Fixed point theory (locally (α,β,ψ) dominated contractive condition)

post by muzammil · 2022-09-01T17:56:17.978Z · LW · GW · 2 comments

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locally (α,β,ψ) dominated contractive condition

   Theorem 2.1 Let α,β:X×X→[0,+∞), r>0, x₀∈X, ψ∈Ψ, (X,d) be an (α,β)-complete metric space and S,T:X→X. If the following conditions hold:
   1) S and T are (α,β)-continuous,
   2) The pair (S,T) satisfies the locally (α,β,ψ) dominated contractive condition on B(x₀,r),
   3) If x and y belongs to set of common fixed points of S and T, then α(x,y)≥β(x,y).
   Then S and T have a unique common fixed point.

I have to solve this theorem in families of mappings.

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answer by muzammil · 2022-10-04T09:49:49.756Z · LW(p) · GW(p)

     (Locally (α,β,ψ) dominated contractive condition)
   Let α,β:X×X→[0,+∞), r>0, x₀∈X, ψ∈Ψ, (X,d) be an (α,β)-complete metric space and S,T:X→X are locally (α,β)-continuous and (α,β)-dominated triangular mappings on B(x₀,r). We say that the pair (S,T) satisfies the locally (α,β,ψ) dominated contractive condition on B(x₀,r), if

d(Sx,Ty)≤ψ(max{d(x,y),d(x,Sx),d(y,Ty),((d(x,Ty)+d(y,Sx))/2)}),   #1.1

for all x,y∈B(x₀,r) with α(x,y)≥β(x,y) or α(y,x)≥β(y,x), and

∑_{i=0}^{j}ψ^{i}(d(x₀,Sx₀))≤r for all j∈ℕ∪{0}.   #1.2

   2. Result for locally (α,β,ψ) dominated contractive condition
   Theorem 2.1 Let α,β:X×X→[0,+∞), r>0, x₀∈X, ψ∈Ψ, (X,d) be an (α,β)-complete metric space and S,T:X→X. If the following conditions hold:
   1) S and T are (α,β)-continuous,
   2) The pair (S,T) satisfies the locally (α,β,ψ) dominated contractive condition on B(x₀,r),
   3) If x and y belongs to set of common fixed points of S and T, then α(x,y)≥β(x,y).
   Then S and T have a unique common fixed point.

According to 1.1 condition I want to solve this theorem in families of mappings.

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comment by Raemon · 2022-09-01T18:08:42.656Z · LW(p) · GW(p)

Is this a homework problem? Can you say more about why you are working on this?

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comment by muzammil · 2022-10-04T09:47:44.819Z · LW(p) · GW(p)

this is assignment from uni Professor