Question #170557

Let {fn} be a sequence of functions defined on S .show.that there exists a.function f such that fn converges to f uniformly on S if and.only if the caught condition is satisfied


1
Expert's answer
2021-03-31T07:19:41-0400

The sequence of functions {fn} defined on S [a, b] {let} converges uniformly on S [a, b] if and only if for every ε > 0 and for all x ∈ [a, b], there exists an integer N such that  


fn+p(x)fn(x)<ϵ,     nN|f_{n+p}(x)-f_n(x)|<\epsilon,\ \ \ \ \forall \ n\geq N



Proof. Let the sequence {fn} uniformly converge on [a, b] to the limit function f, so that for a given ε > 0, and for all x ∈ [a, b], there exist integers n1,n2n_1,n_2 such that 


fn(x)f(x)<ϵ/2,     nn1|f_{n}(x)-f(x)|<\epsilon/2,\ \ \ \ \forall \ n\geq n_1

and


fn+p(x)f(x)<ϵ/2,     nn2|f_{n+p}(x)-f(x)|<\epsilon/2,\ \ \ \ \forall \ n\geq n_2

Let N= max(n1,n2n_1,n_2)


fn+p(x)fn(x)fn+p(x)f(x)+fn(x)f(x)\Rightarrow|f_{n+p}(x)-f_n(x)|\leq|f_{n+p}(x)-f(x)|+ |f_{n}(x)-f(x)|

fn+p(x)fn(x)<ϵ/2+ϵ/2=ϵ,     nN\Rightarrow|f_{n+p}(x)-f_n(x)|< \epsilon/2+\epsilon/2=\epsilon, \ \ \ \ \forall\ n\geq N

Hence   fn+p(x)fn(x)<ϵ,     nN\text{Hence}\ \ \ |f_{n+p}(x)-f_n(x)|<\epsilon,\ \ \ \ \forall \ n\geq N

(Cauchy’s Criterion for Uniform Convergence)

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