Each result cited is universally quantified over the data in its own statement. Throughout, s s s satisfies 0 < s ≤ 1 0<s\le1 0 < s ≤ 1 , and we write ψ s ( z ) = exp ( − ∥ z ∥ 2 / ( 2 s ) ) \psi_{s}(z)=\exp(-\lVert z\rVert^{2}/(2s)) ψ s ( z ) = exp ( − ∥ z ∥ 2 / ( 2 s )) , so that g s = c s ψ s g_{s}=c_{s}\psi_{s} g s = c s ψ s , with ∫ R q ψ s d λ q = 1 / c s \int_{\mathbb{R}^{q}}\psi_{s}\,d\lambda_{q}=1/c_{s} ∫ R q ψ s d λ 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 = q m=q m = q with η = s \eta=s η = s ) and the homogeneity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral with the nonnegative factor 1 / c s 1/c_{s} 1/ c s ; that claim also gives 0 < ψ s ≤ 1 0<\psi_{s}\le1 0 < ψ s ≤ 1 , hence 0 < g s ≤ c s 0<g_{s}\le c_{s} 0 < g s ≤ c s , the evenness g s ( − z ) = g s ( z ) g_{s}(-z)=g_{s}(z) g s ( − z ) = g s ( z ) , and, for every a ∈ R q a\in\mathbb{R}^{q} a ∈ R q , that z ↦ g s ( z − a ) z\mapsto g_{s}(z-a) z ↦ g s ( z − a ) is Borel with ∫ R q g s ( z − a ) d z = 1 \int_{\mathbb{R}^{q}}g_{s}(z-a)\,dz=1 ∫ R q g s ( z − a ) d z = 1 . By that lemma ψ s \psi_{s} ψ s and g s g_{s} g s are sequentially continuous, hence continuous on R q \mathbb{R}^{q} 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} ( s ) m obey ( s ) m ( s ) n = ( s ) m + n (\sqrt{s})^{m}(\sqrt{s})^{n}=(\sqrt{s})^{m+n} ( s ) m ( s ) n = ( 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 = s s = s (\sqrt{s})^{2}=\sqrt{s}\sqrt{s}=s ( s ) 2 = s s = s (claim 1 of Properties of Natural Number Powers in a Field and Existence and Uniqueness of the Nonnegative Square Root ), so that s − m / 2 s − n / 2 = s − ( m + n ) / 2 s^{-m/2}s^{-n/2}=s^{-(m+n)/2} s − m /2 s − n /2 = s − ( m + n ) /2 and s − 2 / 2 = s − 1 s^{-2/2}=s^{-1} s − 2/2 = s − 1 ; and a b = a b \sqrt{a}\sqrt{b}=\sqrt{ab} a b = ab for nonnegative a , b a,b a , 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 + v ) = exp ( u ) exp ( v ) , exp ( u ) > 0 \exp(u)>0 exp ( u ) > 0 , and exp ( u ) ≥ 1 + u \exp(u)\ge1+u exp ( u ) ≥ 1 + u for u ≥ 0 u\ge0 u ≥ 0 (claims 1, 2 and 4 of Basic Properties of the Exponential Function ); in particular, for u ≥ 0 u\ge0 u ≥ 0 , exp ( u ) ≥ 1 \exp(u)\ge1 exp ( u ) ≥ 1 and exp ( − u ) = 1 / exp ( u ) ≤ 1 \exp(-u)=1/\exp(u)\le1 exp ( − u ) = 1/ exp ( u ) ≤ 1 (the inverse of a number ≥ 1 \ge1 ≥ 1 is ≤ 1 \le1 ≤ 1 , by claim 5 of Elementary Arithmetic in an Ordered Field applied to 1 ≤ exp ( u ) 1\le\exp(u) 1 ≤ exp ( u ) with the nonnegative multiplier 1 / exp ( u ) 1/\exp(u) 1/ exp ( u ) , claim 4 there).
