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Proof of The One-Dimensional Fejer Kernel: Regularity, Nonnegativity, Mass and Far-Field Decay

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· 10,344 chars · 28 deps · depth 27 Reason: Proof of the properties of the one-dimensional Fejer kernel, including nonnegativity at the zeros of the sine by a continuity argument and the far-field bound via the extreme value theorem.

Regularity comes from the cosine maps, the kernel identity is the Fejer identity rescaled, nonnegativity follows off the zeros of the sine and extends to them by continuity along a null sequence, the mass is computed term by term, and the far-field bound uses the extreme value theorem to bound the squared sine away from zero on a closed interval.

Proof

Each result cited is universally quantified over the data in its own statement. Fix NNN\in\mathbb{N} and write, for m[N]m\in[N],

cm=2(Nm)N,c_{m}=\frac{2(N-m)}{N},

so that, by the homogeneity of finite sums, claim 3 of Properties of Finite Sums,

FN(t)=C0(t)+m=1NcmCm(t)(tR),F_{N}(t)=C_{0}(t)+\sum_{m=1}^{N}c_{m}C_{m}(t)\qquad(t\in\mathbb{R}),

the constant term being C0C_{0} because C0(t)=1C_{0}(t)=1 for every tt by Cell Integrals of the Trigonometric Monomials §calculus.

Claim 1. By Cell Integrals of the Trigonometric Monomials §calculus each CmC_{m} is continuous on R\mathbb{R}. Let PP be the set of pNp\in\mathbb{N} for which the map tm=1pcmCm(t)t\mapsto\sum_{m=1}^{p}c_{m}C_{m}(t) is continuous on R\mathbb{R}, where cmc_{m} is extended to every mNm\in\mathbb{N} by the same formula. A sum with one summand equals that summand, by claim 1 of Properties of Finite Sums, so the map for p=1p=1 is c1C1c_{1}C_{1}, continuous by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, which gives scalar multiples; hence 1P1\in P. If pPp\in P, the recursion in claim 1 of Properties of Finite Sums writes the map for p+1p+1 as the sum of the map for pp and cp+1Cp+1c_{p+1}C_{p+1}, continuous by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, which gives sums and scalar multiples; hence p+1Pp+1\in P. By Principle of Induction for the Natural Numbers, P=NP=\mathbb{N}, so the map tm=1NcmCm(t)t\mapsto\sum_{m=1}^{N}c_{m}C_{m}(t) is continuous on R\mathbb{R}.

The constant map C0C_{0} is continuous at every point of R\mathbb{R} by claim 1 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, the point being arbitrary, hence continuous on R\mathbb{R}; so FNF_{N} is continuous on R\mathbb{R} by claim 5 of that theorem. Consequently FNJF_{N}|_{J} is λJ\lambda_{J}-integrable by Zero Extension of a Real-Valued Function, and the Unit-Cell Integral of a Continuous Function §continuous.

For periodicity, let tRt\in\mathbb{R} and jZj\in\mathbb{Z}, and let m[N]m\in[N]. The natural number mm is an integer, so mjZmj\in\mathbb{Z} by claim 2 of Arithmetic, Order and Discreteness of the Integers; since 2πm(t+j)=2πmt+2π(mj)2\pi m(t+j)=2\pi mt+2\pi(mj), Quarter-Turn Identities and Periodicity of Sine and Cosine §integer gives Cm(t+j)=Cm(t)C_{m}(t+j)=C_{m}(t). As this holds for every m[N]m\in[N], and C0C_{0} is constant, the displayed formula for FNF_{N} gives FN(t+j)=FN(t)F_{N}(t+j)=F_{N}(t).

Claim 2. Multiplying the defining formula for FN(t)F_{N}(t) by NN and using N1N=1N\tfrac{1}{N}=1 gives

NFN(t)=N+2m=1N(Nm)cos(2πmt).N\,F_{N}(t)=N+2\sum_{m=1}^{N}(N-m)\cos(2\pi mt).

