ReLU approximation rates, depth separation, and linear regions

Before any hidden layer, the input is affine on $I$. Suppose all values through layer $\mathrm{\ell }-1$ are affine on each of $M$ subintervals. On any one such interval, every preactivation in layer $\mathrm{\ell }$ is affine, and each has at most one zero unless it is identically zero. The at most $n_{\mathrm{\ell }}$ zeros divide that interval into at most $n_{\mathrm{\ell }}+1$ subintervals on each of which all ReLU outputs are affine.

Induction over the hidden layers gives at most ${\prod }_{\mathrm{\ell }=1}^H(n_{\mathrm{\ell }}+1)$ intervals; the affine output adds no breakpoints. For $H\ge 1$, AM–GM gives $\prod (n_{\mathrm{\ell }}+1)\le (1+H^{-1}\sum n_{\mathrm{\ell }}{)}^H=(1+U\mathrm{/}H{)}^H$. With no hidden layer the output is affine on the entire interval.

On $[0,1\mathrm{/}2]$, $\tau (x)=2x$, and on $[1\mathrm{/}2,1]$, $\tau (x)=2-2x$. Thus $\tau$ maps each half interval bijectively onto $[0,1]$ with slope magnitude $2$. If ${\tau }^q$ has the stated property, each of its dyadic intervals is split at its midpoint by the preimage of $1\mathrm{/}2$; composition with $\tau$ makes two dyadic intervals, each mapped bijectively onto $[0,1]$, with slopes multiplied by $2$. Induction proves the interval and slope claims.

A layer with two ReLU units computes $u=\rho (2v)$ and $w=\rho (2v-1)$ from the preceding scalar value $v\in [0,1]$, and the affine output $u-2w$ equals $\tau (v)$. For each next block substitute the preceding block's affine output expression for $v$ in its two preactivations, so one hidden layer suffices per iterate. Forward each block's affine output to the final affine unit; these skip edges add only $O(q)$ parameters. The network uses $2q$ hidden units and $O(q+1)$ weights; counting input and output gives depth $q+2$.

Let $I_q$ be the linear interpolant of $x^2$ at the dyadic nodes $j{2}^{-q}$. On each interval of $I_{q-1}$, the difference $I_{q-1}-I_q$ is zero at its endpoints and has height $4^{-q}$ at the midpoint. By [F1], ${\tau }^q$ is the unit tent on each such interval, so $I_{q-1}-I_q=4^{-q}{\tau }^q$. Since $I_0(x)=x$, telescoping gives $I_m(x)=x-{\sum }_{q=1}^m{4}^{-q}{\tau }^q(x)$. To implement this with shared computation, set $t_0=x$ and for each $q=1,\dots ,m$ use one hidden layer with $u_q=\rho (2t_{q-1})$ and $v_q=\rho (2t_{q-1}-1)$, then define the affine expression $t_q=u_q-2v_q=\tau (t_{q-1})$. Thus $t_q={\tau }^q(x)$ and the $m$ iterates use only $2m$ hidden units in $m$ sequential layers. The final affine output reads the input and all hidden units, with coefficient $1$ on $x$ and contribution $-4^{-q}{u}_q+2\cdot 4^{-q}{v}_q$ from layer $q$, so it computes $I_m$ with only $O(m+1)$ additional skip connections.

On a dyadic interval $[a,b]$ of length $2^{-m}$, linear interpolation of $x^2$ has error $(x-a)(b-x)$, whose maximum is $(b-a{)}^2\mathrm{/}4=4^{-(m+1)}$. Choose $m$ with $4^{-(m+1)}\le \delta$. Then $I_m$ approximates $x^2$ within $\delta$ and $I_m(0)=0$.

The shared sequential construction in step 1.1 computes all required iterates in $m$ hidden layers, and its affine skip readout computes $I_m$ using $O(m+1)$ weights. The resulting network has $O(m+1)$ units, weights, and depth, and $m\le C^{\mathrm{\prime }}(1+\log (1\mathrm{/}\delta ))$ by the choice of $m$. Taking $Q_{\delta }=I_m$ proves the claim.

