288 CHAPTER 12. SERIES AND TRANSFORMS
Then using this fact as needed, consider the following computation in which I will try tochoose bk to make ∫
π
−π
∣∣∣∣∣ f (θ)− n
∑k=−n
bkeikθ
∣∣∣∣∣2
dθ (12.13)
as small as possible. Remember that |z|2 = zz̄ whenever z is a complex number. Using thisand doing routine computations,∫
π
−π
∣∣∣∣∣ f (θ)− n
∑k=−n
bkeikθ
∣∣∣∣∣2
dθ
=∫
π
−π
| f (θ)|2 dθ −2Re∫
π
−π
n
∑k=−n
f (θ)bkeiθ dθ +2π
n
∑k=−n|bk|2
=∫
π
−π
| f (θ)|2 dθ −2(2π)Ren
∑k=−n
akbk +2π
n
∑k=−n|bk|2
Note that if bk = ak, this would equal∫π
−π
| f (θ)|2 dθ −2(2π)n
∑k=−n|ak|2 +2π
n
∑k=−n|ak|2 =
∫π
−π
| f (θ)|2 dθ −2π
n
∑k=−n|ak|2
In the general case, it follows from the Cauchy Schwarz inequality,
≥∫
π
−π
| f (θ)|2 dθ −2(2π)
(n
∑k=−n|ak|2
)1/2( n
∑k=−n|bk|2
)1/2
+2π
n
∑k=−n|bk|2
≥∫
π
−π
| f (θ)|2 dθ −2π
(n
∑k=−n|ak|2 +
n
∑k=−n|bk|2
)+2π
n
∑k=−n|bk|2
=∫
π
−π
| f (θ)|2 dθ −2π
n
∑k=−n|ak|2
Therefore, the expression in 12.13 is minimized when bk = ak. We also observe the fol-lowing fundamental inequality. For ak the Fourier coefficients,∫
π
−π
∣∣∣∣∣ f (θ)− n
∑k=−n
akeikθ
∣∣∣∣∣2
dθ ≡∫
π
−π
| f (θ)−Sn f (θ)|2 dθ
=∫
π
−π
| f (θ)|2 dθ −2π
n
∑k=−n|ak|2 ≥ 0
so this yields Parseval’s inequality, an important inequality involving the Fourier coeffi-cients, 1
2π
∫π
−π| f (θ)|2 dθ ≥ ∑
nk=−n |ak|2. This has proved most of the following approxi-
mation theorem.
Theorem 12.6.1 Let αn f (x) denote any linear combination of the functions θ →eikθ for −n≤ k ≤ n. Then∫
π
−π
| f −αn f |2 dx≥∫
π
−π
| f −Sn f |2 dx
Also,∫
π
−π|Sn f |2 dx≤
∫π
−π| f |2 dx.