Multiplying by (sin(πt))2(\sin(\pi t))^{2} and applying The Dirichlet and Fejer Kernel Identities §fejer gives the assertion.

Claim 3. First suppose sin(πt)0\sin(\pi t)\ne0. Then (sin(πt))2(\sin(\pi t))^{2} is nonzero by claim 3 of Zero Products and Elementary Identities in a Field and nonnegative by claim 2 of Nonnegativity of Squares in an Ordered Field, hence positive. Also (sin(Nπt))2(\sin(N\pi t))^{2} is nonnegative, by that same claim 2. If NFN(t)N\,F_{N}(t) were negative, then NFN(t)(sin(πt))2N\,F_{N}(t)(\sin(\pi t))^{2} would be negative, by claim 5 of Elementary Order Arithmetic in an Ordered Field applied to the positive numbers NFN(t)-N\,F_{N}(t) and (sin(πt))2(\sin(\pi t))^{2} together with claim 2 of Zero Products and Elementary Identities in a Field, contradicting claim 2 of the present lemma. Hence 0NFN(t)0\le N\,F_{N}(t), and multiplying by the nonnegative 1N\tfrac{1}{N}, by claim 5 of Elementary Arithmetic in an Ordered Field together with claim 1 of Zero Products and Elementary Identities in a Field, gives 0FN(t)0\le F_{N}(t).

Now suppose sin(πt)=0\sin(\pi t)=0, and let uRu\in\mathbb{R} satisfy 0<u<10<u<1. By Values at Zero, Parity, Addition and Double-Angle Identities for Sine and Cosine §addition,

sin(π(t+u))=sin(πt)cos(πu)+cos(πt)sin(πu)=cos(πt)sin(πu),\sin\bigl(\pi(t+u)\bigr)=\sin(\pi t)\cos(\pi u)+\cos(\pi t)\sin(\pi u)=\cos(\pi t)\sin(\pi u),

using π(t+u)=πt+πu\pi(t+u)=\pi t+\pi u and claim 1 of Zero Products and Elementary Identities in a Field. Here cos(πt)0\cos(\pi t)\ne0: The Pythagorean Identity for Sine and Cosine §identity gives (cos(πt))2+(sin(πt))2=1(\cos(\pi t))^{2}+(\sin(\pi t))^{2}=1, and sin(πt)=0\sin(\pi t)=0 in the case under consideration, so (sin(πt))2=0(\sin(\pi t))^{2}=0 by claim 1 of Zero Products and Elementary Identities in a Field and therefore (cos(πt))2=1(\cos(\pi t))^{2}=1. This is nonzero because 0<10<1 by claim 6 of Elementary Order Arithmetic in an Ordered Field, so cos(πt)0\cos(\pi t)\ne0, again by claim 1 of Zero Products and Elementary Identities in a Field. And sin(πu)0\sin(\pi u)\ne0: the number π\pi is positive, since π=2x0\pi=2x_{0} with 0<x00<x_{0} by The Number Pi §pi and The Least Positive Zero of the Cosine §least-zero, while 0<20<2 by claim 8 of Elementary Order Arithmetic in an Ordered Field, so that claim 5 of that lemma gives 0<π0<\pi; hence 0<πu0<\pi u and πu<π\pi u<\pi by claim 10 of Elementary Order Arithmetic in an Ordered Field applied to 0<u0<u and to u<1u<1, whence 0<sin(πu)0<\sin(\pi u) by Quarter-Turn Identities and Periodicity of Sine and Cosine §positive. So sin(π(t+u))0\sin(\pi(t+u))\ne0 by claim 3 of Zero Products and Elementary Identities in a Field, and the case already treated gives 0FN(t+u)0\le F_{N}(t+u).