Claim 1 (scaling). Let c = s − 1 / 2 c=s^{-1/2} c = s − 1/2 . For z ∈ R q z\in\mathbb{R}^{q} z ∈ R q , ∥ c z ∥ 2 = c 2 ∥ z ∥ 2 = s − 1 ∥ z ∥ 2 \lVert cz\rVert^{2}=c^{2}\lVert z\rVert^{2}=s^{-1}\lVert z\rVert^{2} ∥ cz ∥ 2 = c 2 ∥ z ∥ 2 = s − 1 ∥ z ∥ 2 by claim 5 of Elementary Properties of the Euclidean Norm on R n \mathbb{R}^n R n , so ψ 1 ( c z ) = exp ( − ∥ z ∥ 2 / ( 2 s ) ) = ψ s ( z ) \psi_{1}(cz)=\exp(-\lVert z\rVert^{2}/(2s))=\psi_{s}(z) ψ 1 ( cz ) = exp ( − ∥ z ∥ 2 / ( 2 s )) = ψ s ( z ) . By claim 2 of Scaling of Lebesgue Measure and the Lebesgue Integral on R n \mathbb{R}^n R n , applied to the nonnegative Borel function ψ 1 \psi_{1} ψ 1 ,
1 c s = ∫ R q ψ s d λ q = ∫ R q ψ 1 ( c z ) d z = ∣ c ∣ − q ∫ R q ψ 1 d λ q = ( s ) q c 1 , \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}}, c s 1 = ∫ R q ψ s d λ q = ∫ R q ψ 1 ( cz ) d z = ∣ c ∣ − q ∫ R q ψ 1 d λ q = c 1 ( s ) q ,
since ∣ c ∣ q = ( s − 1 / 2 ) q |c|^{q}=(s^{-1/2})^{q} ∣ c ∣ q = ( s − 1/2 ) q is the inverse of ( s ) q (\sqrt{s})^{q} ( s ) q . Hence c s = c 1 s − q / 2 c_{s}=c_{1}\,s^{-q/2} c s = c 1 s − q /2 , and g s ( z ) = c s ψ s ( z ) = s − q / 2 c 1 ψ 1 ( s − 1 / 2 z ) = s − q / 2 g 1 ( s − 1 / 2 z ) 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) g s ( z ) = c s ψ s ( z ) = s − q /2 c 1 ψ 1 ( s − 1/2 z ) = s − q /2 g 1 ( s − 1/2 z ) .
Claim 2 (smoothness and derivatives). Let C \mathcal{C} C be the set of functions F : R q → R F:\mathbb{R}^{q}\to\mathbb{R} F : R q → R of the form F = h g s F=h\,g_{s} F = h g s (pointwise product) with h : R q → R h:\mathbb{R}^{q}\to\mathbb{R} h : R q → R smooth on R q \mathbb{R}^{q} R q . Every F ∈ C F\in\mathcal{C} F ∈ 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 h h h is of class C 1 C^{1} C 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 ∂ i g s ( z ) = − z i s g s ( z ) \partial_{i}g_{s}(z)=-\frac{z_{i}}{s}g_{s}(z) ∂ i g s ( z ) = − s z i g s ( z ) exists at every z z z , so by the product rule in claim 1 of Constants, Coordinate Functions, Sums and Products of C k C^k C k Functions on a Euclidean Open Set , for F = h g s ∈ C F=h\,g_{s}\in\mathcal{C} F = h g s ∈ C ,
∂ i F ( z ) = ∂ i h ( z ) g s ( z ) + h ( z ) ∂ i g s ( z ) = ( ∂ i h ( z ) − z i s h ( z ) ) g s ( 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), ∂ i F ( z ) = ∂ i h ( z ) g s ( z ) + h ( z ) ∂ i g s ( z ) = ( ∂ i h ( z ) − s z i h ( z ) ) g s ( z ) ,
and ∂ i h − z i s h \partial_{i}h-\frac{z_{i}}{s}h ∂ i h − s z i h is smooth: ∂ i h \partial_{i}h ∂ i h is of class C k C^{k} C k for every k k k by clause 2 of C^k Maps on a Euclidean Open Set applied to h ∈ C k + 1 h\in C^{k+1} h ∈ C k + 1 , hence smooth; the coordinate function z ↦ z i z\mapsto z_{i} z ↦ z i and the constant − 1 s -\frac1s − s 1 are smooth by claim 2 of Constants, Coordinate Functions, Sums and Products