Analysis:Why does Weierstrass Aprox. Thm require compactness?

[mscudder3]

The Weierstrass Approximation Theorem states "Let f be a continuous function on a compact set K in R. Therefore, f can be uniformly approximated by polynomials.

This profound statement was testing my logic earlier today. I was pondering why K must be compact. I figure if it is not bounded, say R itself, then p(sub n) would continue to converge to a value, while say some f=sinx was slowly oscillating about the x axis? (i.e. while p reaches its limit, f goes beyond it)

I was hoping someone could explain why it does not converge on say (0.1), a bounded and closed set. How do the functions act at these boundary points?

Thanks!

 

[disregardthat]

The point is that continuous functions on compact sets are bounded. Unbounded continuous functions may not be so easy to approximate with polynomials.

On (0,1) you could have the function f(x) = 1/x + 1/(x-1). This function will have the vertical asymptotes x = 0 and x = 1, and no polynomial can uniformally approximate that to any degree. (0,1) is not a closed set however. If it were, as you say, closed and bounded, it would imply that it was compact.

Even bounded functions may not be approximated, your example sin(x) is such a function. The reason is simply that any non-constant polynomial will approach infinity as x gets large, hence no non-constant polynomial will uniformally approximate sin(x) to any degree.

 

[mscudder3]

Thanks but I was wondering about the detail of one of the points you made.
From the above:
"The function will have the vertical asymptotes x = 0 and x = 1, and no polynomial can uniformally approximate that to any degree."

Why exactly is this? Sorry if this is overly thorough, I just cannot see the argument. TO rephrase, what property of polynomials informs us that they may not aprox. a function on an open set (given that x is bounded).

 

[disregardthat]

Of course, some functions may be approximated, a trivial example being a constant function on (0,1). But in general, the theorem cannot extended to open sets. I brought a counter-example, and it relies on the simple fact that no polynomial can have vertical asymptotes. Note that this would be required, as (if I name my function f(x)) for any polynomial P(x), |P(x)-f(x)| being bounded on (0,1) would require that P(x) to approach infinity as x approaches 0.