3. The Polynomial of Best Approximation
Let f be a continuous function on the interval [a,b]
. We have previously defined:
dn=inf{‖f-p‖:p is a polynomial of degree ≤n}
Our goal is to show that this minimum is actually achieved: there exists a polynomial pn
for which ‖f-pn‖=dn
. Such a polynomial pn
is called a polynomial of best approximation of degree n to f. Our proof follows the argument in Isaacson and Keller (1994). Later we will show that for every f and n, the polynomial of best approximation is unique.
Let f, g be continuous real-valued functions on the interval [a,b]
. Then
- ‖f‖≥0
and ‖f‖=0
iff f=0
.
- ‖f+g‖≤‖f‖+‖g‖
- For any constant c, ‖cf‖=|c|⋅‖f‖
The proof is easy. The lemma says that the supremum norm satisfies the requirements of a vector space norm of the space of all continuous real-valued functions on [a,b]
. Thus, we can define a metric on this space: the distance between f
and g
is d(f,g)=‖f-g‖
.
Let a=(a0,a1,...,an)
be a vector in ℝn+1
. We define the polynomial associated with a
:
pa(x)=a0+a1x1+⋯+anxn
and the ordinary Euclidean length of a
:
|a|=a02+a12+⋯+an2
The Euclidean length defines a metric on ℝn+1
where the distance between a
and b
is d(a,b)=|b-a|
.
The function mapping each a∈ℝn+1
to ‖pa‖
is continuous.
Proof
We will show this when the interval is [0,1]
. Given ε>0
, let δ=ε/(n+1)
. Let a=(a0,…,an)
and b=(b0,…,bn)
, and suppose that |b-a|<δ
. Then for every i=0,…,n
,
|bi-ai|≤|b-a|<ε/(n+1)
.
Hence for any x∈[0,1]
,
|pb(x)-pa(x)|≤|b0-a0|+|b1-a1|⋅|x|+⋯+|bn-an|⋅|xn|<ε/(n+1)+⋯+ε/(n+1)=ε
.
We now can state and prove the main result of this section:
Let f be a continuous function on [a,b]
. Then for every n, there is a polynomial pn(x)
of degree ≤n
such that:
‖f-pn‖=dn=inf{‖f-q‖:q is a polynomial of degree ≤n}
Proof
Let S={a∈ℝn+1:|a|=1}
and let ε=inf{‖pa‖:a∈S}
. Note that ε
is the minimum value of a continuous function on S, a closed bounded set in ℝn+1
. Hence there is an a∈S
such that ε=‖pa(x)‖
. Further, we must have ε>0
, for otherwise pa=0
and a=0
, and so a
cannot be in S
.
Next, observe that ‖pa‖
tends to infinity as |a|
tends to infinity, for
‖pa‖=|a|⋅‖pa/|a|‖≥|a|⋅ε
which grows arbitrarily large as |a|
tends to infinity.
Now choose M so large that ‖pa‖≥dn+1+‖f‖
whenever |a|≥M
. Since
‖f-p‖≥‖p‖-‖f‖≥dn+1+‖f‖-‖f‖=dn+1
it follows that a polynomial pc
of best approximation, if it exists, must satisfy |c|≤M
.
Therefore
dn=inf{‖f-pa‖:a∈ℝn+1}=inf{‖f-pa‖:|a|≤M}
and so dn
is the minimum of a continuous function defined on the closed ball {a∈ℝn+1:|a|≤M}
. By compactness, there is a vector c
within the closed ball that achieves the minimum value. For that vector c
, we have dn=‖f-pc‖
and thus pc
is a polynomial of best approximation.