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Proof of The Gaussian Kernels on Euclidean Space: Scaling, Derivatives up to Order Three, the Convolution Identity, Moments and Tails

lemmalem:gaussian-kernel-euclidean-2026a
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· 13,642 chars · 30 deps · depth 17 Reason: First publication of the proof of the Gaussian kernel lemma (Goal 3F, batch F0).

Scaling follows from the change of variables z -> z/sqrt(s) in the normalisation integral; smoothness and the derivative formulas from the product rule applied to polynomial multiples of the Gaussian; the bounds from the elementary inequality umu^m exp(-u^2/4) <= DmD_m; the convolution identity from completing the square; and the moments and tails from the moment identities of the smoothing weight and Markov's inequality.

Proof

Each result cited is universally quantified over the data in its own statement. Throughout, ss satisfies 0<s10<s\le1, and we write ψs(z)=exp(z2/(2s))\psi_{s}(z)=\exp(-\lVert z\rVert^{2}/(2s)), so that gs=csψsg_{s}=c_{s}\psi_{s}, with Rqψsdλq=1/cs\int_{\mathbb{R}^{q}}\psi_{s}\,d\lambda_{q}=1/c_{s} by claim 1 of The Gaussian Smoothing Weight: Normalization, Derivatives, Exponential Tilting, Moments, and First-Order Remainder (read in dimension m=qm=q with η=s\eta=s) and the homogeneity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral with the nonnegative factor 1/cs1/c_{s}; that claim also gives 0<ψs10<\psi_{s}\le1, hence 0<gscs0<g_{s}\le c_{s}, the evenness gs(z)=gs(z)g_{s}(-z)=g_{s}(z), and, for every aRqa\in\mathbb{R}^{q}, that zgs(za)z\mapsto g_{s}(z-a) is Borel with Rqgs(za)dz=1\int_{\mathbb{R}^{q}}g_{s}(z-a)\,dz=1. By that lemma ψs\psi_{s} and gsg_{s} are sequentially continuous, hence continuous on Rq\mathbb{R}^{q} by Continuity Between Metric Spaces is Equivalent to Sequential Continuity, hence Borel (Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets, preamble). Powers (s)m(\sqrt{s})^{m} obey (s)m(s)n=(s)m+n(\sqrt{s})^{m}(\sqrt{s})^{n}=(\sqrt{s})^{m+n} (claim 1 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities, real and natural powers agreeing) and (s)2=ss=s(\sqrt{s})^{2}=\sqrt{s}\sqrt{s}=s (claim 1 of Properties of Natural Number Powers in a Field and Existence and Uniqueness of the Nonnegative Square Root), so that sm/2sn/2=s(m+n)/2s^{-m/2}s^{-n/2}=s^{-(m+n)/2} and s2/2=s1s^{-2/2}=s^{-1}; and ab=ab\sqrt{a}\sqrt{b}=\sqrt{ab} for nonnegative a,ba,b by claim 5 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities. The exponential satisfies exp(u+v)=exp(u)exp(v)\exp(u+v)=\exp(u)\exp(v), exp(u)>0\exp(u)>0, and exp(u)1+u\exp(u)\ge1+u for u0u\ge0 (claims 1, 2 and 4 of Basic Properties of the Exponential Function); in particular, for u0u\ge0, exp(u)1\exp(u)\ge1 and exp(u)=1/exp(u)1\exp(-u)=1/\exp(u)\le1 (the inverse of a number 1\ge1 is 1\le1, by claim 5 of Elementary Arithmetic in an Ordered Field applied to 1exp(u)1\le\exp(u) with the nonnegative multiplier 1/exp(u)1/\exp(u), claim 4 there).

