We use the notation of the statement. By Metric Space the metric d satisfies 0β€d(u,v), d(u,u)=0, d(u,v)=d(v,u) and d(u,w)β€d(u,v)+d(v,w) for all u,v,wβX.
Claim 1. Fix xβX. Since A is nonempty we may fix a0ββA, and then f(a0β)+Ld(x,a0β)βS(x), so S(x) is nonempty.
We show that f(a0β)βLd(x,a0β) is a lower bound for S(x). Let aβA. By claim 3 of Properties of the Absolute Value in an Ordered Field and the Lipschitz hypothesis,
f(a0β)βf(a)β€β£f(a0β)βf(a)β£β€Ld(a0β,a).
The triangle inequality gives d(a0β,a)β€d(a0β,x)+d(x,a)=d(x,a0β)+d(x,a), and multiplying this inequality by the nonnegative number L preserves it, so
f(a0β)βf(a)β€Ld(x,a0β)+Ld(x,a),thatΒ isf(a0β)βLd(x,a0β)β€f(a)+Ld(x,a).
As aβA was arbitrary, f(a0β)βLd(x,a0β) is a lower bound for S(x). Being nonempty and bounded below, S(x) has a greatest lower bound in R by Existence of the Infimum of a Nonempty Subset of R Bounded Below, so F(x)=infS(x) is a well-defined real number.
Claim 2. Let aβA. Taking the point a itself in the definition of S(a) and using d(a,a)=0 shows f(a)=f(a)+Ld(a,a)βS(a), whence F(a)β€f(a). Applying the lower bound established in claim 1 with x=a and a0β=a shows that f(a)βLd(a,a)=f(a) is a lower bound for S(a); since F(a) is the greatest lower bound, f(a)β€F(a). Therefore F(a)=f(a).
Claim 3. Let x,yβX and let aβA. By the triangle inequality d(x,a)β€d(x,y)+d(y,a), and multiplying by Lβ₯0 and adding f(a) gives
F(x)β€f(a)+Ld(x,a)β€(f(a)+Ld(y,a))+Ld(x,y),
the first inequality because f(a)+Ld(x,a)βS(x) and F(x) is a lower bound for S(x). Hence
F(x)βLd(x,y)β€f(a)+Ld(y,a)forΒ everyΒ aβA,
so F(x)βLd(x,y) is a lower bound for S(y) and therefore F(x)βLd(x,y)β€F(y), that is F(x)βF(y)β€Ld(x,y). Exchanging the roles of x and y and using d(y,x)=d(x,y) gives F(y)βF(x)β€Ld(x,y). By claim 6 of Properties of the Absolute Value in an Ordered Field the two inequalities together give β£F(x)βF(y)β£β€Ld(x,y), so F is Lipschitz with constant L in the sense of Lipschitz Map Between Metric Spaces.