Tests For Uniform Convergence

Tests For Uniform Convergence :

Theorem 1 (Mₙ _ Test) :

                 Let <fₙ> be a sequence of function defined on a metric space X .

     Let  lim  fₙ(x) = f(x)   ∀ x ∈ X and let

           n-->∞ 

        Mₙ = Sup {|fₙ(x) - f(x)| : x∈ X}

  Then <fₙ> converges uniformly to f iff     Mₙ-->0 as n-->∞.

Proof Of Necessary Part :

         Let us suppose the sequence <fₙ> of functions converges uniformly to f on X . Then by definition , for a given ε > 0 ∃ a positive integer m (independent of x)     such that  n≥ m ⇒|fₙ(x) - f(x)| < ε ∀ x∈X

   Also , Mₙ is the supremum of |fₙ(x) - f(x)|.

Therefore |fₙ(x) - f(x)| < ε ∀ n≥m  ∀ x∈X

        ⇒ Mₙ = Sup |fₙ(x) - f(x)| < ε  ∀ n≥ m

                      x∈X  

       ⇒   |fₙ(x) - f(x)| ≤ Mₙ <ε  ∀ n≥m ,∀ x∈X

       ⇒   <fₙ> converges uniformly to f on X .

For Example :

            Show that the sequence <fₙ(x)> where  fₙ(x) = nx (1-x)ⁿ does not converge uniformly on [0,1] .

Solution :

           Here , we have 

                   f(x) = lim fₙ(x) = lim  nx/(1-x)⁻ⁿ 

                           n-->∞          n-->∞   

                            = lim        x/-(1-x)⁻ⁿlog(1-x)

                             n-->∞

        [By L. Hospital's Rule  i.e lim f(x)/g(x) 

                                                 = lim f'(x)/g'(x) 

when the form is indefinite or ∞/∞ or 0/0 ]

                      = lim    -x(1-x)ⁿ / log(1-x)

                       n-->∞

                      = 0     [∵ (1-x)ⁿ --> 0 ∀ x∈ [0,1]  ]

 ⇒ f(x) = 0 ∀ x∈[0,1]

Now Mₙ = Sup{|fₙ(x) - f(x)|: x∈ [0,1]}

                = Sup { nx (1-x)ⁿ : x ∈[0,1]}

                = Sup (1-x)ⁿ nx   ∀ x∈ [0,1]

Therefore , Mₙ ≥ n. 1/n (1- 1/n)ⁿ  

                                   ( Taking x = 1/n ∈ [0,1])

                       = (1- 1/n)ⁿ --> 1/e  as n-->∞

Hence by Mₙ _ test <fₙ> does not converge uniformly on [0,1] . Therefore , 0 is a point of non uniform convergence , since 

x= 1/n --> 0 as n--> ∞ .