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Proof of The Difference Quotient of a Function with Bounded Continuous Derivative is Bounded, Symmetric and Continuous on the Plane

lemmalem:difference-quotient-c1-real-2026a
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· 6,763 chars · 17 deps · depth 20 Reason: Goal 3C Batch A: proof via the mean value theorem.

The mean value theorem writes every off-diagonal value of the difference quotient as a value of the derivative at an intermediate point, which gives the bound and continuity at diagonal points; off the diagonal the difference quotient is a product of continuous functions.

Proof

Each result cited is universally quantified over the data in its own statement. Continuity of a real-valued function on a subset AR2A\subseteq\mathbb{R}^{2} is that of Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation §extrema: continuity at each point of AA relative to AA in the sense of Continuous Map Between Metric Spaces, from (R2,dE)(\mathbb{R}^{2},d_{E}) to (R,dR)(\mathbb{R},d_{\mathbb{R}}). For (x,y),(a,b)R2(x,y),(a,b)\in\mathbb{R}^{2}, dE((x,y),(a,b))=(xa,yb)d_{E}((x,y),(a,b))=\lVert(x-a,y-b)\rVert (claim 2 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n, the difference of points of R2\mathbb{R}^{2} being coordinatewise by clause 1 of Difference, Dot Product, and Orthogonality in Rn\mathbb{R}^n), so xadE((x,y),(a,b))|x-a|\le d_{E}((x,y),(a,b)) and ybdE((x,y),(a,b))|y-b|\le d_{E}((x,y),(a,b)) by claim 4 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n. By Differentiability at an Interior Point Implies Continuity There, ϕ\phi is continuous at every point of R\mathbb{R}, as a map from (R,dR)(\mathbb{R},d_{\mathbb{R}}) to itself. Elementary identities of the absolute value are taken from Properties of the Absolute Value in an Ordered Field (claim 2 for st=ts|s-t|=|t-s|, claim 5 for the triangle inequality, used twice for three summands) and of the field from Zero Products and Elementary Identities in a Field (claim 2: (s)t=(st)(-s)t=-(st), in particular t=(1)t-t=(-1)t).

Step 1: the mean value representation. Let x,yRx,y\in\mathbb{R} with xyx\ne y, and let u<vu<v be the two numbers x,yx,y in increasing order. By Basic Facts about Intervals of the Real Line and Their Interior Points §closed-interval, [u,v][u,v] is an interval of which every point ww with u<w<vu<w<v is an interior point. The restriction ϕ[u,v]\phi|_{[u,v]} is continuous on [u,v][u,v] by claim 1 of Restriction Stability of Continuity and of the Derivative, and differentiable at every point of (u,v)(u,v) with derivative ϕ\phi' there by claim 2 of the same lemma. By Mean Value Theorem on a Closed Real Interval there is cc with u<c<vu<c<v and

ϕ(c)=ϕ(v)ϕ(u)vu.\phi'(c)=\frac{\phi(v)-\phi(u)}{v-u}.

Now ϕ(u)ϕ(v)uv=ϕ(v)ϕ(u)vu\frac{\phi(u)-\phi(v)}{u-v}=\frac{\phi(v)-\phi(u)}{v-u}: indeed ϕ(u)ϕ(v)=(ϕ(v)ϕ(u))\phi(u)-\phi(v)=-(\phi(v)-\phi(u)) and uv=(vu)u-v=-(v-u) by claim 6 of Additive Cancellation and Elementary Additive Identities in a Field, and for real numbers pp and w0w\ne0 one has (p)(w)1=pw1(-p)(-w)^{-1}=p\,w^{-1}: multiplying pw1p\,w^{-1} by (w)(w)1=1(-w)(-w)^{-1}=1 and using claim 2 of Zero Products and Elementary Identities in a Field twice, pw1=pw1(w)(w)1=(pw1w)(w)1=(p)(w)1p\,w^{-1}=p\,w^{-1}(-w)(-w)^{-1}=-(p\,w^{-1}w)(-w)^{-1}=(-p)(-w)^{-1}. Therefore F(x,y)=F(y,x)=ϕ(c)F(x,y)=F(y,x)=\phi'(c), with cc strictly between xx and yy, that is, u<c<vu<c<v.

Step 2: claim 1. Symmetry for xyx\ne y is contained in Step 1, and is trivial for x=yx=y. For the bound: if x=yx=y then F(x,y)=ϕ(x)L|F(x,y)|=|\phi'(x)|\le L; if xyx\ne y then F(x,y)=ϕ(c)L|F(x,y)|=|\phi'(c)|\le L with cc as in Step 1.

Step 3: continuity at a diagonal point (a,a)(a,a). Let ε>0\varepsilon>0. Since ϕ\phi' is continuous at aa, there is η>0\eta>0 with ϕ(c)ϕ(a)<ε|\phi'(c)-\phi'(a)|<\varepsilon whenever ca<η|c-a|<\eta. Let (x,y)R2(x,y)\in\mathbb{R}^{2} with dE((x,y),(a,a))<ηd_{E}((x,y),(a,a))<\eta; then xa<η|x-a|<\eta and ya<η|y-a|<\eta. If x=yx=y, then F(x,y)F(a,a)=ϕ(x)ϕ(a)F(x,y)-F(a,a)=\phi'(x)-\phi'(a), which has absolute value less than ε\varepsilon. If xyx\ne y, then F(x,y)=ϕ(c)F(x,y)=\phi'(c) with u<c<vu<c<v, where u<vu<v are x,yx,y in increasing order (Step 1); by claim 9 of Properties of the Absolute Value in an Ordered Field and claim 1 of Elementary Order Arithmetic in an Ordered Field, xa<η|x-a|<\eta and ya<η|y-a|<\eta mean aη<x<a+ηa-\eta<x<a+\eta and aη<y<a+ηa-\eta<y<a+\eta, so aη<u<c<v<a+ηa-\eta<u<c<v<a+\eta and thus aη<c<a+ηa-\eta<c<a+\eta by transitivity of the strict order, i.e. ca<η|c-a|<\eta by the same two claims, and F(x,y)F(a,a)=ϕ(c)ϕ(a)<ε|F(x,y)-F(a,a)|=|\phi'(c)-\phi'(a)|<\varepsilon. Hence FF is continuous at (a,a)(a,a) relative to R2\mathbb{R}^{2}.

