherical average of $u$ about $x$. The outer boundary contributes zero because $u$ has compact support. Letting $\epsilon \downarrow 0$ provesThe absolute value is at most $C{I}_1(|\nabla u|)(x)$. ThusFor a general member of the first-order Sobolev space $W^{1,p}({\mathbb{R}}^d)=L_1^p({\mathbb{R}}^d)$, approximate in the Sobolev norm by compactly supported smooth functions. This density of smooth functions in a Sobolev space follows from cutoffs tending to one and subsequent mollification: the cutoff-gradient term tends to zero in $L^p$, and a mollifier approximates both the function and each weak derivative. The displayed estimate makes the approximants Cauchy in $L^{q_1}$. Their $L^p$ limit is the original function, so subsequences converging almost everywhere identify the $L^{q_1}$ limit with it. This proves continuous embedding. The range $1<p<d$ is essential for this consequence; the formula is not a general critical $p=d$ embedding into $L^{\infty }$.

Let ${\Phi }_t$ be the local flow of a vector field $V$. The Lie derivative of a differential form is ${\mathcal{L}}_V\eta ={\frac{d}{dt}|}_{t=0}{\Phi }_t^{\ast }\eta$. We prove Cartan's magic formulaBoth sides are degree-zero derivations of an algebra on the exterior algebra of differential forms. They agree on a function $f$, since ${\iota }_{V}df=V(f)$. They also agree on $df$: pullback commutes with the exterior derivative, so ${\mathcal{L}}_{V}df=d(Vf)$, while $(d{\iota }_V+{\iota }_{V}d)df=d(Vf)$. Locally every differential form is a sum of products of functions and coordinate differentials, proving the identity in every degree.

For a compactly supported Hamiltonian vector field $V_H$, take the convention ${\iota }_{V_H}\omega =-dH$. Since $d\omega =0$, Cartan's magic formula gives ${\mathcal{L}}_{V_H}\omega =0$, so its complete flow of a vector field consists of symplectomorphisms. Given $p$ and $v\in T_{p}M$, a bump function lets us choose a compactly supported $H$ with $d{H}_p=-{\iota }_v{\omega }_p$, and hence $V_H(p)=v$. Choose such fields for a basis of $T_{p}M$. Their successive small-time flows give a map with invertible derivative at the origin. The inverse function theorem makes the orbit of $p$ open. Every orbit is open; because $M$ is connected, there is only one orbit. Thus the compactly supported symplectomorphisms act transitively on points, by Hamiltonian transitivity on a connected symplectic manifold.

On the unit sphere, the standard symplectic area is $4\pi$. The equator has complementary disks of areas $2\pi ,2\pi$, whereas the latitude at height $1/2$ has complementary disks of areas $\pi ,3\pi$: the cap above height $h$ has area $2\pi (1-h)$. A symplectomorphism must preserve the unordered pair of complementary areas. Therefore no such symplectomorphism exists, by complementary areas obstruct symplectic equivalence of separating curves.

Use ${\mathbb{T}}^n=(\mathbb{R}/2\pi \mathbb{Z}{)}^n$ and the uniform metric $d(f,g)=\Vert f-g{\Vert }_{\infty }$. Let $N(\epsilon ,E)$ be the least number of radius-$\epsilon$ balls in this metric covering $E$. The Kolmogorov entropy, equivalently metric entropy, isIt is infinite if no finite cover exists; set the entropy of the empty set to zero. Changing the logarithm base or requiring centres in $E$ changes only the constants and resolution factors in the estimates below.

For an explicit appropriate norm, takeAn equivalent standard norm gives the same exponents, with different constants. We prove the two bounds for the entropy of a smooth periodic function ball separately.