of C k C^k C k Functions on a Euclidean Open Set ; and products and sums of smooth functions are smooth by claim 3 there. Thus ∂ i F ∈ C \partial_{i}F\in\mathcal{C} ∂ i F ∈ C , in particular ∂ i F \partial_{i}F ∂ i F is continuous, and so every F ∈ C F\in\mathcal{C} F ∈ C is of class C 1 C^{1} C 1 (clauses 1 and 3 of C^k Maps on a Euclidean Open Set ) with all partial derivatives in C \mathcal{C} C . Let T T T be the set of k ∈ N k\in\mathbb{N} k ∈ N such that every member of C \mathcal{C} C is of class C k C^{k} C k ; then 1 ∈ T 1\in T 1 ∈ T , and if k ∈ T k\in T k ∈ T then every F ∈ C F\in\mathcal{C} F ∈ C is of class C 1 C^{1} C 1 with partial derivatives of class C k C^{k} C k , hence of class C k + 1 C^{k+1} C k + 1 by clause 2 of C^k Maps on a Euclidean Open Set , so k + 1 ∈ T k+1\in T k + 1 ∈ T ; by Principle of Induction for the Natural Numbers , T = N T=\mathbb{N} T = N , and every member of C \mathcal{C} C is smooth. In particular g s = 1 ⋅ g s ∈ C g_{s}=1\cdot g_{s}\in\mathcal{C} g s = 1 ⋅ g s ∈ 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 = ∂ i g s F=\partial_{i}g_{s} F = ∂ i g s , that is h ( z ) = − z i s h(z)=-\frac{z_{i}}{s} h ( z ) = − s z i , with ∂ j h = − δ i j s \partial_{j}h=-\frac{\delta_{ij}}{s} ∂ j h = − s δ ij (the partial derivative of a coordinate function is 1 1 1 or 0 0 0 according as the indices agree or not, by Partial Derivative on a Euclidean Open Set ), gives
∂ j ∂ i g s ( z ) = ( − δ i j s + z j z i s 2 ) g s ( z ) ; \partial_{j}\partial_{i}g_{s}(z)=\Bigl(-\frac{\delta_{ij}}{s}+\frac{z_{j}z_{i}}{s^{2}}\Bigr)g_{s}(z); ∂ j ∂ i g s ( z ) = ( − s δ ij + s 2 z j z i ) g s ( z ) ;
and applying it once more to h ( z ) = z i z j s 2 − δ i j s h(z)=\frac{z_{i}z_{j}}{s^{2}}-\frac{\delta_{ij}}{s} h ( z ) = s 2 z i z j − s δ ij , with ∂ k h ( z ) = δ i k z j + δ j k z i s 2 \partial_{k}h(z)=\frac{\delta_{ik}z_{j}+\delta_{jk}z_{i}}{s^{2}} ∂ k h ( z ) = s 2 δ ik z j + δ jk z i (product rule, claim 1 of Constants, Coordinate Functions, Sums and Products of C k C^k C k Functions on a Euclidean Open Set ), gives
∂ k ∂ j ∂ i g s ( z ) = ( δ i k z j + δ j k z i s 2 − z k s ( z i z j s 2 − δ i j s ) ) g s ( z ) = ( − z i z j z k s 3 + δ i j z k + δ i k z j + δ j k z i s 2 ) g s ( 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), ∂ k ∂ j ∂ i g s ( z ) = ( s 2 δ ik z j + δ jk z i − s z k ( s 2 z i z j − s δ ij ) ) g s ( z ) = ( − s 3 z i z j z k + s 2 δ ij z k + δ ik z j + δ jk z i ) g s ( z ) ,
the stated formulas.
Claim 3 (bounds). Step (a): an elementary inequality. For every real u ≥ 0 u\ge0 u ≥ 0 ,
u exp ( − u 2 / 4 ) ≤ 1 , u 2 exp ( − u 2 / 4 ) ≤ 4 , u 3 exp ( − u 2 / 4 ) ≤ 512. u\,\exp(-u^{2}/4)\le1,\qquad u^{2}\exp(-u^{2}/4)\le4,\qquad u^{3}\exp(-u^{2}/4)\le512 . u exp ( − u 2 /4 ) ≤ 1 , u 2 exp ( − u 2 /4 ) ≤ 4 , u 3 exp ( − u 2 /4 ) ≤ 512.