Set $\delta =\epsilon \mathrm{/}(6M^2)$ and, for $\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }y\mathrm{\mid }\le M$, put $P_{M,\epsilon }(x,y)=2M^2[Q_{\delta }\text{\hspace{0.17em}⁣}\left(\frac{\mathrm{\mid }x+y\mathrm{\mid }}{2M}\right)-Q_{\delta }\text{\hspace{0.17em}⁣}\left(\frac{\mathrm{\mid }x\mathrm{\mid }}{2M}\right)-Q_{\delta }\text{\hspace{0.17em}⁣}\left(\frac{\mathrm{\mid }y\mathrm{\mid }}{2M}\right)]\mathrm{.}$ All three arguments lie in $[0,1]$. Replacing $Q_{\delta }(u)$ by $u^2$ gives $2M^2[((x+y{)}^2-x^2-y^2)\mathrm{/}(4M^2)]=xy$, so the absolute error is at most $2M^2(3\delta )=\epsilon$.

Since $Q_{\delta }(0)=0$, substituting $x=0$ makes the first and third terms equal and the second zero; substituting $y=0$ makes the first and second terms equal and the third zero. The formula is globally defined, so this cancellation holds for every real value of the other coordinate. Absolute value is represented by $\mathrm{\mid }z\mathrm{\mid }=\rho (z)+\rho (-z)$, and the three square networks run in parallel followed by an affine combination, so the depth, weights, and units are $O(1+\log (1\mathrm{/}\delta ))=O(1+\log (M^2\mathrm{/}\epsilon ))$, since $0<\delta <1\mathrm{/}6$ and $\log (1\mathrm{/}\delta )=\log 6+\log (M^2\mathrm{/}\epsilon )$.

For each $\mathrm{\mid }\alpha \mathrm{\mid }<n$, $D^{\alpha }f\in W^{1,\mathrm{\infty }}(Q)$ by [F1], since its first weak derivatives are among the derivatives of $f$ of order at most $n$. By [F1], choose their Lipschitz representatives; their Lipschitz constants are at most $\sqrt{d}$, because each coordinate weak derivative has essential supremum at most $1$. Since $Q$ has finite measure and every weak derivative of $f$ through order $n$ is essentially bounded, the same weak-derivative identities give $f\in W^{n,1}(Q)$, hence $f\in W^{n-1,1}(Q)$. By [A1] and [F2], Countable Choice holds, so the interior approximation theorem applies with $k=n-1$ and $p=1$, while [F3] applies with $k=n$ and $p=1$. For every compact segment in $Q$, convolution of each chosen Lipschitz representative differs from it uniformly by at most $\sqrt{d}$ times the mollifier radius, so the mollified derivatives through order $n-1$ converge uniformly there.

For an interior grid center $a_m$ and interior $x$, choose the mollifier radius smaller than the distance from their segment to $\mathrm{\partial }Q$. By [F3] and step 1.1, $D^{\alpha }{f}_{\delta }={\rho }_{\delta }\ast D^{\alpha }f$ on that segment; nonnegativity and unit mass of ${\rho }_{\delta }$ imply $\mathrm{\mid }{D}^{\alpha }{f}_{\delta }\mathrm{\mid }\le \mathrm{\parallel }{D}^{\alpha }f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ for $\mathrm{\mid }\alpha \mathrm{\mid }=n$. The one-variable Taylor formula applied to $t\mapsto f_{\delta }(a_m+t(x-a_m))$ therefore gives remainder at most $\frac{1}{n!}{\sup }_t\mathrm{\mid }\frac{d^n}{d{t}^n}{f}_{\delta }(a_m+t(x-a_m))\mathrm{\mid }\le C(d,n)\mathrm{\parallel }x-a_m{\mathrm{\parallel }}_{\mathrm{\infty }}^n$, since ${\sum }_i\mathrm{\mid }{x}_i-a_{m,i}\mathrm{\mid }\le d\mathrm{\parallel }x-a_m{\mathrm{\parallel }}_{\mathrm{\infty }}$. Passing to the uniform limits from step 1.1 proves the estimate for interior points. Approximate any boundary grid center and any boundary input by interior points; the continuous extensions specified in the statement and the finite Taylor polynomial make the same estimate hold on the closed cube.