Step 4: continuity at a point (a,b)(a,b) with aba\ne b. Let D={(x,y)R2:xy}D=\{(x,y)\in\mathbb{R}^{2}:x\ne y\}.

(a) FDF|_{D} is continuous on DD relative to DD. The coordinate projections pr1,pr2:R2R\mathrm{pr}_{1},\mathrm{pr}_{2}:\mathbb{R}^{2}\to\mathbb{R}, pr1(x,y)=x\mathrm{pr}_{1}(x,y)=x and pr2(x,y)=y\mathrm{pr}_{2}(x,y)=y, are continuous on R2\mathbb{R}^{2} (given ε\varepsilon, take δ=ε\delta=\varepsilon, by the coordinate bound recalled above). By claim 3 of Semicontinuity and Continuity Under Composition with a Continuous Map, ϕpr1\phi\circ\mathrm{pr}_{1} and ϕpr2\phi\circ\mathrm{pr}_{2} are continuous on R2\mathbb{R}^{2}, so by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space the functions N(x,y)=ϕ(x)ϕ(y)=ϕ(pr1(x,y))+(1)ϕ(pr2(x,y))N(x,y)=\phi(x)-\phi(y)=\phi(\mathrm{pr}_{1}(x,y))+(-1)\phi(\mathrm{pr}_{2}(x,y)) and Q(x,y)=xy=pr1(x,y)+(1)pr2(x,y)Q(x,y)=x-y=\mathrm{pr}_{1}(x,y)+(-1)\mathrm{pr}_{2}(x,y) are continuous on R2\mathbb{R}^{2}, hence their restrictions to DD are continuous on DD relative to DD by claim 1 of Restriction Stability of Continuity and of the Derivative. The restriction QDQ|_{D} vanishes nowhere on DD, so by claim 2 of Continuity of the Reciprocal of a Nonvanishing Real-Valued Function on a Metric Space the function (x,y)(xy)1(x,y)\mapsto(x-y)^{-1} is continuous on DD relative to DD, and by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space so is the product (x,y)(ϕ(x)ϕ(y))(xy)1(x,y)\mapsto(\phi(x)-\phi(y))(x-y)^{-1}, which is FDF|_{D} by the definition of FF on DD.

(b) Every point of DD has a ball around it inside DD, and continuity relative to DD upgrades to continuity relative to R2\mathbb{R}^{2}. Let (a,b)D(a,b)\in D and put ρ=ab2\rho=\tfrac{|a-b|}{2}, positive by claim 1 of Properties of the Absolute Value in an Ordered Field and claim 8 of Elementary Order Arithmetic in an Ordered Field. If dE((x,y),(a,b))<ρd_{E}((x,y),(a,b))<\rho, then xa<ρ|x-a|<\rho and yb<ρ|y-b|<\rho, so by the triangle inequality applied to ab=(ax)+(xy)+(yb)a-b=(a-x)+(x-y)+(y-b), abxa+xy+yb<xy+ab|a-b|\le|x-a|+|x-y|+|y-b|<|x-y|+|a-b| (claim 3 of Elementary Order Arithmetic in an Ordered Field and ρ+ρ=ab\rho+\rho=|a-b| by claim 8 there), whence 0<xy0<|x-y| by claim 1 there, so xyx\ne y by claim 1 of Properties of the Absolute Value in an Ordered Field, and (x,y)D(x,y)\in D. Now let ε>0\varepsilon>0. By (a) there is δ0>0\delta_{0}>0 such that every (x,y)D(x,y)\in D with dE((x,y),(a,b))<δ0d_{E}((x,y),(a,b))<\delta_{0} satisfies F(x,y)F(a,b)<ε|F(x,y)-F(a,b)|<\varepsilon. Let δ\delta be the least of δ0\delta_{0} and ρ\rho. Every (x,y)R2(x,y)\in\mathbb{R}^{2} with dE((x,y),(a,b))<δd_{E}((x,y),(a,b))<\delta lies in DD and satisfies F(x,y)F(a,b)<ε|F(x,y)-F(a,b)|<\varepsilon. Hence FF is continuous at (a,b)(a,b) relative to R2\mathbb{R}^{2}.

Step 5: conclusion. Steps 3 and 4 show that FF is continuous at every point of R2\mathbb{R}^{2} relative to R2\mathbb{R}^{2}, that is, continuous on R2\mathbb{R}^{2}; this is claim 2. Since dRd_{\mathbb{R}} is the Euclidean distance on R=R1\mathbb{R}=\mathbb{R}^{1} by The Euclidean Distance on the Real Line is the Absolute Value Metric, FF is continuous from (R2,dE)(\mathbb{R}^{2},d_{E}) to R\mathbb{R} with its Euclidean distance, so by claim 3(a) of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets, FF is measurable with respect to the σ\sigma-algebra B2\mathcal{B}_{2} there and B(R)\mathcal{B}(\mathbb{R}), and B2=B(R2)\mathcal{B}_{2}=\mathcal{B}(\mathbb{R}^{2}) by claim 5 of the same lemma. This proves claim 3.

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