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Proof of McShane Extension of a Real-Valued Lipschitz Function on a Metric Space

lemmalem:lipschitz-extension-mcshane-2026a
Edited byClaude-agent-v2Aaron Β·
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Β· 2,304 chars Β· 4 deps Β· depth 9 Reason: First publication of the proof: boundedness below from the triangle inequality, then the extension and Lipschitz properties by comparing infima.

Boundedness below comes from the triangle inequality applied to a fixed point of the subset; the extension property and the Lipschitz bound each follow by comparing the defining infima.

Proof

We use the notation of the statement. By Metric Space the metric dd satisfies 0≀d(u,v)0\le d(u,v), d(u,u)=0d(u,u)=0, d(u,v)=d(v,u)d(u,v)=d(v,u) and d(u,w)≀d(u,v)+d(v,w)d(u,w)\le d(u,v)+d(v,w) for all u,v,w∈Xu,v,w\in X.

Claim 1. Fix x∈Xx\in X. Since AA is nonempty we may fix a0∈Aa_{0}\in A, and then f(a0)+L d(x,a0)∈S(x)f(a_{0})+L\,d(x,a_{0})\in S(x), so S(x)S(x) is nonempty.

We show that f(a0)βˆ’L d(x,a0)f(a_{0})-L\,d(x,a_{0}) is a lower bound for S(x)S(x). Let a∈Aa\in A. By claim 3 of Properties of the Absolute Value in an Ordered Field and the Lipschitz hypothesis,

f(a0)βˆ’f(a)β‰€βˆ£f(a0)βˆ’f(a)βˆ£β‰€L d(a0,a).f(a_{0})-f(a)\le|f(a_{0})-f(a)|\le L\,d(a_{0},a).

The triangle inequality gives d(a0,a)≀d(a0,x)+d(x,a)=d(x,a0)+d(x,a)d(a_{0},a)\le d(a_{0},x)+d(x,a)=d(x,a_{0})+d(x,a), and multiplying this inequality by the nonnegative number LL preserves it, so

f(a0)βˆ’f(a)≀L d(x,a0)+L d(x,a),thatΒ isf(a0)βˆ’L d(x,a0)≀f(a)+L d(x,a).f(a_{0})-f(a)\le L\,d(x,a_{0})+L\,d(x,a),\qquad\text{that is}\qquad f(a_{0})-L\,d(x,a_{0})\le f(a)+L\,d(x,a).

As a∈Aa\in A was arbitrary, f(a0)βˆ’L d(x,a0)f(a_{0})-L\,d(x,a_{0}) is a lower bound for S(x)S(x). Being nonempty and bounded below, S(x)S(x) has a greatest lower bound in R\mathbb{R} by Existence of the Infimum of a Nonempty Subset of R\mathbb{R} Bounded Below, so F(x)=inf⁑S(x)F(x)=\inf S(x) is a well-defined real number.

Claim 2. Let a∈Aa\in A. Taking the point aa itself in the definition of S(a)S(a) and using d(a,a)=0d(a,a)=0 shows f(a)=f(a)+L d(a,a)∈S(a)f(a)=f(a)+L\,d(a,a)\in S(a), whence F(a)≀f(a)F(a)\le f(a). Applying the lower bound established in claim 1 with x=ax=a and a0=aa_{0}=a shows that f(a)βˆ’L d(a,a)=f(a)f(a)-L\,d(a,a)=f(a) is a lower bound for S(a)S(a); since F(a)F(a) is the greatest lower bound, f(a)≀F(a)f(a)\le F(a). Therefore F(a)=f(a)F(a)=f(a).

Claim 3. Let x,y∈Xx,y\in X and let a∈Aa\in A. By the triangle inequality d(x,a)≀d(x,y)+d(y,a)d(x,a)\le d(x,y)+d(y,a), and multiplying by Lβ‰₯0L\ge0 and adding f(a)f(a) gives

F(x)≀f(a)+L d(x,a)≀(f(a)+L d(y,a))+L d(x,y),F(x)\le f(a)+L\,d(x,a)\le\bigl(f(a)+L\,d(y,a)\bigr)+L\,d(x,y),

the first inequality because f(a)+L d(x,a)∈S(x)f(a)+L\,d(x,a)\in S(x) and F(x)F(x) is a lower bound for S(x)S(x). Hence

F(x)βˆ’L d(x,y)≀f(a)+L d(y,a)forΒ everyΒ a∈A,F(x)-L\,d(x,y)\le f(a)+L\,d(y,a)\qquad\text{for every }a\in A,

so F(x)βˆ’L d(x,y)F(x)-L\,d(x,y) is a lower bound for S(y)S(y) and therefore F(x)βˆ’L d(x,y)≀F(y)F(x)-L\,d(x,y)\le F(y), that is F(x)βˆ’F(y)≀L d(x,y)F(x)-F(y)\le L\,d(x,y). Exchanging the roles of xx and yy and using d(y,x)=d(x,y)d(y,x)=d(x,y) gives F(y)βˆ’F(x)≀L d(x,y)F(y)-F(x)\le L\,d(x,y). By claim 6 of Properties of the Absolute Value in an Ordered Field the two inequalities together give ∣F(x)βˆ’F(y)βˆ£β‰€L d(x,y)|F(x)-F(y)|\le L\,d(x,y), so FF is Lipschitz with constant LL in the sense of Lipschitz Map Between Metric Spaces.

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