11.6.1 Uniform Approximation With Trig. Polynomials
It turns out that if you don’t insist the ak be the Fourier coefficients, then every continuous
2π periodic function θ → f
can be approximated uniformly with a Trig. polynomial of the
This means that for all ε > 0 there exists a pn
Definition 11.6.2 Recall the nth partial sum of the Fourier series Snf
is the Dirichlet kernel,
The nth Fejer mean, σnf
is the average of the first n of the Snf
The Fejer kernel is
As was the case with the Dirichlet kernel, the Fejer kernel has some properties.
Lemma 11.6.3 The Fejer kernel has the following properties.
- Fn+1 =
dt = 1
dt = ∑
k=−nnbkeikθ for a suitable choice of bk.
- Fn+1 =
≥ 0,Fn =
- For every δ > 0,
In fact, for
Proof: Part 1.) is obvious because Fn+1 is the average of functions for which this is
Part 2.) is also obvious for the same reason as Part 1.). Part 3.) is obvious because it is
true for Dn in place of Fn+1 and then taking the average yields the same sort of
The last statements in 4.) are obvious from the formula which is the only hard part of
Using the identity sin
which completes the demonstration of 4.).
Next consider 5.). Since Fn+1 is even it suffices to show
For the given t,
which shows 5.). This proves the lemma.
Here is a picture of the Fejer kernels for n=2,4,6.
Note how these kernels are nonnegative, unlike the Dirichlet kernels. Also there is a large
bump in the center which gets increasingly large as n gets larger. The fact these kernels are
nonnegative is what is responsible for the superior ability of the Fejer means to approximate a
Theorem 11.6.4 Let f be a continuous and 2π periodic function. Then
Proof: Let ε > 0 be given. Then by part 2. of Lemma 11.6.3,
is even and
is continuous and periodic, hence bounded by some constant
the above is dominated by
Now choose δ such that for all x, it follows that if
This can be done because f is uniformly continuous on
. Since it is periodic, it must also be uniformly continuous on ℝ
. (why?) Therefore, for this
this has shown that for all x
and now by Lemma 11.6.3 it follows
provided n is large enough. This proves the theorem.