Indeed, exp ( u 2 / 4 ) ≥ 1 + u 2 / 4 ≥ u \exp(u^{2}/4)\ge1+u^{2}/4\ge u exp ( u 2 /4 ) ≥ 1 + u 2 /4 ≥ u , the last by 0 ≤ ( u / 2 − 1 ) 2 = u 2 / 4 − u + 1 0\le(u/2-1)^{2}=u^{2}/4-u+1 0 ≤ ( 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 ( u 2 / 4 ) ≥ 1 + u 2 / 4 ≥ u 2 / 4 \exp(u^{2}/4)\ge1+u^{2}/4\ge u^{2}/4 exp ( u 2 /4 ) ≥ 1 + u 2 /4 ≥ u 2 /4 ; multiplying by exp ( − u 2 / 4 ) > 0 \exp(-u^{2}/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 u ≤ 8 u\le8 u ≤ 8 then u 3 ≤ 8 3 = 512 u^{3}\le8^{3}=512 u 3 ≤ 8 3 = 512 (claim 5 of Properties of Natural Number Powers in a Field , monotonicity of powers of nonnegative numbers) and exp ( − u 2 / 4 ) ≤ 1 \exp(-u^{2}/4)\le1 exp ( − u 2 /4 ) ≤ 1 ; if 8 ≤ u 8\le u 8 ≤ u then exp ( u 2 / 4 ) = exp ( u 2 / 8 ) 2 ≥ ( 1 + u 2 / 8 ) 2 ≥ ( u 2 / 8 ) 2 = u 4 / 64 \exp(u^{2}/4)=\exp(u^{2}/8)^{2}\ge(1+u^{2}/8)^{2}\ge(u^{2}/8)^{2}=u^{4}/64 exp ( u 2 /4 ) = exp ( u 2 /8 ) 2 ≥ ( 1 + u 2 /8 ) 2 ≥ ( u 2 /8 ) 2 = u 4 /64 , so u 3 exp ( − u 2 / 4 ) ≤ 64 u 3 / u 4 = 64 / u ≤ 8 ≤ 512 u^{3}\exp(-u^{2}/4)\le64u^{3}/u^{4}=64/u\le8\le512 u 3 exp ( − u 2 /4 ) ≤ 64 u 3 / u 4 = 64/ u ≤ 8 ≤ 512 .
Step (b): pointwise bounds with a Gaussian factor. Fix z z z , put u = ∥ z ∥ / s ≥ 0 u=\lVert z\rVert/\sqrt{s}\ge0 u = ∥ z ∥ / s ≥ 0 , so that ∥ z ∥ = s u \lVert z\rVert=\sqrt{s}\,u ∥ z ∥ = s u and ∥ z ∥ 2 / ( 2 s ) = u 2 / 2 = u 2 / 4 + u 2 / 4 \lVert z\rVert^{2}/(2s)=u^{2}/2=u^{2}/4+u^{2}/4 ∥ z ∥ 2 / ( 2 s ) = u 2 /2 = u 2 /4 + u 2 /4 ; hence ψ s ( z ) = exp ( − u 2 / 4 ) exp ( − u 2 / 4 ) \psi_{s}(z)=\exp(-u^{2}/4)\exp(-u^{2}/4) ψ s ( z ) = exp ( − u 2 /4 ) exp ( − u 2 /4 ) and, by step (a), for m ∈ { 1 , 2 , 3 } m\in\{1,2,3\} m ∈ { 1 , 2 , 3 } ,
∥ z ∥ m ψ s ( z ) = ( s ) m u m exp ( − u 2 / 4 ) exp ( − u 2 / 4 ) ≤ D m ( s ) m exp ( − ∥ z ∥ 2 / ( 4 s ) ) , \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), ∥ z ∥ m ψ s ( z ) = ( s ) m u m exp ( − u 2 /4 ) exp ( − u 2 /4 ) ≤ D m ( s ) m exp ( − ∥ z ∥ 2 / ( 4 s ) ) ,
