Advanced Calculus Single Variable

11.6.2 Mean Square Approximation

The partial sums of the Fourier series of f do a better job approximating f in the mean square sense than any other linear combination of the functions, e^ikθ for

≤ n. This will be shown next. It is nothing but a simple computation. Recall the Fourier coefficients are

∫ π ak = 1-- f (θ)e−ikθ 2π −π

Then using this fact as needed, consider the following computation.

∫ | |2 π || ∑n ikθ|| −π ||f (θ)− bke || dθ k= −n

∫ π ( n ) ( n ) = f (θ) − ∑ bkeikθ f-(θ)− ∑ ale−ilθ dθ −π k=− n l=−n

∫ π 2 ∑ --∫ π ikθ− ilθ ∑ -- ∑ -- = − π|f (θ)| dθ+ bkbl −πe e dθ − 2π blal − 2π bkak kl l k

Then adding and subtracting 2π ∑ _k

²,

∫ π ∑ ( ∑ (-- )) = |f (θ)|2dθ− 2π |ak|2 + 2π (bk − ak) bk − ak −π k k

∫ π ∑ ∑ = |f (θ)|2dθ− 2π |ak|2 + 2π |bk − ak|2 −π k k

Therefore, to make

∫ π || n ||2 ||f (θ)− ∑ b eikθ|| dθ −π | k= −n k |

as small as possible for all choices of b_k, one should let b_k = a_k, the k^th Fourier coefficient. Stated another way,

| | ∫ π|| ∑n ||2 ∫ π 2 ||f (θ)− bkeikθ||dθ ≥ |f (θ)− Snf (θ)| dθ −π k=−n −π

for any choice of b_k. In particular,

∫ π ∫ π |f (θ)− σn+1f (θ)|2dθ ≥ |f (θ)− Snf (θ)|2dθ. (11.23) − π −π

(11.23)

Also,

Similarly,

∫ π ∑n f (θ)Snf (θ)dθ = 2π |ak|2 −π k=−n

and a simple computation of the above sort shows that also

∫ π ------- n∑ Snf (θ)Snf (θ)dθ = 2π |ak|2. −π k=−n

Therefore,

∫ 0 ≤ π (f (θ)− S f (θ))(f-(θ) − S-f (θ))dθ − π n n

showing

n ∫ π ∫ π 2π ∑ |a |2 ≤ |S f (θ)|2dθ ≤ |f (θ)|2 dθ (11.24) k=−n k −π n −π

(11.24)

Now it is easy to prove the following fundamental theorem.

Proof: First assume f is continuous and 2π periodic. Then by 11.23

and the last expression converges to 0 by Theorem 11.6.4.

Next suppose f ∈ R

and < M for all x. Then the construction used in Lemma 11.3.2 yields a continuous function h which vanishes off some closed interval contained in such that

∫ ∫ --ε-- π -1-- π 2 32M 2 > −π|f (x)− h (x)|dx ≥ 2M − π|f (x)− h(x)| dx. (11.25)

(11.25)

Since h vanishes off some closed interval contained in

, if h is extended off to be 2π periodic, it follows the resulting function, still denoted by h, is continuous. Then using the inequality (For a better inequality, see Problem 2.)

( ) (a+ b +c)2 ≤ 4 a2 + b2 + c2

∫ ∫ π |f − S f|2dx = π (|f − h|+ |h − S h|+ |S h− S f|)2dx −π n − π n n n

∫ π ( ) ≤ 4 |f − h|2 + |h − Snh|2 + |Snh − Snf|2 dx −π

and from 11.24 and 11.25 this is no larger than

and by the first part, this last term converges to 0 as n →∞. Therefore, since ε is arbitrary, this shows that for n large enough,

∫ π 2 −π|f − Snf |dx < ε

This proves the theorem.