Claim 1 (scaling). Let c=s1/2c=s^{-1/2}. For zRqz\in\mathbb{R}^{q}, cz2=c2z2=s1z2\lVert cz\rVert^{2}=c^{2}\lVert z\rVert^{2}=s^{-1}\lVert z\rVert^{2} by claim 5 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n, so ψ1(cz)=exp(z2/(2s))=ψs(z)\psi_{1}(cz)=\exp(-\lVert z\rVert^{2}/(2s))=\psi_{s}(z). By claim 2 of Scaling of Lebesgue Measure and the Lebesgue Integral on Rn\mathbb{R}^n, applied to the nonnegative Borel function ψ1\psi_{1},

1cs=Rqψsdλq=Rqψ1(cz)dz=cqRqψ1dλq=(s)qc1,\frac{1}{c_{s}}=\int_{\mathbb{R}^{q}}\psi_{s}\,d\lambda_{q}=\int_{\mathbb{R}^{q}}\psi_{1}(cz)\,dz=|c|^{-q}\int_{\mathbb{R}^{q}}\psi_{1}\,d\lambda_{q}=\frac{(\sqrt{s})^{q}}{c_{1}},

since cq=(s1/2)q|c|^{q}=(s^{-1/2})^{q} is the inverse of (s)q(\sqrt{s})^{q}. Hence cs=c1sq/2c_{s}=c_{1}\,s^{-q/2}, and gs(z)=csψs(z)=sq/2c1ψ1(s1/2z)=sq/2g1(s1/2z)g_{s}(z)=c_{s}\psi_{s}(z)=s^{-q/2}\,c_{1}\psi_{1}(s^{-1/2}z)=s^{-q/2}g_{1}(s^{-1/2}z).

Claim 2 (smoothness and derivatives). Let C\mathcal{C} be the set of functions F:RqRF:\mathbb{R}^{q}\to\mathbb{R} of the form F=hgsF=h\,g_{s} (pointwise product) with h:RqRh:\mathbb{R}^{q}\to\mathbb{R} smooth on Rq\mathbb{R}^{q}. Every FCF\in\mathcal{C} is continuous, as a product of continuous functions (claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space; a smooth hh is of class C1C^{1}, hence continuous, by Smooth Map on a Euclidean Open Set and clause 1 of C^k Maps on a Euclidean Open Set). By claim 2 of The Gaussian Smoothing Weight: Normalization, Derivatives, Exponential Tilting, Moments, and First-Order Remainder the partial derivative igs(z)=zisgs(z)\partial_{i}g_{s}(z)=-\frac{z_{i}}{s}g_{s}(z) exists at every zz, so by the product rule in claim 1 of Constants, Coordinate Functions, Sums and Products of CkC^k Functions on a Euclidean Open Set, for F=hgsCF=h\,g_{s}\in\mathcal{C},

iF(z)=ih(z)gs(z)+h(z)igs(z)=(ih(z)zish(z))gs(z),\partial_{i}F(z)=\partial_{i}h(z)\,g_{s}(z)+h(z)\,\partial_{i}g_{s}(z)=\Bigl(\partial_{i}h(z)-\frac{z_{i}}{s}h(z)\Bigr)g_{s}(z),

and ihzish\partial_{i}h-\frac{z_{i}}{s}h is smooth: ih\partial_{i}h is of class CkC^{k} for every kk by clause 2 of C^k Maps on a Euclidean Open Set applied to hCk+1h\in C^{k+1}, hence smooth; the coordinate function zziz\mapsto z_{i} and the constant 1s-\frac1s are smooth by claim 2 of Constants, Coordinate Functions, Sums and Products of CkC^k Functions on a Euclidean Open Set; and products and sums of smooth functions are smooth by claim 3 there. Thus iFC\partial_{i}F\in\mathcal{C}, in particular iF\partial_{i}F is continuous, and so every FCF\in\mathcal{C} is of class C1C^{1} (clauses 1 and 3 of C^k Maps on a Euclidean Open Set) with all partial derivatives in C\mathcal{C}. Let TT be the set of kNk\in\mathbb{N} such that every member of C\mathcal{C} is of class CkC^{k}; then 1T1\in T, and if kTk\in T then every FCF\in\mathcal{C} is of class C1C^{1} with partial derivatives of class CkC^{k}, hence of class Ck+1C^{k+1} by clause 2 of C^k Maps on a Euclidean Open Set, so k+1Tk+1\in T; by Principle of Induction for the Natural Numbers, T=NT=\mathbb{N}, and every member of C\mathcal{C} is smooth. In particular gs=1gsCg_{s}=1\cdot g_{s}\in\mathcal{C} is smooth. The first formula is claim 2 of The Gaussian Smoothing Weight: Normalization, Derivatives, Exponential Tilting, Moments, and First-Order Remainder. Applying the displayed rule to F=igsF=\partial_{i}g_{s}, that is h(z)=zish(z)=-\frac{z_{i}}{s}, with jh=δijs\partial_{j}h=-\frac{\delta_{ij}}{s} (the partial derivative of a coordinate function is 11 or 00 according as the indices agree or not, by Partial Derivative on a Euclidean Open Set), gives