The one-dimensional hats sum to $1$ on each grid interval, so their tensor products satisfy ${\sum }_m{\varphi }_m=1$ on the cube; at most $2^d$ are nonzero at any point, and each nonzero term has $\mathrm{\parallel }x-a_m{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1\mathrm{/}N$. Therefore $\mathrm{\mid }f(x)-g_N(x)\mathrm{\mid }\le {\sum }_{m:{\varphi }_m(x)>0}{\varphi }_m(x)C(d,n)N^{-n}=C(d,n)N^{-n}$. Each extended Lipschitz representative and each hat is continuous on the closed cube, so the finite sum $g_N$ is continuous there.

Choose an integer $N\ge 2$ with $C(d,n)N^{-n}\le \epsilon \mathrm{/}2$ and form $g_N$ from [F1]. Then $N\le C^{\mathrm{\prime }}(d,n){\epsilon }^{-1\mathrm{/}n}$, so the number of grid centers is $(N+1{)}^d=O_{d,n}({\epsilon }^{-d\mathrm{/}n})$.

For each center $a_m$, write its Taylor polynomial as a finite sum of monomials in the scaled variables $z_i=N(x_i-a_{m,i})$, multiplied by coefficients $D^{\alpha }f(a_m)N^{-\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{/}\alpha !$. On the support of ${\varphi }_m$, $\mathrm{\mid }{z}_i\mathrm{\mid }\le 1$, and all coefficients are bounded by a constant depending only on $d,n$. The hat ${\varphi }_m$ is the product of $d$ one-dimensional ReLU hats, since $\mathrm{\mid }z\mathrm{\mid }=\rho (z)+\rho (-z)$ and $u_{+}=\rho (u)$. There are only a number depending on $d,n$ of products and summands. Choose the error of each product operation to be $\eta \mathrm{/}C(d,n)$; inductively, multiplying by a factor of magnitude at most $1$ adds at most that error to the previous product error, and summing the finitely many terms multiplies it by at most another $C(d,n)$. On the hat support, all intermediate inputs stay in fixed bounded intervals depending only on $d,n$, so the bounded-domain error guarantee in [F2] applies there and the monomial sum followed by multiplication by the hat has term error at most $\eta$. Off the hat support, the final product is exactly zero: although the error guarantee is only used on the bounded square, the explicit product formula in [F2] defines a network globally, and the axis cancellation proved there holds for every value of the other input. Thus the local error is zero there. The resulting subnetwork has $O_{d,n}(1+\log (1\mathrm{/}\eta ))$ weights, units, and depth.

At any $x$, at most $2^d$ of the exact hats are nonzero by [F1], and the constructed approximate term is zero outside the same support. Thus summing all local subnetworks gives a network ${\mathrm{\Phi }}_f$ with $\mathrm{\parallel }{g}_N-{\mathrm{\Phi }}_f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2^d\eta$. Choose $\eta =\epsilon \mathrm{/}{2}^{d+1}$. Then the triangle inequality and step 1.1 give $\mathrm{\parallel }f-{\mathrm{\Phi }}_f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \epsilon$.

The local polynomial degree and number of terms depend only on $d,n$, while the grid size depends only on $N$. All local subnetworks have depth $O_{d,n}(1+\log (1\mathrm{/}\eta ))$ and size of the same order; they run in parallel and their outputs are summed affinely. Hence the total depth is $O_{d,n}(1+\log (1\mathrm{/}\epsilon ))$ and the total weights and units are $O_{d,n}(N^d(1+\log (1\mathrm{/}\epsilon )))=O_{d,n}({\epsilon }^{-d\mathrm{/}n}(1+\log (1\mathrm{/}\epsilon )))$. The edge pattern depends on $N,d,n,\epsilon$ but not on $f$; only Taylor coefficients and output weights depend on $f$.

By [A1] and [F2], the Countable Choice convention in [F1] holds. Choose the Lipschitz representative on $(0,1)$ and extend it uniquely to $[0,1]$: values along any sequence approaching an endpoint are Cauchy and any two such limits agree by the Lipschitz bound. The continuous representative satisfies $\mathrm{\mid }f\mathrm{\mid }\le 1$ everywhere, since any interior value with modulus greater than $1$ would by continuity violate the almost-everywhere bound on an interval of positive measure. Let $T\ge 1$ be an integer and let $f_1$ linearly interpolate $f$ at $t\mathrm{/}T$, $0\le t\le T$. Then $f_1$ is $1$-Lipschitz, and $f_2=f-f_1$ vanishes at every grid endpoint and is $2$-Lipschitz. Indeed, both summands are $1$-Lipschitz.

