HOMEWORK EXERCISE NO. 3

      Please submit this by December 17.


First you are asked to answer questions 2,3,4,5 and 6
on page 16 of the book by Yitzhak Katznelson (First edition)

Then one additional question, let us call it 3A:

3A. Use the results obtained in our lectures (and some other basic facts
that you should know from previous courses) to prove the following
theorem, usually known as the Weierstrass Approximation Theorem.

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Theorem:
Let [a,b] be a bounded closed interval and let f:[a,b] ---> C be a
continuous complex valued function on [a,b].
Let  \epsilon be a positive number.
Then there exists an algebraic polynomial P such that
|P(x)-f(x)| < \epsilon
for every x in [a,b].

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Some remarks about question 3A:

By "algebraic polynomial" we of course mean a function P(x) which is a
finite sum of functions of the form a_n x^n  where n is a positive integer
and a_n is a constant.

Here x denotes a real variable, ranging over [a,b].

Since f may be complex then we have to allow for the possibility that the
coefficients a_n of the polynomial may also be complex.

You should be able to prove quite easily that if f takes only real values
then we can choose P so that all its coefficients are real.

In contrast to the corresponding result for trigonometric polynomials
on the interval [-\pi, \pi], your proof should also work in the case
where f(a) and f(b) are not equal to each other.

At first sight it might seem that Taylor's theorem and Taylor's formula
might be good starting points for proving Weierstrass' theorem. But the
Taylor polynomial for some continuous functions f can be very different
and far away from the function f. Suppose for example that [a,b]=[-1,1]
and f is the function defined by f(0)=0, and f(x)=1000000000 exp(-1/x^2)
for non zero x.
Then the function f is continuous, and in fact all of its
derivatives of all orders exist and are continuous. But, if you calculate
the Taylor polynomial of f of any order, expanding about x=0, it is
identically zero. This is a very bad approximation to f .