jigs(z)=(δijs+zjzis2)gs(z);\partial_{j}\partial_{i}g_{s}(z)=\Bigl(-\frac{\delta_{ij}}{s}+\frac{z_{j}z_{i}}{s^{2}}\Bigr)g_{s}(z);

and applying it once more to h(z)=zizjs2δijsh(z)=\frac{z_{i}z_{j}}{s^{2}}-\frac{\delta_{ij}}{s}, with kh(z)=δikzj+δjkzis2\partial_{k}h(z)=\frac{\delta_{ik}z_{j}+\delta_{jk}z_{i}}{s^{2}} (product rule, claim 1 of Constants, Coordinate Functions, Sums and Products of CkC^k Functions on a Euclidean Open Set), gives

kjigs(z)=(δikzj+δjkzis2zks(zizjs2δijs))gs(z)=(zizjzks3+δijzk+δikzj+δjkzis2)gs(z),\partial_{k}\partial_{j}\partial_{i}g_{s}(z)=\Bigl(\frac{\delta_{ik}z_{j}+\delta_{jk}z_{i}}{s^{2}}-\frac{z_{k}}{s}\Bigl(\frac{z_{i}z_{j}}{s^{2}}-\frac{\delta_{ij}}{s}\Bigr)\Bigr)g_{s}(z)=\Bigl(-\frac{z_{i}z_{j}z_{k}}{s^{3}}+\frac{\delta_{ij}z_{k}+\delta_{ik}z_{j}+\delta_{jk}z_{i}}{s^{2}}\Bigr)g_{s}(z),

the stated formulas.

Claim 3 (bounds). Step (a): an elementary inequality. For every real u0u\ge0,

uexp(u2/4)1,u2exp(u2/4)4,u3exp(u2/4)512.u\,\exp(-u^{2}/4)\le1,\qquad u^{2}\exp(-u^{2}/4)\le4,\qquad u^{3}\exp(-u^{2}/4)\le512 .

Indeed, exp(u2/4)1+u2/4u\exp(u^{2}/4)\ge1+u^{2}/4\ge u, the last by 0(u/21)2=u2/4u+10\le(u/2-1)^{2}=u^{2}/4-u+1 (claim 5 of Zero Products and Elementary Identities in a Field and claim 2 of Nonnegativity of Squares in an Ordered Field), and exp(u2/4)1+u2/4u2/4\exp(u^{2}/4)\ge1+u^{2}/4\ge u^{2}/4; multiplying by exp(u2/4)>0\exp(-u^{2}/4)>0 (claim 5 of Elementary Arithmetic in an Ordered Field) gives the first two bounds. For the third, if u8u\le8 then u383=512u^{3}\le8^{3}=512 (claim 5 of Properties of Natural Number Powers in a Field, monotonicity of powers of nonnegative numbers) and exp(u2/4)1\exp(-u^{2}/4)\le1; if 8u8\le u then exp(u2/4)=exp(u2/8)2(1+u2/8)2(u2/8)2=u4/64\exp(u^{2}/4)=\exp(u^{2}/8)^{2}\ge(1+u^{2}/8)^{2}\ge(u^{2}/8)^{2}=u^{4}/64, so u3exp(u2/4)64u3/u4=64/u8512u^{3}\exp(-u^{2}/4)\le64u^{3}/u^{4}=64/u\le8\le512.

