Question #170550

Define uniform convergence of sequence of functions. Give an example


1
Expert's answer
2021-03-30T10:44:16-0400

Define uniform convergence of sequence of functions. Give an example

Defifinition. A sequence of functions fn:XRf_n:X\to R converges uniformly to the function f:XRf:X\to R if and only if limn+sup{fn(x)f(x):xX}=0\lim\limits_{n\to+\infty}\sup\{|f_n(x)-f(x)|:x\in X\}=0 or, equivalently, if and only if for any ε>0\varepsilon>0 there exists a sufficiently large integer N such that for all xXx\in X and n>Nn>N fn(x)f(x)<ε|f_n(x)-f(x)|<\varepsilon.


Example 1. X=[0, q], where 0<q<1, fn(x)=xnf_n(x)=x^n, f(x)=0f(x)=0. Then for all xXx\in X we have

fn(x)f(x)=xnqn0|f_n(x)-f(x)|=x^n\leq q^n\to0, hence limn+sup{fn(x)f(x):xX}=0\lim\limits_{n\to+\infty}\sup\{|f_n(x)-f(x)|:x\in X\}=0 and the sequence of functions fn(x)f_n(x) converges to zero uniformly.


Example 2. X=[0, 1), fn(x)=xnf_n(x)=x^n, f(x)=0f(x)=0. Then for all xXx\in X we have fn(x)f(x)=xn|f_n(x)-f(x)|=x^n, sup{xn:xX}=1\sup\{x^n:x\in X\}=1, hence limn+sup{fn(x)f(x):xX}=1\lim\limits_{n\to+\infty}\sup\{|f_n(x)-f(x)|:x\in X\}=1 and the sequence of functions fn(x)f_n(x) does not converge to f(x) uniformly.


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