For the lower bound, choose a nonnegative bump function $\phi \in C_c^{\infty }((-1,1{)}^n)$ with $0\le \phi \le 1$ and $\phi (0)=1$. Let $K$ bound all its derivatives of orders at most $p$, and choose $c>0$ with $cK\le 1$. For sufficiently small $h\le 1$, place $L\ge c_n{h}^{-n}$ translates of its scaled support disjointly on the torus, for instance by taking grid centres separated by $3h$ in a fundamental cube. Define the real functionsusing periodic extension. Disjointness means that at any point only one summand contributes. For every $|\alpha |\le p$, its derivative is bounded by $c{h}^{p-|\alpha |}K\le 1$. Thus every $f_{\sigma }$ lies in $B_{n,p}$. If $\sigma \ne \tau$, evaluate at a centre where the binary coefficients differ to obtainChoose $h=(4\epsilon /c{)}^{1/p}$ for sufficiently small $\epsilon$. The $2^L$ functions are more than $2\epsilon$ apart, so no radius-$\epsilon$ ball covers two of them. The metric covering number consequently satisfiesTo include the remaining $\epsilon <1/2$, observe that the two constant functions $1,-1$ belong to the ball and require distinct covering balls. Reducing the positive lower-bound constant extends the estimate over that fixed remaining interval.

For the upper bound, use this precise version of multivariable Jackson approximation: for every $f\in C^p({\mathbb{T}}^n)$ and integer $N\ge 1$, there exists a real trigonometric polynomial $P_N$ with $|k_j|\le N$ in every coordinate andChoose $N=\lceil (4J_{n,p}/\epsilon {)}^{1/p}\rceil$. Then the approximation error is at most $\epsilon /4$, and $\Vert P_N{\Vert }_{\infty }\le 2$. Write $P_N(x)={\sum }_{k\in \{-N,\dots ,N{\}}^n}{c}_k{e}^{ik\cdot x}$. There are $D=(2N+1{)}^n$ coefficients, and normalized integration gives $|c_k|\le \Vert P_N{\Vert }_{\infty }\le 2$.

Round their real and imaginary parts with mesh $\delta =\epsilon /(8D)$, respecting $c_{-k}=\overline{c_k}$ so the rounded polynomial remains real. Each coefficient changes by at most $2\delta$, and hence the uniform change in the polynomial is at most $2D\delta =\epsilon /4$. At most $(C/\delta {)}^{2D}$ choices are needed; allowing redundant choices only enlarges this bound. These rounded polynomials form an $\epsilon$-net for the ball, andHere $D\le C_{n,p}{\epsilon }^{-n/p}$, and the last logarithm absorbs the constant and $\log D$ for $\epsilon <1/2$. Together the bounds give

For the unit ball of $C({\mathbb{T}}^n)$ alone, take infinitely many disjoint small neighbourhoods and a height-one continuous bump supported in each. Distinct such functions have uniform distance one. Thus no finite $\epsilon$-net exists when $\epsilon <1/2$, andThe derivative bound was essential for total boundedness; mere continuity gives no common small-scale oscillation bound.

For the final composition estimate, interpret outer periodic functions on ${\mathbb{T}}^m$ by their $2\pi$-periodic extensions to ${\mathbb{R}}^m$. Since $p\ge 1$, each $f\in B_{m,p}$ satisfiesby integrating its first partial derivatives along coordinate segments. This uniform Lipschitz continuity controls errors in the inner functions.

Choose an outer net of radius $\eta =\epsilon /2$ for $B_{m,p}$ and an inner net of radius $\delta =\epsilon /(2m)$ for $E$. For each $g=f(u_1,\dots ,u_m)$ choose corresponding approximants $f_0,v_1,\dots ,v_m$. No derivative bound for $f_0$ is needed: use the derivative bound of $f$ before replacing the outer function. Pointwise,All composed centres are continuous. Multiplying the numbers of outer and inner choices gives the entropy of Lipschitz compositions inequalityApply the proved upper bound with dimension $m$, and the assumed estimate for $E$. Fixed rescalings of $\epsilon$ preserve its exponent and change $\log (1/\epsilon )$ by only an additive constant. Therefore

In this context a nowhere locally homogeneous metric, also called a bumpy metric, is a smooth Riemannian metric for which no two distinct nonempty open subsets are isometric with their induced metrics. Equivalently every local isometry between open subsets is the identity wherever defined: a nonidentity local isometry sends some point $x$ to a distinct point, and restriction to sufficiently small disjoint neighborhoods would violate the first formulation. This is the local-isometry meaning of the terminology here; degeneracy of periodic geodesics is a different use of the word bumpy.