Step (b): pointwise bounds with a Gaussian factor. Fix zz, put u=z/s0u=\lVert z\rVert/\sqrt{s}\ge0, so that z=su\lVert z\rVert=\sqrt{s}\,u and z2/(2s)=u2/2=u2/4+u2/4\lVert z\rVert^{2}/(2s)=u^{2}/2=u^{2}/4+u^{2}/4; hence ψs(z)=exp(u2/4)exp(u2/4)\psi_{s}(z)=\exp(-u^{2}/4)\exp(-u^{2}/4) and, by step (a), for m{1,2,3}m\in\{1,2,3\},

zmψs(z)=(s)mumexp(u2/4)exp(u2/4)Dm(s)mexp(z2/(4s)),\lVert z\rVert^{m}\,\psi_{s}(z)=(\sqrt{s})^{m}\,u^{m}\exp(-u^{2}/4)\,\exp(-u^{2}/4)\le D_{m}(\sqrt{s})^{m}\exp\bigl(-\lVert z\rVert^{2}/(4s)\bigr),

with D1=1D_{1}=1, D2=4D_{2}=4, D3=512D_{3}=512. Since ziz|z_{i}|\le\lVert z\rVert for every ii (claim 4 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n) and δij1|\delta_{ij}|\le1, the formulas of claim 2 and the triangle inequality (claims 4 and 5 of Properties of the Absolute Value in an Ordered Field) give, with Es(z)=exp(z2/(4s))1E_{s}(z)=\exp(-\lVert z\rVert^{2}/(4s))\le1,

gs(z)csEs(z),igs(z)zscsψs(z)csD1ssEs(z)=c1D1s(q+1)/2Es(z),g_{s}(z)\le c_{s}E_{s}(z),\qquad|\partial_{i}g_{s}(z)|\le\frac{\lVert z\rVert}{s}c_{s}\psi_{s}(z)\le c_{s}\,D_{1}\frac{\sqrt{s}}{s}E_{s}(z)=c_{1}D_{1}\,s^{-(q+1)/2}E_{s}(z), jigs(z)(z2s2+1s)csψs(z)c1(D2+1)s(q+2)/2Es(z),kjigs(z)(z3s3+3zs2)csψs(z)c1(D3+3D1)s(q+3)/2Es(z),|\partial_{j}\partial_{i}g_{s}(z)|\le\Bigl(\frac{\lVert z\rVert^{2}}{s^{2}}+\frac1s\Bigr)c_{s}\psi_{s}(z)\le c_{1}(D_{2}+1)\,s^{-(q+2)/2}E_{s}(z),\qquad |\partial_{k}\partial_{j}\partial_{i}g_{s}(z)|\le\Bigl(\frac{\lVert z\rVert^{3}}{s^{3}}+\frac{3\lVert z\rVert}{s^{2}}\Bigr)c_{s}\psi_{s}(z)\le c_{1}(D_{3}+3D_{1})\,s^{-(q+3)/2}E_{s}(z),

using cs=c1sq/2c_{s}=c_{1}s^{-q/2} (claim 1), ψs=Es2Es\psi_{s}=E_{s}^{2}\le E_{s} (since 0<Es10<E_{s}\le1) in the first bound and in the δij/s\delta_{ij}/s term of the third, and the power rules of the preamble (for instance (s)3/s3=s3/2(\sqrt{s})^{3}/s^{3}=s^{-3/2} and s/s2=s3/2\sqrt{s}/s^{2}=s^{-3/2}, so both terms of the last bound carry the same power of ss). Dropping the factor Es1E_{s}\le1 gives the uniform bounds with A0=c1A_{0}=c_{1}, A1=c1D1A_{1}=c_{1}D_{1}, A2=c1(D2+1)A_{2}=c_{1}(D_{2}+1), A3=c1(D3+3D1)A_{3}=c_{1}(D_{3}+3D_{1}), which depend only on qq (through c1c_{1}).