with D 1 = 1 D_{1}=1 D 1 = 1 , D 2 = 4 D_{2}=4 D 2 = 4 , D 3 = 512 D_{3}=512 D 3 = 512 . Since ∣ z i ∣ ≤ ∥ z ∥ |z_{i}|\le\lVert z\rVert ∣ z i ∣ ≤ ∥ z ∥ for every i i i (claim 4 of Elementary Properties of the Euclidean Norm on R n \mathbb{R}^n R n ) and ∣ δ i j ∣ ≤ 1 |\delta_{ij}|\le1 ∣ δ ij ∣ ≤ 1 , 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 E s ( z ) = exp ( − ∥ z ∥ 2 / ( 4 s ) ) ≤ 1 E_{s}(z)=\exp(-\lVert z\rVert^{2}/(4s))\le1 E s ( z ) = exp ( − ∥ z ∥ 2 / ( 4 s )) ≤ 1 ,
g s ( z ) ≤ c s E s ( z ) , ∣ ∂ i g s ( z ) ∣ ≤ ∥ z ∥ s c s ψ s ( z ) ≤ c s D 1 s s E s ( z ) = c 1 D 1 s − ( q + 1 ) / 2 E s ( 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), g s ( z ) ≤ c s E s ( z ) , ∣ ∂ i g s ( z ) ∣ ≤ s ∥ z ∥ c s ψ s ( z ) ≤ c s D 1 s s E s ( z ) = c 1 D 1 s − ( q + 1 ) /2 E s ( z ) ,
∣ ∂ j ∂ i g s ( z ) ∣ ≤ ( ∥ z ∥ 2 s 2 + 1 s ) c s ψ s ( z ) ≤ c 1 ( D 2 + 1 ) s − ( q + 2 ) / 2 E s ( z ) , ∣ ∂ k ∂ j ∂ i g s ( z ) ∣ ≤ ( ∥ z ∥ 3 s 3 + 3 ∥ z ∥ s 2 ) c s ψ s ( z ) ≤ c 1 ( D 3 + 3 D 1 ) s − ( q + 3 ) / 2 E s ( 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), ∣ ∂ j ∂ i g s ( z ) ∣ ≤ ( s 2 ∥ z ∥ 2 + s 1 ) c s ψ s ( z ) ≤ c 1 ( D 2 + 1 ) s − ( q + 2 ) /2 E s ( z ) , ∣ ∂ k ∂ j ∂ i g s ( z ) ∣ ≤ ( s 3 ∥ z ∥ 3 + s 2 3 ∥ z ∥ ) c s ψ s ( z ) ≤ c 1 ( D 3 + 3 D 1 ) s − ( q + 3 ) /2 E s ( z ) ,
using c s = c 1 s − q / 2 c_{s}=c_{1}s^{-q/2} c s = c 1 s − q /2 (claim 1), ψ s = E s 2 ≤ E s \psi_{s}=E_{s}^{2}\le E_{s} ψ s = E s 2 ≤ E s (since 0 < E s ≤ 1 0<E_{s}\le1 0 < E s ≤ 1 ) in the first bound and in the δ i j / s \delta_{ij}/s δ ij / s term of the third, and the power rules of the preamble (for instance ( s ) 3 / s 3 = s − 3 / 2 (\sqrt{s})^{3}/s^{3}=s^{-3/2} ( s ) 3 / s 3 = s − 3/2 and s / s 2 = s − 3 / 2 \sqrt{s}/s^{2}=s^{-3/2} s / s 2 = s − 3/2 , so both terms of the last bound carry the same power of s s s ). Dropping the factor E s ≤ 1 E_{s}\le1 E s ≤ 1 gives the uniform bounds with A 0 = c 1 A_{0}=c_{1} A 0 = c 1 , A 1 = c 1 D 1 A_{1}=c_{1}D_{1} A 1 = c 1 D 1 , A 2 = c 1 ( D 2 + 1 ) A_{2}=c_{1}(D_{2}+1) A 2 = c 1 ( D 2 + 1 ) , A 3 = c 1 ( D 3 + 3 D 1 ) A_{3}=c_{1}(D_{3}+3D_{1}) A 3 = c 1 ( D 3 + 3 D 1 ) , which depend only on q q q (through c 1 c_{1} c 1 ).