Sunada's local isometry lemma states that on a compact smooth manifold without boundary of dimension $m\ge 2$ the nowhere locally homogeneous metrics contain a residual set in the space of smooth Riemannian metrics with its $C^{\infty }$ topology. In particular they are dense, by the Baire category theorem. A residual set is a countable intersection of open dense sets. The dimension assumption matters: every one-dimensional Riemannian metric is locally $d{s}^2$ in arclength coordinates and has local translations.

The jet bundle of maps $J^k(M,N)$ consists of equivalence classes $j_x^{k}f$ of smooth maps near $x$, where two maps agree to order $k$ at $x$ in coordinate charts. Its projection to $M\times N$ sends $j_x^{k}f$ to $(x,f(x))$. For a multi-index $\alpha$ in $m$ variables there are $\left(\genfrac{}{}{0ex}{}{m+r-1}{r}\right)$ derivatives of order $r$. Thus a coordinate chart consists of the source point, the target point, and $n$ coefficients for every derivative order $1$ through $k$. Summing the counts givesThis includes $J^0(M,N)=M\times N$. The fibre over $(x,y)$ is the space of truncated Taylor maps with fixed constant term $y$, locally modeled onFor $k=1$ its identification with $\mathrm{Hom}(T_{x}M,T_{y}N)$ is canonical. For higher $k$, changes of target coordinates mix derivatives of different orders, so this is a coordinate or connection-dependent description, not a canonical vector-bundle identification. The truncation $J^k\to J^{k-1}$ is an affine bundle modeled on the pullback of ${\mathrm{Sym}}^k(T^{\ast }M)\otimes TN$ over $M\times N$.

For the density assertion put $A={\overline{U}}_i$ and $B={\overline{U}}_j$. Their closures must be distinct. If $A=B$, the identity is an isometry for every Riemannian metric, and the requested complement is empty. Under the intended distinct-domain assumption, the smooth closed-ball hypothesis makes $A$ and $B$ regular closed domains, equal to the closures of their interiors. After interchanging them if necessary, there is a nonempty open set $W$ with compact closure in $\mathrm{int}(A)\setminus B$. Otherwise each interior would be contained in the other closure, forcing $A=B$.

Take any smooth Riemannian metric $g$. If ${\mathrm{vol}}_g(A)\ne {\mathrm{vol}}_g(B)$, it already lies outside ${\mathcal{S}}_{ij}$, because an isometry preserves the Riemannian volume form. If the volumes agree, choose a nonzero nonnegative smooth bump function $\eta$ supported in $W$ and setThese are positive definite Riemannian metrics, they agree with $g$ on $B$, and $g_t\to g$ in $C^{\infty }$ as $t\downarrow 0$, since every derivative of $g_t-g$ is $t$ times a fixed compactly supported smooth tensor. Their Riemannian volume forms satisfyHence ${\mathrm{vol}}_{g_t}(B)={\mathrm{vol}}_g(B)$ whereas ${\mathrm{vol}}_{g_t}(A)>{\mathrm{vol}}_g(A)$ for every $t>0$. In particularThe two domains cannot be isometric for $g_t$. Every $C^{\infty }$ neighborhood of $g$ therefore meets the complement of ${\mathcal{S}}_{ij}$, provingThis localized volume perturbation proves the requested density even when the two domains overlap, and works in every positive dimension. It does not by itself prove the stronger residual local-isometry statement: fixed isometric closures and arbitrary isometric open subsets are different conditions.

A smooth cutoff function is a smooth function used to localize an argument, typically equal to one on one region and zero outside a slightly larger region. Such functions can be built from bump functions and smooth partitions of unity.