Step (c): square-integral bounds. Each of gsg_{s}, igs\partial_{i}g_{s}, jigs\partial_{j}\partial_{i}g_{s}, kjigs\partial_{k}\partial_{j}\partial_{i}g_{s} is continuous (claim 2), hence Borel, and so is its square (claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions). Since Es(z)2=exp(z2/(2s))=ψs(z)E_{s}(z)^{2}=\exp(-\lVert z\rVert^{2}/(2s))=\psi_{s}(z), writing βgs\partial^{\beta}g_{s} for any one of these four functions and β{0,1,2,3}|\beta|\in\{0,1,2,3\} for its order, squaring the bounds of step (b) gives (βgs)2Aβ2s(q+β)ψs(\partial^{\beta}g_{s})^{2}\le A_{|\beta|}^{2}\,s^{-(q+|\beta|)}\,\psi_{s} and s(q+β)=(s(q+β)/2)2s^{-(q+|\beta|)}=(s^{-(q+|\beta|)/2})^{2}; by claim 1 of Linearity and Monotonicity of the Lebesgue Integral and ψs=1/cs=sq/2/c1\int\psi_{s}=1/c_{s}=s^{q/2}/c_{1},

Rq(βgs)2dλqAβ2s(q+β)(s)qc1=Aβ2c1s(q+2β)/2,\int_{\mathbb{R}^{q}}(\partial^{\beta}g_{s})^{2}\,d\lambda_{q}\le A_{|\beta|}^{2}\,s^{-(q+|\beta|)}\,\frac{(\sqrt{s})^{q}}{c_{1}}=\frac{A_{|\beta|}^{2}}{c_{1}}\,s^{-(q+2|\beta|)/2},

a finite number, so the square is integrable (Integrable Function and the Lebesgue Integral), and Bβ=Aβ2/c1B_{|\beta|}=A_{|\beta|}^{2}/c_{1} serves.

Claim 4 (convolution identity). Let 0<s120<s\le\tfrac12, so that s/21s/2\le1 and 2s12s\le1, and let x,xRqx,x'\in\mathbb{R}^{q}, m=12(x+x)m=\tfrac12(x+x'). For yRqy\in\mathbb{R}^{q} put a=yxa=y-x and b=yxb=y-x', so a+b=2(ym)a+b=2(y-m) and ab=xxa-b=x'-x. Expanding a±b2=(a±b)(a±b)\lVert a\pm b\rVert^{2}=(a\pm b)\cdot(a\pm b) by the bilinearity and symmetry of the dot product (Bilinearity and Symmetry of the Dot Product on Rn\mathbb{R}^n, with v2=vv\lVert v\rVert^{2}=v\cdot v by Euclidean Norm on Rn\mathbb{R}^n) gives a2+b2=12a+b2+12ab2=2ym2+12xx2\lVert a\rVert^{2}+\lVert b\rVert^{2}=\tfrac12\lVert a+b\rVert^{2}+\tfrac12\lVert a-b\rVert^{2}=2\lVert y-m\rVert^{2}+\tfrac12\lVert x-x'\rVert^{2}, using claim 5 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n and xx=xx\lVert x'-x\rVert=\lVert x-x'\rVert (claim 2 there with Euclidean Distance is a Metric on Rn\mathbb{R}^n). Hence

gs(yx)gs(yx)=cs2exp(ym22(s/2))exp(xx22(2s))=cs2cs/2c2sgs/2(ym)g2s(xx),g_{s}(y-x)\,g_{s}(y-x')=c_{s}^{2}\exp\Bigl(-\frac{\lVert y-m\rVert^{2}}{2(s/2)}\Bigr)\exp\Bigl(-\frac{\lVert x-x'\rVert^{2}}{2(2s)}\Bigr)=\frac{c_{s}^{2}}{c_{s/2}\,c_{2s}}\,g_{s/2}(y-m)\,g_{2s}(x-x'),

and cs/2c2s=c12(s/22s)q=c12(s2)q=c12sq=cs2c_{s/2}c_{2s}=c_{1}^{2}\bigl(\sqrt{s/2}\,\sqrt{2s}\bigr)^{-q}=c_{1}^{2}(\sqrt{s^{2}})^{-q}=c_{1}^{2}s^{-q}=c_{s}^{2} by claim 1 and ab=ab\sqrt{a}\sqrt{b}=\sqrt{ab}. So ygs(yx)gs(yx)y\mapsto g_{s}(y-x)g_{s}(y-x') equals g2s(xx)gs/2(ym)g_{2s}(x-x')\cdot g_{s/2}(y-m), a nonnegative multiple of a Borel function with integral 11; it is integrable with integral g2s(xx)g_{2s}(x-x') by claim 1 of Linearity and Monotonicity of the Lebesgue Integral.