Let $m\ge 1$ and let $\mathrm{\Gamma }$ be the set of piecewise-linear functions on the grid $r\mathrm{/}m$, $0\le r\le m$, with endpoint values zero and successive increments in $\{-2\mathrm{/}m,0,2\mathrm{/}m\}$. Thus $\mathrm{\mid }\mathrm{\Gamma }\mathrm{\mid }\le 3^m$. For each coarse interval $I_t=[t\mathrm{/}T,(t+1)\mathrm{/}T)$, rescale $g_t(y)=T{f}_2((t+y)\mathrm{/}T)$. This function is $2$-Lipschitz and vanishes at both endpoints. Quantize each node value down to the nearest multiple of $2\mathrm{/}m$; adjacent quantized values differ by one of the three allowed increments. At a fraction $\theta$ of a grid subinterval, a $2$-Lipschitz function differs from its endpoint chord by at most $4\theta (1-\theta )\mathrm{/}m\le 1\mathrm{/}m$; quantizing both endpoint values changes that chord by at most $2\mathrm{/}m$. Hence some ${\gamma }_t\in \mathrm{\Gamma }$ satisfies ${\sup }_{x\in I_t}\mathrm{\mid }{f}_2(x)-T^{-1}{\gamma }_t(Tx-t)\mathrm{\mid }\le 3\mathrm{/}(Tm)$.

Define the discontinuous selector $\chi (z)=z$ for $0\le z<1$ and $0$ otherwise. For $x\in [0,1)$, exactly the index $t$ with $x\in I_t$ has $\chi (Tx-t)=Tx-t$; all other terms vanish. Therefore ${\tilde{f}}_2(x)=\frac{1}{T}{\sum }_{\gamma \in \mathrm{\Gamma }}\gamma \text{\hspace{0.17em}⁣}({\sum }_{t:{\gamma }_t=\gamma }\chi (Tx-t))$ equals the local interpolant $T^{-1}{\gamma }_t(Tx-t)$ and has error at most $3\mathrm{/}(Tm)$. If $s_{\gamma ,r}$ is the slope of $\gamma$ on $[r\mathrm{/}m,(r+1)\mathrm{/}m]$, then its zero value at $0$ gives $\gamma (z)=s_{\gamma ,0}\rho (z)+{\sum }_{r=1}^{m-1}(s_{\gamma ,r}-s_{\gamma ,r-1})\rho (z-r\mathrm{/}m),0\le z\le 1,$ by matching the initial slope and each slope jump. This uses $m$ ReLU hinges. The cache is a depth-five mixed-gate network with $O(T+m{3}^m)$ units and weights: $T$ selector units, at most $3^m$ grouped sums, $m{3}^m$ ReLU hinge units, and an affine output. For the coarse interpolant, set $s_t=T(f((t+1)\mathrm{/}T)-f(t\mathrm{/}T))$ for $0\le t<T$; then $f_1(x)=f(0)+s_0\rho (x)+{\sum }_{t=1}^{T-1}(s_t-s_{t-1})\rho (x-t\mathrm{/}T)$ on $[0,1]$. This slope-jump identity has $T$ ReLU units and one affine output, so $f_1$ is a depth-three ReLU network of size $O(T)$.

Replace each discontinuous selector by ${\chi }_{\delta }(z)=\rho (z)-{\delta }^{-1}\rho (z-(1-\delta ))+({\delta }^{-1}-1)\rho (z-1)\mathrm{.}$ This continuous piecewise-linear function agrees with $\chi$ on $[0,1-\delta ]$, decreases linearly to zero on $[1-\delta ,1]$, and is zero outside $[0,1]$. Let ${\phi }_{\delta }$ be the piecewise-linear tent that is zero outside $[0,1-\delta ]$, rises from $0$ to $1$ on $[0,\delta ]$, equals $1$ on $[\delta ,1-2\delta ]$, and falls to $0$ on $[1-2\delta ,1-\delta ]$. Its four slope jumps give a representation by at most four ReLU hinges as in step 3.1. Put ${\mathrm{\Phi }}_{\delta }(x)={\sum }_{t=0}^{T-1}{\phi }_{\delta }(Tx-t)$; this comb is a ReLU spline with $O(T)$ units and weights. Replacing the $T$ selectors by three ReLU units each and adding the comb keeps the cache size $O(T+m{3}^m)$ independently of $\delta$.