Let uj=12ju_{j}=\tfrac{1}{2j} for jNj\in\mathbb{N}. Each uju_{j} is positive, and 22j2\le 2j, because 1j1\le j by claim 4 of Properties of the Order on the Natural Numbers and multiplication on the left by the nonnegative 22 preserves the order, by claim 5 of Elementary Arithmetic in an Ordered Field; so uj12u_{j}\le\tfrac{1}{2} by Order Reversal under Reciprocals, and Summability of the Reciprocals of the Squares §reciprocal and hence uj<1u_{j}<1, since 12<1\tfrac{1}{2}<1 by claim 8 of Elementary Order Arithmetic in an Ordered Field. Therefore 0FN(t+uj)0\le F_{N}(t+u_{j}) for every jNj\in\mathbb{N}, by the previous paragraph.

The sequence (1j)jN(\tfrac{1}{j})_{j\in\mathbb{N}} converges to 00 by claim 3 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, taken with the exponent 11, its claim 1 identifying the real power with the natural power; hence (uj)jN(u_{j})_{j\in\mathbb{N}} converges to 00 by Arithmetic of Limits of Real Sequences, being the multiple of that sequence by 12\tfrac{1}{2}. The constant sequence with value tt converges to tt by Constant Sequences and Index-Shifted Sequences of Real Numbers §constant, so (t+uj)jN(t+u_{j})_{j\in\mathbb{N}} converges to tt by Arithmetic of Limits of Real Sequences. Since FNF_{N} is continuous at tt by claim 1, Continuity Between Metric Spaces is Equivalent to Sequential Continuity §sequential gives that (FN(t+uj))jN(F_{N}(t+u_{j}))_{j\in\mathbb{N}} converges to FN(t)F_{N}(t); and every term of that sequence is nonnegative, so 0FN(t)0\le F_{N}(t) by Order Properties of Limits of Real Sequences.

Claim 4. Each restriction CmJC_{m}|_{J} for m[N]m\in[N] is λJ\lambda_{J}-integrable by Cell Integrals of the Trigonometric Monomials §calculus, so Linearity of the Lebesgue Integral over a Finite Sum of Integrable Functions §integrable and Linearity of the Lebesgue Integral over a Finite Sum of Integrable Functions §linear, applied on the measure space (J,BJ,λJ)(J,\mathcal{B}_{J},\lambda_{J}) to these restrictions with the coefficients cmc_{m}, give that the restriction of tm=1NcmCm(t)t\mapsto\sum_{m=1}^{N}c_{m}C_{m}(t) to JJ is integrable with

Jm=1NcmCmdλJ=m=1NcmJCmdλJ.\int_{J}\sum_{m=1}^{N}c_{m}C_{m}\,d\lambda_{J}=\sum_{m=1}^{N}c_{m}\int_{J}C_{m}\,d\lambda_{J}.

For m[N]m\in[N] one has 1m1\le m by claim 4 of Properties of the Order on the Natural Numbers, so m0m\ne0 because 0<10<1 by claim 6 of Elementary Order Arithmetic in an Ordered Field; hence JCmdλJ=0\int_{J}C_{m}\,d\lambda_{J}=0 by Cell Integrals of the Trigonometric Monomials §single. Every summand on the right is therefore 00 by claim 1 of Zero Products and Elementary Identities in a Field, and the sum is 00 by claim 3 of Properties of Finite Sums applied with the scalar 00.

The same clause Cell Integrals of the Trigonometric Monomials §single, taken with m=0m=0, gives JC0dλJ=1\int_{J}C_{0}\,d\lambda_{J}=1. Since FN=C0+m=1NcmCmF_{N}=C_{0}+\sum_{m=1}^{N}c_{m}C_{m} and both restrictions to JJ are integrable, claim 2 of Linearity and Monotonicity of the Lebesgue Integral gives

JFNdλJ=JC0dλJ+Jm=1NcmCmdλJ=1+0=1.\int_{J}F_{N}\,d\lambda_{J}=\int_{J}C_{0}\,d\lambda_{J}+\int_{J}\sum_{m=1}^{N}c_{m}C_{m}\,d\lambda_{J}=1+0=1 .