Claim 5 (moments and tails). For i[q]i\in[q] apply claim 4 of The Gaussian Smoothing Weight: Normalization, Derivatives, Exponential Tilting, Moments, and First-Order Remainder with a=eia=e_{i}, the iith standard basis vector: Za(z)=zi/sZ_{a}(z)=z_{i}/s and κa=ei2/s=1/s\kappa_{a}=\lVert e_{i}\rVert^{2}/s=1/s, so Rqzi2s2gs(z)dz=1s\int_{\mathbb{R}^{q}}\frac{z_{i}^{2}}{s^{2}}g_{s}(z)\,dz=\frac1s, whence Rqzi2gs(z)dz=s\int_{\mathbb{R}^{q}}z_{i}^{2}g_{s}(z)\,dz=s by claim 1 of Linearity and Monotonicity of the Lebesgue Integral (multiplication by the nonnegative constant s2s^{2}). Since z2=i=1qzi2\lVert z\rVert^{2}=\sum_{i=1}^{q}z_{i}^{2} (Euclidean Norm on Rn\mathbb{R}^n), the additivity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral, applied q1q-1 times by induction on the number of summands (the inductive set being the set of nqn\le q for which the integral of i=1nzi2gs\sum_{i=1}^{n}z_{i}^{2}g_{s} is nsns, with Principle of Induction for the Natural Numbers and claim 1 of Properties of Finite Sums), gives z2gs=qs\int\lVert z\rVert^{2}g_{s}=qs, finite, so zz2gs(z)z\mapsto\lVert z\rVert^{2}g_{s}(z) (Borel by claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions and Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §functions) is integrable. Next, let t=qst=\sqrt{qs}, which is positive since qs>0qs>0 (claim 5 of Elementary Order Arithmetic in an Ordered Field with claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field) and t2=qst^{2}=qs. For every zz, 0(zt)2=z22tz+t20\le(\lVert z\rVert-t)^{2}=\lVert z\rVert^{2}-2t\lVert z\rVert+t^{2} (claim 5 of Zero Products and Elementary Identities in a Field and claim 2 of Nonnegativity of Squares in an Ordered Field), so 2tzz2+t22t\lVert z\rVert\le\lVert z\rVert^{2}+t^{2} (claim 3 of Elementary Arithmetic in an Ordered Field). Multiplying by gs(z)0g_{s}(z)\ge0 (claim 5 of Elementary Arithmetic in an Ordered Field) and integrating, by claim 1 of Linearity and Monotonicity of the Lebesgue Integral with gs=1\int g_{s}=1 and the first moment identity,

2tRqzgs(z)dzqs+t2=2t2,2t\int_{\mathbb{R}^{q}}\lVert z\rVert g_{s}(z)\,dz\le qs+t^{2}=2t^{2},

so the integral of the nonnegative Borel function zzgs(z)z\mapsto\lVert z\rVert g_{s}(z) is finite, the function is integrable, and zgs(z)dzt=qs\int\lVert z\rVert g_{s}(z)\,dz\le t=\sqrt{qs} (multiplying by the positive (2t)1(2t)^{-1}, claim 5 of Elementary Arithmetic in an Ordered Field). Finally let R>0R>0. For zARz\in A_{R} one has R<zR<\lVert z\rVert, hence R2<z2R^{2}<\lVert z\rVert^{2} (claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field) and 1<z2/R21<\lVert z\rVert^{2}/R^{2} (claim 10 of Elementary Order Arithmetic in an Ordered Field with the positive factor R2R^{-2}); for zARz\notin A_{R} the indicator vanishes. Thus 1AR(z)gs(z)R2z2gs(z)\mathbf{1}_{A_{R}}(z)\,g_{s}(z)\le R^{-2}\lVert z\rVert^{2}g_{s}(z) for every zz, and the monotonicity and homogeneity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral give 1ARgsdλqR2qs\int\mathbf{1}_{A_{R}}g_{s}\,d\lambda_{q}\le R^{-2}\,qs, as claimed.

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