Theorem 6 (The Chebyshev Equioscillation Theorem)
Let f be a continuous real-valued function on [a,b]
. Then pn(x)
is a polynomial of best approximation of degree n
if and only if f,pn
has an alternating set of length n+2
. Furthermore, the polynomial of best approximation is unique.
Proof
The theorem is trivially true if f is itself a polynomial of degree ≤n
. We assume not, and so dn>0
.
Step 1
Suppose that f,pn
has an alternating set of length n+2
. By Theorem 4, we have ‖f-pn‖≤dn
. As dn≤‖f-pn‖
by the definition of dn
, it follows that pn
is a polynomial of best approximation to f
.
Step 2
Now suppose that pn
is a polynomial of best approximation to f. By Lemma 4, f,pn
has an alternating set of length 2, and by Theorem 5, it can be extended into a sectioned alternating set of length m. We must have m≥n+2
, for if m≤n+1
then by Lemma 6, we could add a polynomial q of degree ≤n
to pn
and get a better approximation than pn
, which is impossible. Thus every polynomial of best approximation has an alternating set of length at least n+2
.
Step 3
To show uniqueness, suppose that pn
and qn
are both polynomials of best approximation, and we will show that they are equal.
Note that (pn+qn)/2
is a polynomial of best approximation, as:
||f-pn+qn2||=||f-pn2+f-qn2||≤12‖f-pn‖+12‖f-qn‖=dn
Therefore, there are n+2
alternating points at which (f-pn)/2+(f-qn)/2=±dn
.
At each of these alternating points, f-pn
and f-qn
are both dn
or both -dn
. So f-pn
and f-qn
agree on n+2
points, and so (f-pn)-(f-qn)=qn-pn=0
at these n+2
points. Since qn-pn
is a polynomial of degree ≤n
, qn
and pn
must be identical. Therefore the polynomial pn
of best approximation is unique.