Let ${\tilde{f}}_{2,\delta }$ be the cache network using ${\chi }_{\delta }$, and set $f_{2,\delta }(x)=\rho ({\tilde{f}}_{2,\delta }(x)+2{\mathrm{\Phi }}_{\delta }(x)-1)-\rho (2{\mathrm{\Phi }}_{\delta }(x)-1)\mathrm{.}$ At most one selector is nonzero at any $x$, its value lies in $[0,1]$, and each rescaled ${\gamma }_t$ vanishes at both endpoints and is $2$-Lipschitz; hence $\mathrm{\mid }{\tilde{f}}_{2,\delta }(x)\mathrm{\mid }\le 1$. On each plateau where ${\mathrm{\Phi }}_{\delta }=1$, the selector is unchanged and the gate outputs the cache. Where ${\mathrm{\Phi }}_{\delta }=0$, the two ReLU terms cancel. On the transition strips $y\in [0,\delta ]$ or $y\in [1-2\delta ,1-\delta ]$ in a coarse coordinate $y=Tx-t$, ${\chi }_{\delta }(y)=\chi (y)$, so the cache equals the continuous piecewise-linear spline ${\tilde{f}}_2$ from step 3.1. For $0\le p\le 1$, the map $a\mapsto \rho (a+2p-1)-\rho (2p-1)$ is $1$-Lipschitz and vanishes at $a=0$, hence its output has modulus at most $\mathrm{\mid }a\mathrm{\mid }$. Every point where $0<{\mathrm{\Phi }}_{\delta }<1$ is within $2\delta \mathrm{/}T$ of a coarse endpoint; both $f_2$ and ${\tilde{f}}_2$ vanish there and are $2$-Lipschitz, so on those strips $\mathrm{\mid }{f}_2-f_{2,\delta }\mathrm{\mid }\le 8\delta \mathrm{/}T$. Where ${\mathrm{\Phi }}_{\delta }=0$, $f_{2,\delta }=0$ and $\mathrm{\mid }{f}_2\mathrm{\mid }\le 4\delta \mathrm{/}T$, which is also within this bound. On plateaus the quantization error is at most $3\mathrm{/}(Tm)$ by step 2.1. Thus $\mathrm{\parallel }{f}_2-f_{2,\delta }{\mathrm{\parallel }}_{\mathrm{\infty }}\le 8\delta \mathrm{/}T+3\mathrm{/}(Tm)\mathrm{.}$ The two final ReLU preactivations absorb the former affine cache output, followed by a shared affine output, so depth increases from five to six; complexity remains $O(T+m{3}^m)$.

Choose $m=\lceil \frac{1}{2}{\log }_3(1\mathrm{/}\epsilon )\rceil$, $T=\lceil 6\mathrm{/}(m\epsilon )\rceil$, and $0<\delta \le \min \{1\mathrm{/}4,\epsilon T\mathrm{/}16\}$. Then both error terms in step 5.1 are at most $\epsilon \mathrm{/}2$, while $T+m{3}^m\le C\mathrm{/}[\epsilon \log (1\mathrm{/}\epsilon )]$: here $T=O(1\mathrm{/}(\epsilon m))$, $3^m=O({\epsilon }^{-1\mathrm{/}2})$, and $m{\epsilon }^{-1\mathrm{/}2}=O(1\mathrm{/}(\epsilon m))$ because ${\epsilon }^{1\mathrm{/}2}(\log (1\mathrm{/}\epsilon ){)}^2$ is bounded on $(0,1\mathrm{/}2)$. The sum $f_1+f_{2,\delta }$ is the required ReLU approximation.

We first rule out a continuous odd map $g:S^m\to S^{m-1}$ for $m\ge 1$. Such a map descends to $b:{\mathbb{R}\mathbb{P}}^m\to {\mathbb{R}\mathbb{P}}^{m-1}$, $b([x])=[g(x)]$. This is continuous by the quotient topology, since the composite with the sphere quotient is continuous and $g(-x)=-g(x)$.

