Measure, Integration & Real Analysis, 2021a

248 Chapter 8 Hilbert Spaces
8.73 Example best approximation
Find the polynomial g of degree at most 10 that minimizes
∫ 1
−1
√
∣ |x|−g(x) ∣ 2 dx.
Solution We will work in the real Hilbert space L 2 ([−1, 1]) with the usual inner
product 〈g, h〉 = ∫ 1
−1 gh. Fork ∈{0, 1, . . . , 10}, let f k ∈ L 2 ([−1, 1]) be defined by
f k (x) =x k . Let U be the subspace of L 2 ([−1, 1]) defined by
U = span{ f k } k∈{0, ..., 10} .
Apply the Gram–Schmidt process from the proof of 8.67 to { f k } k∈{0, ..., 10} , producing
an orthonormal basis {e k } k∈{0,...,10} of U, which is a closed subspace of
L 2 ([−1, 1]) (see Exercise 8). The point here is that {e k } k∈{0, ..., 10} can be computed
explicitly and exactly by using 8.70 and evaluating some integrals (using software that
can do exact rational arithmetic will make the process easier), getting e 0 (x) =1/ √ 2,
e 1 (x) = √ 6x/2, . . . up to
e 10 (x) =
√
42
512 (−63 + 3465x2 − 30030x 4 + 90090x 6 − 109395x 8 + 46189x 10 ).
Define f ∈ L 2 ([−1, 1]) by f (x) = √ |x|. Because U is the subspace of
L 2 ([−1, 1]) consisting of polynomials of degree at most 10 and P U f equals the
element of U closest to f (see 8.34), the formula in 8.71 tells us that the solution g to
our minimization problem is given by the formula
g =
10
∑
k=0
〈 f , e k 〉e k .
Using the explicit expressions for e 0 ,...,e 10 and again evaluating some integrals,
this gives
g(x) = 693 + 15015x2 − 64350x 4 + 139230x 6 − 138567x 8 + 51051x 10
.
2944
The figure here shows the graph of
f (x) = √ |x| (red) and the graph of
its closest polynomial g (blue) of degree
at most 10; here closest means as
measured in the norm of L 2 ([−1, 1]).
The approximation of f by g is
pretty good, especially considering
that f is not differentiable at 0 and thus
a Taylor series expansion for f does
not make sense.
Measure, Integration & Real Analysis, by Sheldon Axler