Claim 5. By Values at Zero, Parity, Addition and Double-Angle Identities for Sine and Cosine §double-angle, taken with x=πtx=\pi t and using 2(πt)=2πt2(\pi t)=2\pi t, one has cos(2πt)=12(sin(πt))2\cos(2\pi t)=1-2(\sin(\pi t))^{2}, so that

(sin(πt))2=1C1(t)2(tR),\bigl(\sin(\pi t)\bigr)^{2}=\frac{1-C_{1}(t)}{2}\qquad(t\in\mathbb{R}),

the number 22 being positive, hence invertible, by claim 8 of Elementary Order Arithmetic in an Ordered Field. Write gg for the map on the right-hand side; it is continuous on R\mathbb{R} by claims 1 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, since C1C_{1} is continuous by Cell Integrals of the Trigonometric Monomials §calculus, and hence continuous on KδK_{\delta} by claim 4 of Semicontinuity and Continuity Under Composition with a Continuous Map, which gives restrictions.

From 2δ<12\delta<1 and the compatibility of the order with addition one gets δ<1δ\delta<1-\delta, so KδK_{\delta} is the closed interval determined by δ\delta and 1δ1-\delta and is nonempty. By Extreme Value Theorem on a Closed Real Interval there is tδKδt_{\delta}\in K_{\delta} with g(tδ)g(t)g(t_{\delta})\le g(t) for every tKδt\in K_{\delta}; put cδ=g(tδ)=(sin(πtδ))2c_{\delta}=g(t_{\delta})=(\sin(\pi t_{\delta}))^{2}. Since 0<δtδ0<\delta\le t_{\delta} and tδ1δ<1t_{\delta}\le1-\delta<1, and π\pi is positive, claim 10 of Elementary Order Arithmetic in an Ordered Field gives 0<πtδ0<\pi t_{\delta} and πtδ<π\pi t_{\delta}<\pi; so 0<sin(πtδ)0<\sin(\pi t_{\delta}) by Quarter-Turn Identities and Periodicity of Sine and Cosine §positive, and cδc_{\delta} is positive by claim 5 of Elementary Order Arithmetic in an Ordered Field. Note that cδc_{\delta} depends only on δ\delta.

Let NNN\in\mathbb{N} and tKδt\in K_{\delta}. By The Pythagorean Identity for Sine and Cosine §bounds one has sin(Nπt)1|\sin(N\pi t)|\le1, so

(sin(Nπt))2=sin(Nπt)21,\bigl(\sin(N\pi t)\bigr)^{2}=\bigl|\sin(N\pi t)\bigr|^{2}\le1,

by claim 1 of Nonnegativity of Squares in an Ordered Field and claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, whose arguments are nonnegative, together with 11=11\cdot1=1, an axiom of the field R\mathbb{R}. The number NFN(t)N\,F_{N}(t) is nonnegative by claim 3 and the positivity of NN, and cδg(t)=(sin(πt))2c_{\delta}\le g(t)=(\sin(\pi t))^{2}; so claim 5 of Elementary Arithmetic in an Ordered Field, applied with the nonnegative multiplier NFN(t)N\,F_{N}(t), and claim 2 of the present lemma give

NFN(t)cδNFN(t)(sin(πt))2=(sin(Nπt))21.N\,F_{N}(t)\,c_{\delta}\le N\,F_{N}(t)\bigl(\sin(\pi t)\bigr)^{2}=\bigl(\sin(N\pi t)\bigr)^{2}\le1 .

Finally NcδN\,c_{\delta} is positive by claim 5 of Elementary Order Arithmetic in an Ordered Field, so its inverse is positive by claim 7 of that lemma; multiplying the two ends of the display by that inverse, by claim 5 of Elementary Arithmetic in an Ordered Field, and using associativity and commutativity of multiplication gives FN(t)1NcδF_{N}(t)\le\tfrac{1}{N\,c_{\delta}}.

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