Weiertrass Approximation Theorem

Weiertrass Approximation Theorem :

 Statement :

               If f is a real continuous function defined on a closed interval [a,b] then there exists a sequence of real polynomials {Pₙ} which converges uniformly to f(x) on [a,b] i.e lim Pₙ(x)=f(x)

                                                   n-->∞

converges uniformly on [a,b] .

Proof :

       If a=b , the conclusion follows by taking  Pₙ(x) to be a constant polynomial ,
defined by   Pₙ(x) = f(a) for all n .

We may thus assume that a<b . We next observe that a linear transformation

                         x' = (x-a) / (b-a)

is a continuous mapping of  [a,b] onto [1,0] . Accordingly , we assume without loss of generality that a=0  , b= 1 .

      Consider

    F(x) = f(x) - f(0) - x[f(1) - f(0)] , for 0≤x≤1

   Hence F(0) = 0= F(1) , and if F can be expressed as a limit of uniformly convergent sequence  of polynomials , then the same is true for f , since f-F is a polynomial . So may assume that f(1)=f(0)=0 .

    Let us further define f(x) to be zero for x outside [0,1] . Thus f is now uniformly continuous on the whole real line .

    Let us consider the polynomial (non_ negative for |x|≤1)

    Bₙ(x) = Cₙ(1-x²)ⁿ , n= 1,2,3..... .....(1)

Where Cₙ is independent of x , is so chosen that
        1
        ∫ Bₙ(x) dx = 1 for n= 1,2,3.... ....(2)
      -1      1                                  1
∴       1 = ∫ Cₙ(1-x²)ⁿ dx = 2Cₙ ∫ (1-x²)ⁿ dx
              -1                                  0
                       1/√(n)
             ≥ 2Cₙ ∫  (1-x²)ⁿ dx
                       0
                       1/√(n)
             ≥ 2Cₙ ∫  (1-nx²) dx
                       0
             =  4Cₙ / 3√(n) > Cₙ / √(n)

    ⇒   Cₙ < √(n)                                 .......(3)

 Which gives some information about the order of magnitude of Cₙ

         Therefore , for any δ> 0 equation(3) gives

   Bₙ (x) ≤ √(n)(1 - δ²)ⁿ , when δ≤|x|≤1.....(4)

so that Bₙ --> 0 uniformly for δ≤|x|≤1

  Again let
                    1
       Pₙ(x) = ∫ f(x+t) Bₙ(t) dt 0≤x≤1  .......(5)
                  -1
            -x                            1-x
         = ∫ f(x+t) Bₙ(t) dt + ∫ f(x+t) Bₙ(t) dt
             -1                           -x
              1
         +  ∫  f(x+t) Bₙ(t) dt
            1-x

   For |x| ≤ 1, -1+x ≤ x+t ≤ 0 , for -1≤t≤-x , so that x+t lies outside [0,1] and therefore f(x+t) =0 , and hence the first integral on the right vanishes . Similarly , the third integral is also equal to zero . Hence
                     1-x                          1
         Pₙ(x) = ∫ f(x+t) Bₙ(t) dt =∫f(t)Bₙ(t-x)dt
                      -x                           0

which is a real polynomial .

        We now proceed to show that the sequence {Pₙ(x)} converges uniformly to f on [0,1] .

    Continuity of f on the closed interval [0,1] .

      Therefore , there exist M such that

                    M = sup |f(x)|             ......(6)

and for any ε >0 , we can choose δ >0    such that for any two points x₁, x₂ in [0,1].

               |f(x₁) - f(x₂)| < ε/2 , when

                                 |x₁-x₂| < δ≤ 1 .....(7)

      For 0≤ x≤ 1 , we have
                              1
 | Pₙ(x) - f(x)|= |∫ f(x+t) Bₙ(t) dt - f(x)|
                             -1
                            1
                      = |∫ {f(x+t) - f(x)} Bₙ(t) dt|
                           -1
                                                       [using (2)]
                    1
                 ≤ ∫ |f(x+t) - f(x)| Bₙ(x) dt
                   -1                 

                -δ                                    ( ∵ Bₙ(t)≥0)
             = ∫ |f(x+t) - f(x)|Bₙ(t) dt
               -1  δ
             +     ∫ |f(x+t) - f(x)|Bₙ(t) dt
                   -δ
                   1
             +   ∫ |f(x+t) - f(x)|Bₙ(t) dt
                   δ
                  -δ                     δ
       ≤ 2M ∫ Bₙ(t) dt + ε/2∫ Bₙ(t) dt
                 -1                    -δ
                       1
             + 2M ∫ Bₙ(t) dt
                      δ

                    [Using equations(6) and (7)]
                                    -δ        1
     ≤ 2M √(n)(1-δ²)ⁿ {∫ dt + ∫ dt }+ ε/2
                                    -1       δ

               [Using equations(2) and (4)]

      ≤ 4M √(n)(1-δ²)ⁿ + ε/2

      < ε , for large values of n.

                 Thus for ε > 0 , there exists N (independent of x) such that

           |Pₙ(x) - f(x)|< ε , ∀ n≥ N

 ⇒      lim   Pₙ(x) = f(x) , uniformly on[0,1].
         n-->∞

My Querry :

                  At last I want to say that if you have any querry or doubt regarding this you can ask me by comment me . I will try my best to clarify your doubts .