In the coordinate subspace ${\mathbb{R}}^{m+1}\oplus {\mathbb{R}}^m\subset {\mathbb{R}}^{\mathrm{\infty }}$, define $H([x],t)=[\cos (\pi t\mathrm{/}2)x,\sin (\pi t\mathrm{/}2)g(x)],0\le t\le 1.$ Its representing vector has norm one. Replacing $x$ by $-x$ negates the whole vector, so the projective point is independent of representative. Continuity can be checked on each chart where a chosen coordinate of $x$ is nonzero: choose the unit representative with that coordinate positive, which varies continuously in the ratios of the remaining coordinates, and use the displayed formula. These charts cover projective space, proving joint continuity in $([x],t)$. Thus $H$ is a homotopy from $i_m$ to $j\circ b$, where $j([y])=[0,y]$ is the inclusion in the second block. The same formula with $y$ in the first $m$ coordinates and $y$ in the second block, namely $[\cos (\pi t\mathrm{/}2)y,0,\sin (\pi t\mathrm{/}2)y]$, is a homotopy from $i_{m-1}$ to $j$. Both homotopies lie in the finite projective subspace ${\mathbb{R}\mathbb{P}}^{2m}$ of the infinite space and therefore are continuous there as well.

Put $a_k=i_k^{\ast }a$. By [F3] and step 2.1, $a_m=b^{\ast }{a}_{m-1}$, including $m=1$ when $a_0=0$. Consequently $a_m^m=b^{\ast }(a_{m-1}^m)=0,$ since $H^m({\mathbb{R}\mathbb{P}}^{m-1};{\mathbb{F}}_2)=0$ by [F2]. On the other hand, [F1] says $a^m\mathrm{\ne }0$ and that $i_m^{\ast }$ is an isomorphism in degree $m$, so $a_m^m=i_m^{\ast }(a^m)\mathrm{\ne }0$. These incompatible conclusions rule out every such odd map.

For $m\ge 1$, if $h(x)-h(-x)$ never vanished, then $g(x)=\frac{h(x)-h(-x)}{\mathrm{\parallel }h(x)-h(-x){\mathrm{\parallel }}_2}$ would be continuous, take values in $S^{m-1}$ and satisfy $g(-x)=-g(x)$. Step 3.1 rules this out, so an antipodal pair has equal images. When $m=0$, the codomain ${\mathbb{R}}^0$ consists of one point and the two points of $S^0$ have equal images directly. AC is inherited exactly as stated in [A1].

First establish an elementary zero-counting fact. If $D\ge 1$ and $r_i(x)=x_i^D+s_i(x)$ with $\mathrm{deg}{s}_i<D$ for $1\le i\le k$, then these $k$ polynomials have at most $D^k$ distinct common real zeros. In the quotient vector space $V=\mathbb{R}[x_1,\dots ,x_k]\mathrm{/}(r_1,\dots ,r_k)$, every monomial is a linear combination of the $D^k$ monomials $x_1^{a_1}\cdots x_k^{a_k}$ with $0\le a_i<D$. Indeed, if some exponent is at least $D$, replace its factor $x_i^D$ by $-s_i$ modulo the ideal. Every resulting monomial has strictly smaller total degree, so induction on total degree terminates and proves the spanning assertion. Thus $\dim V\le D^k$.

If no strict sign vector occurs, its bound is immediate. Otherwise choose one representative $a_{\sigma }$ for each attained strict vector $\sigma$. Only finitely many choices are involved, since there are at most $2^m$ vectors. Set $P={\prod }_{i=1}^m{p}_i$ and $D=2md+1$. Then $P(a_{\sigma })\mathrm{\ne }0$. Choose a single $\delta >0$ so small that $H(a_{\sigma })>0$ for every representative, where $H(x)=P(x{)}^2-\delta {\sum }_{i=1}^k{x}_i^{D+1}$. Such a choice exists: there are finitely many positive numbers $P(a_{\sigma }{)}^2$, and the other term has a fixed finite value at each representative. Since $D+1$ is even and greater than $2\mathrm{deg}P$, $H(x)\to -\mathrm{\infty }$ as $\mathrm{\parallel }x\mathrm{\parallel }\to \mathrm{\infty }$. Explicitly, $\mathrm{\mid }P(x){\mathrm{\mid }}^2\le C(1+\mathrm{\parallel }x{\mathrm{\parallel }}^{2md})$ and ${\sum }_i{x}_i^{D+1}\ge k^{-(D+1)\mathrm{/}2}\mathrm{\parallel }x{\mathrm{\parallel }}^{D+1}$.