What are they doing here? (polynomials-functions)

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Main Question or Discussion Point

I am currently studying Stone–Weierstrass approximation theorem, and some parts really confuse me.
basically, the Weierstrass approximation theorem states that every continuous function defined on an interval [a,b] can be uniformly approximated as closely as desired by a polynomial function.

In the proof,
at first they say on [a,b]=[0,1]
f(0)=f(1)=0. Why do they want to say that at the end points function must be zero?

second, if the theorem is proved for this case, consider
g(x)= f(x)- f(0) - x[f(1)-f(0)] (0<= x <= 1)
Here g(0)=g(1)=0

Can anyone explain the meaning and purpose of these transformation?

My guess is that they want to define function f and then find a g(x) which looks like polynomials. Is that right? and also, is there any relationship between these transformation and uniformly convergence?
 

Answers and Replies

  • #2
HallsofIvy
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I am currently studying Stone–Weierstrass approximation theorem, and some parts really confuse me.
basically, the Weierstrass approximation theorem states that every continuous function defined on an interval [a,b] can be uniformly approximated as closely as desired by a polynomial function.

In the proof,
at first they say on [a,b]=[0,1]
f(0)=f(1)=0. Why do they want to say that at the end points function must be zero?
I don't understand what you are saying. What is "f"? What are the hypotheses here?

second, if the theorem is proved for this case, consider
g(x)= f(x)- f(0) - x[f(1)-f(0)] (0<= x <= 1)
Here g(0)=g(1)=0

Can anyone explain the meaning and purpose of these transformation?

My guess is that they want to define function f and then find a g(x) which looks like polynomials. Is that right? and also, is there any relationship between these transformation and uniformly convergence?
 
  • #3
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I don't understand what you are saying. What is "f"? What are the hypotheses here?
f is any continuous function f(x) on an interval [a,b] . in the proof they chose [0,1] .
they want to prove that for any continuous function, there is a polynomial that can be approximated to this function as much as desired.
i couldn't understand the change between those function.
 
  • #4
mathman
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Any function f(x) on a closed interval [a,b] can be represented as the sum of two functions. h(x)=f(a) + [(x-a)/(b-a)]f(b) and g(x)=f(x)-h(x). g(a)=g(b)=0. The proof is then done in two steps: first prove it for any continuous function g(x), wich is 0 at the two ends, then use the decomposition to prove it for any continuous function f(x) by adding h(x) to the polynomial approximations for g(x).
 
  • #5
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thanks mathman,
now i think i got the idea...
 

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