Step (c): square-integral bounds. Each of g s g_{s} g s , ∂ i g s \partial_{i}g_{s} ∂ i g s , ∂ j ∂ i g s \partial_{j}\partial_{i}g_{s} ∂ j ∂ i g s , ∂ k ∂ j ∂ i g s \partial_{k}\partial_{j}\partial_{i}g_{s} ∂ k ∂ j ∂ 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 E s ( z ) 2 = exp ( − ∥ z ∥ 2 / ( 2 s ) ) = ψ s ( z ) E_{s}(z)^{2}=\exp(-\lVert z\rVert^{2}/(2s))=\psi_{s}(z) E s ( z ) 2 = exp ( − ∥ z ∥ 2 / ( 2 s )) = ψ s ( z ) , writing ∂ β g s \partial^{\beta}g_{s} ∂ β g s for any one of these four functions and ∣ β ∣ ∈ { 0 , 1 , 2 , 3 } |\beta|\in\{0,1,2,3\} ∣ β ∣ ∈ { 0 , 1 , 2 , 3 } for its order, squaring the bounds of step (b) gives ( ∂ β g s ) 2 ≤ A ∣ β ∣ 2 s − ( q + ∣ β ∣ ) ψ s (\partial^{\beta}g_{s})^{2}\le A_{|\beta|}^{2}\,s^{-(q+|\beta|)}\,\psi_{s} ( ∂ β g s ) 2 ≤ A ∣ β ∣ 2 s − ( q + ∣ β ∣ ) ψ s and s − ( q + ∣ β ∣ ) = ( s − ( q + ∣ β ∣ ) / 2 ) 2 s^{-(q+|\beta|)}=(s^{-(q+|\beta|)/2})^{2} s − ( q + ∣ β ∣ ) = ( s − ( q + ∣ β ∣ ) /2 ) 2 ; by claim 1 of Linearity and Monotonicity of the Lebesgue Integral and ∫ ψ s = 1 / c s = s q / 2 / c 1 \int\psi_{s}=1/c_{s}=s^{q/2}/c_{1} ∫ ψ s = 1/ c s = s q /2 / c 1 ,
∫ R q ( ∂ β g s ) 2 d λ q ≤ A ∣ β ∣ 2 s − ( q + ∣ β ∣ ) ( s ) q c 1 = A ∣ β ∣ 2 c 1 s − ( q + 2 ∣ β ∣ ) / 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}, ∫ R q ( ∂ β g s ) 2 d λ q ≤ A ∣ β ∣ 2 s − ( q + ∣ β ∣ ) c 1 ( s ) q = c 1 A ∣ β ∣ 2 s − ( q + 2∣ β ∣ ) /2 ,
a finite number, so the square is integrable (Integrable Function and the Lebesgue Integral ), and B ∣ β ∣ = A ∣ β ∣ 2 / c 1 B_{|\beta|}=A_{|\beta|}^{2}/c_{1} B ∣ β ∣ = A ∣ β ∣ 2 / c 1 serves.
Claim 4 (convolution identity). Let 0 < s ≤ 1 2 0<s\le\tfrac12 0 < s ≤ 2 1 , so that s / 2 ≤ 1 s/2\le1 s /2 ≤ 1 and 2 s ≤ 1 2s\le1 2 s ≤ 1 , and let x , x ′ ∈ R q x,x'\in\mathbb{R}^{q} x , x ′ ∈ R q , m = 1 2 ( x + x ′ ) m=\tfrac12(x+x') m = 2 1 ( x + x ′ ) . For y ∈ R q y\in\mathbb{R}^{q} y ∈ R q put a = y − x a=y-x a = y − x and b = y − x ′ b=y-x' b = y − x ′ , so a + b = 2 ( y − m ) a+b=2(y-m) a + b = 2 ( y − m ) and a − b = x ′ − x a-b=x'-x a − b = x ′ − x . Expanding ∥ a ± b ∥ 2 = ( a ± b ) ⋅ ( a ± b ) \lVert a\pm b\rVert^{2}=(a\pm b)\cdot(a\pm b) ∥ a ± b ∥ 2 = ( a ± b ) ⋅ ( a ± b ) by the bilinearity and symmetry of the dot product (Bilinearity and Symmetry of the Dot Product on R n \mathbb{R}^n R n , with ∥ v ∥ 2 = v ⋅ v \lVert v\rVert^{2}=v\cdot v ∥ v ∥ 2 = v ⋅ v by Euclidean Norm on R n \mathbb{R}^n R n ) gives ∥ a ∥ 2 + ∥ b ∥ 2 = 1 2 ∥ a + b ∥ 2 + 1 2 ∥ a − b ∥ 2 = 2 ∥ y − m ∥ 2 + 1 2 ∥ x − x ′ ∥ 2 \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} ∥ a ∥ 2 + ∥ b ∥ 2 = 2 1 ∥ a + b ∥ 2 + 2 1 ∥ a − b ∥ 2 = 2 ∥ y − m ∥ 2 + 2 1 ∥ x − x ′ ∥ 2 , using claim 5 of Elementary Properties of the Euclidean Norm on R n \mathbb{R}^n R n and ∥ x ′ − x ∥ = ∥ x − x ′ ∥ \lVert x'-x\rVert=\lVert x-x'\rVert ∥ x ′ − x ∥ = ∥ x − x ′ ∥ (claim 2 there with Euclidean Distance is a Metric on R n \mathbb{R}^n R n ). Hence
