Question

LetX = (0, 1) , with the usual distance. Define f:X X as f(x)=x/2 , with the help of Banach contraction principle.
​

Answer 1

Answer:

LetX = (0, 1) , with the usual distance. Define f:X X as f(x)=x/2 , with the help of Banach contraction principle.

Answer 2

Answer:

Let X be a complete metric space and Y be a topological space.

Let f: X × Y → X be a continuous function. Assume that f is a contraction on X

uniformly in Y , that is,

d(f(x1, y), f(x2, y)) ≤ λd(x1, x2), ∀ x1, x2 ∈ X, ∀ y ∈ Y

for some λ < 1. Then, for every fixed y ∈ Y , the map x 7→ f(x, y) has a unique

fixed point ϕ(y). Moreover, the function y 7→ ϕ(y) is continuous from Y to X.

Notice that if f : X × Y → X is continuous on Y and is a contraction on X

uniformly in Y , then f is in fact continuous on X × Y .

proof In light of Theorem 1.3, we only have to prove the continuity of ϕ. For

y, y0 ∈ Y , we have

d(ϕ(y), ϕ(y0)) = d(f(ϕ(y), y), f(ϕ(y0), y0))

≤ d(f(ϕ(y), y), f(ϕ(y0), y)) + d(f(ϕ(y0), y), f(ϕ(y0), y0))

≤ λd(ϕ(y), ϕ(y0)) + d(f(ϕ(y0), y), f(ϕ(y0), y0))

which implies

d(ϕ(y), ϕ(y0)) ≤ 1/1 − λ d(f(ϕ(y0), y), f(ϕ(y0), y0)).

Since the above right-hand side goes to zero as y → y0, we have the desired continuity.