Question #170550

Define uniform convergence of sequence of functions. Give an example


Expert's answer

Define uniform convergence of sequence of functions. Give an example

Defifinition. A sequence of functions fn:X→Rf_n:X\to R converges uniformly to the function f:X→Rf:X\to R if and only if lim⁡n→+∞sup⁡{∣fn(x)−f(x)∣:x∈X}=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 x∈Xx\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 x∈Xx\in X we have

∣fn(x)−f(x)∣=xn≤qn→0|f_n(x)-f(x)|=x^n\leq q^n\to0, hence lim⁡n→+∞sup⁡{∣fn(x)−f(x)∣:x∈X}=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 x∈Xx\in X we have ∣fn(x)−f(x)∣=xn|f_n(x)-f(x)|=x^n, sup⁡{xn:x∈X}=1\sup\{x^n:x\in X\}=1, hence lim⁡n→+∞sup⁡{∣fn(x)−f(x)∣:x∈X}=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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