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Proof of Sequential Form of the Continuity of a Second-Order Equation Operator

lemmalem:operator-sequential-continuity-2026a
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Β· 2,467 chars Β· 8 deps Β· depth 22 Reason: First publication of the proof: beyond the maximum of the four indices supplied by the four convergences, all four arguments lie within the modulus supplied by continuity.

Take the single modulus supplied by continuity at the quadruple and the four indices supplied by the four convergences; beyond their maximum all four arguments are within the modulus.

Proof

Conventions. From the setting we use the real numbers with their order, Euclidean space with its norm and distance, and S(n)\mathcal{S}(n) with its distance. Recall dE(z,w)=βˆ₯zβˆ’wβˆ₯d_{E}(z,w)=\lVert z-w\rVert and, by The Absolute Value Metric on the Real Line, dR(s,t)=∣sβˆ’t∣d_{\mathbb{R}}(s,t)=|s-t|; the symmetry of a metric is one of the metric axioms. The order ≀\le on N\mathbb{N} is a total order by claims 1, 2 and 3 of Properties of the Order on the Natural Numbers, so the maximum of two natural numbers is defined and Elementary Properties of the Maximum of Two Elements applies to it.

Proof. Let Ρ∈R\varepsilon\in\mathbb{R} be positive. By clause Continuity of a Second-Order Equation Operator §at-point there is a positive δ∈R\delta\in\mathbb{R} such that all y∈Uy\in U, s∈Rs\in\mathbb{R}, q∈Rnq\in\mathbb{R}^{n} and Y∈S(n)Y\in\mathcal{S}(n) satisfying

dE(y,x0)<Ξ΄,∣sβˆ’r0∣<Ξ΄,βˆ₯qβˆ’p0βˆ₯<Ξ΄,dS(n)(Y,X0)<Ξ΄d_{E}(y,x_{0})<\delta,\qquad|s-r_{0}|<\delta,\qquad\lVert q-p_{0}\rVert<\delta,\qquad d_{\mathcal{S}(n)}(Y,X_{0})<\delta

also satisfy ∣F(y,s,q,Y)βˆ’F(x0,r0,p0,X0)∣<Ξ΅\bigl|F(y,s,q,Y)-F(x_{0},r_{0},p_{0},X_{0})\bigr|<\varepsilon.

By convergence of the four given sequences there are N1,N2,N3,N4∈NN_{1},N_{2},N_{3},N_{4}\in\mathbb{N} such that: dE(yk,x0)<Ξ΄d_{E}(y_{k},x_{0})<\delta for every kβ‰₯N1k\ge N_{1}; dR(sk,r0)=∣skβˆ’r0∣<Ξ΄d_{\mathbb{R}}(s_{k},r_{0})=|s_{k}-r_{0}|<\delta for every kβ‰₯N2k\ge N_{2}; dE(qk,p0)=βˆ₯qkβˆ’p0βˆ₯<Ξ΄d_{E}(q_{k},p_{0})=\lVert q_{k}-p_{0}\rVert<\delta for every kβ‰₯N3k\ge N_{3}; and dS(n)(Yk,X0)<Ξ΄d_{\mathcal{S}(n)}(Y_{k},X_{0})<\delta for every kβ‰₯N4k\ge N_{4}.

Put N=max⁑{max⁑{N1,N2},max⁑{N3,N4}}N=\max\bigl\{\max\{N_{1},N_{2}\},\max\{N_{3},N_{4}\}\bigr\}. By claim 1 of Elementary Properties of the Maximum of Two Elements, N1≀max⁑{N1,N2}≀NN_{1}\le\max\{N_{1},N_{2}\}\le N and likewise for N2N_{2}, N3N_{3} and N4N_{4}, so Ni≀NN_{i}\le N for i=1,2,3,4i=1,2,3,4 by the transitivity of claim 1 of Properties of the Order on the Natural Numbers. Let k∈Nk\in\mathbb{N} satisfy kβ‰₯Nk\ge N. Then kβ‰₯Nik\ge N_{i} for each ii, again by transitivity, so all four displayed bounds hold with y=yky=y_{k}, s=sks=s_{k}, q=qkq=q_{k} and Y=YkY=Y_{k}, and therefore

dR(F(yk,sk,qk,Yk),F(x0,r0,p0,X0))=∣F(yk,sk,qk,Yk)βˆ’F(x0,r0,p0,X0)∣<Ξ΅.d_{\mathbb{R}}\bigl(F(y_{k},s_{k},q_{k},Y_{k}),F(x_{0},r_{0},p_{0},X_{0})\bigr)=\bigl|F(y_{k},s_{k},q_{k},Y_{k})-F(x_{0},r_{0},p_{0},X_{0})\bigr|<\varepsilon .

Since Ξ΅\varepsilon was an arbitrary positive real, (F(yk,sk,qk,Yk))k∈N\bigl(F(y_{k},s_{k},q_{k},Y_{k})\bigr)_{k\in\mathbb{N}} converges to F(x0,r0,p0,X0)F(x_{0},r_{0},p_{0},X_{0}) in (R,dR)(\mathbb{R},d_{\mathbb{R}}). β– \blacksquare

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