g s ( y − x ) g s ( y − x ′ ) = c s 2 exp ( − ∥ y − m ∥ 2 2 ( s / 2 ) ) exp ( − ∥ x − x ′ ∥ 2 2 ( 2 s ) ) = c s 2 c s / 2 c 2 s g s / 2 ( y − m ) g 2 s ( x − x ′ ) , 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'), g s ( y − x ) g s ( y − x ′ ) = c s 2 exp ( − 2 ( s /2 ) ∥ y − m ∥ 2 ) exp ( − 2 ( 2 s ) ∥ x − x ′ ∥ 2 ) = c s /2 c 2 s c s 2 g s /2 ( y − m ) g 2 s ( x − x ′ ) ,
and c s / 2 c 2 s = c 1 2 ( s / 2 2 s ) − q = c 1 2 ( s 2 ) − q = c 1 2 s − q = c s 2 c_{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} c s /2 c 2 s = c 1 2 ( s /2 2 s ) − q = c 1 2 ( s 2 ) − q = c 1 2 s − q = c s 2 by claim 1 and a b = a b \sqrt{a}\sqrt{b}=\sqrt{ab} a b = ab . So y ↦ g s ( y − x ) g s ( y − x ′ ) y\mapsto g_{s}(y-x)g_{s}(y-x') y ↦ g s ( y − x ) g s ( y − x ′ ) equals g 2 s ( x − x ′ ) ⋅ g s / 2 ( y − m ) g_{2s}(x-x')\cdot g_{s/2}(y-m) g 2 s ( x − x ′ ) ⋅ g s /2 ( y − m ) , a nonnegative multiple of a Borel function with integral 1 1 1 ; it is integrable with integral g 2 s ( x − x ′ ) g_{2s}(x-x') g 2 s ( x − x ′ ) by claim 1 of Linearity and Monotonicity of the Lebesgue Integral .
Claim 5 (moments and tails). For i ∈ [ q ] i\in[q] i ∈ [ q ] apply claim 4 of The Gaussian Smoothing Weight: Normalization, Derivatives, Exponential Tilting, Moments, and First-Order Remainder with a = e i a=e_{i} a = e i , the i i i th standard basis vector: Z a ( z ) = z i / s Z_{a}(z)=z_{i}/s Z a ( z ) = z i / s and κ a = ∥ e i ∥ 2 / s = 1 / s \kappa_{a}=\lVert e_{i}\rVert^{2}/s=1/s κ a = ∥ e i ∥ 2 / s = 1/ s , so ∫ R q z i 2 s 2 g s ( z ) d z = 1 s \int_{\mathbb{R}^{q}}\frac{z_{i}^{2}}{s^{2}}g_{s}(z)\,dz=\frac1s ∫ R q s 2 z i 2 g s ( z ) d z = s 1 , whence ∫ R q z i 2 g s ( z ) d z = s \int_{\mathbb{R}^{q}}z_{i}^{2}g_{s}(z)\,dz=s ∫ R q z i 2 g s ( z ) d z = s by claim 1 of Linearity and Monotonicity of the Lebesgue Integral (multiplication by the nonnegative constant s 2 s^{2} s 2 ). Since ∥ z ∥ 2 = ∑ i = 1 q z i 2 \lVert z\rVert^{2}=\sum_{i=1}^{q}z_{i}^{2} ∥ z ∥ 2 = ∑ i = 1 q z i 2 (Euclidean Norm on R n \mathbb{R}^n R n ), the additivity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral , applied q − 1 q-1 q − 1 times by induction on the number of summands (the inductive set being the set of n ≤ q n\le q n ≤ q for which the integral of ∑ i = 1 n z i 2 g s \sum_{i=1}^{n}z_{i}^{2}g_{s} ∑ i = 1 n z i 2 g s is n s ns n s , with Principle of Induction for the Natural Numbers and claim 1 of Properties of Finite Sums ), gives ∫ ∥ z ∥ 2 g s = q s \int\lVert z\rVert^{2}g_{s}=qs ∫ ∥ z ∥ 2 g s = q s , finite, so z ↦ ∥ z ∥ 2 g s ( z ) z\mapsto\lVert z\rVert^{2}g_{s}(z) z ↦ ∥ z ∥ 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 = q s t=\sqrt{qs} t = q s , which is positive since q s > 0 qs>0 q s > 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 t 2 = q s t^{2}=qs t 2 = q s . For every z z z , 0 ≤ ( ∥ z ∥ − t ) 2 = ∥ z ∥ 2 − 2 t ∥ z ∥ + t 2 0\le(\lVert z\rVert-t)^{2}=\lVert z\rVert^{2}-2t\lVert z\rVert+t^{2} 0 ≤ (∥ z ∥ − t ) 2 = ∥ z ∥ 2 − 2 t ∥ z ∥ + 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 2 t ∥ z ∥ ≤ ∥ z ∥ 2 + t 2 2t\lVert z\rVert\le\lVert z\rVert^{2}+t^{2} 2 t ∥ z ∥ ≤ ∥ z ∥ 2 + t 2 (claim 3 of Elementary Arithmetic in an Ordered Field ). Multiplying by g s ( z ) ≥ 0 g_{s}(z)\ge0 g s ( z ) ≥ 0 (claim 5 of Elementary Arithmetic in an Ordered Field ) and integrating, by claim 1 of Linearity and Monotonicity of the Lebesgue Integral with ∫ g s = 1 \int g_{s}=1 ∫ g s = 1 and the first moment identity,
2 t ∫ R q ∥ z ∥ g s ( z ) d z ≤ q s + t 2 = 2 t 2 , 2t\int_{\mathbb{R}^{q}}\lVert z\rVert g_{s}(z)\,dz\le qs+t^{2}=2t^{2}, 2 t ∫ R q ∥ z ∥ g s ( z ) d z ≤ q s + t 2 = 2 t 2 ,
so the integral of the nonnegative Borel function z ↦ ∥ z ∥ g s ( z ) z\mapsto\lVert z\rVert g_{s}(z) z ↦ ∥ z ∥ g s ( z ) is finite, the function is integrable, and ∫ ∥ z ∥ g s ( z ) d z ≤ t = q s \int\lVert z\rVert g_{s}(z)\,dz\le t=\sqrt{qs} ∫ ∥ z ∥ g s ( z ) d z ≤ t = q s (multiplying by the positive ( 2 t ) − 1 (2t)^{-1} ( 2 t ) − 1 , claim 5 of Elementary Arithmetic in an Ordered Field ). Finally let R > 0 R>0 R > 0 . For z ∈ A R z\in A_{R} z ∈ A R one has R < ∥ z ∥ R<\lVert z\rVert R < ∥ z ∥ , hence R 2 < ∥ z ∥ 2 R^{2}<\lVert z\rVert^{2} R 2 < ∥ z ∥ 2 (claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field ) and 1 < ∥ z ∥ 2 / R 2 1<\lVert z\rVert^{2}/R^{2} 1 < ∥ z ∥ 2 / R 2 (claim 10 of Elementary Order Arithmetic in an Ordered Field with the positive factor R − 2 R^{-2} R − 2 ); for z ∉ A R z\notin A_{R} z ∈ / A R the indicator vanishes. Thus 1 A R ( z ) g s ( z ) ≤ R − 2 ∥ z ∥ 2 g s ( z ) \mathbf{1}_{A_{R}}(z)\,g_{s}(z)\le R^{-2}\lVert z\rVert^{2}g_{s}(z) 1 A R ( z ) g s ( z ) ≤ R − 2 ∥ z ∥ 2 g s ( z ) for every z z z , and the monotonicity and homogeneity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral give ∫ 1 A R g s d λ q ≤ R − 2 q s \int\mathbf{1}_{A_{R}}g_{s}\,d\lambda_{q}\le R^{-2}\,qs ∫ 1 A R g s d λ q ≤ R − 2 q s , as claimed.