12.7. EXERCISES 289

Proof: It only remains to verify the last inequality. However, a short computation showsthat

∫π

−π|Sn f |2 dx = 2π ∑

nk=−n |ak|2 ≤

∫π

−π| f (θ)|2 dθ .

Now it is easy to prove the following fundamental theorem.

Theorem 12.6.2 Let f ∈ R([−π,π]) and it is periodic of period 2π . Then

limn→∞

∫π

−π

| f −Sn f |2 dx = 0.

Proof: First assume f is continuous and 2π periodic. Then by Theorem 12.6.1,∫π

−π

| f −Sn f |2 dx ≤∫

π

−π

| f −σn+1 f |2 dx

≤∫

π

−π

∥ f −σn+1 f∥20 dx = 2π ∥ f −σn+1 f∥2

0

and the last expression converges to 0 by Theorem 12.5.3. Here σn+1 f is the Ceasaro meanof f .

Next suppose f ∈ R([−π,π]). By Lemma 10.1.2, there is h with |h| ≤ | f |, h continuousand h equal to 0 at the ±π and

∫π

−π| f −h|2 dx < ε. Since h vanishes at the endpoints, if

h is extended off [−π,π] to be 2π periodic, it follows the resulting function, still denotedby h, is continuous. Then using the inequality (For a better inequality, see Problem 2.)(a+b+ c)2 ≤ 4

(a2 +b2 + c2

),∫

π

−π

| f −Sn f |2 dx =∫

π

−π

(| f −h|+ |h−Snh|+ |Snh−Sn f |)2 dx

≤ 4∫

π

−π

(| f −h|2 + |h−Snh|2 + |Snh−Sn f |2

)dx

≤ 4ε +4∫

π

−π

|h−Snh|2 dx+4∫

π

−π

|Sn (h− f )|2 dx

By Theorem 12.6.1, this is dominated by ≤ 8ε + 4∫

π

−π|h−Snh|2 dx and by the first part,

the last term converges to 0. Since ε is arbitrary, this proves the theorem.

12.7 Exercises1. Suppose f has infinitely many derivatives and is also periodic with period 2π . Let

the Fourier series of f be ∑∞k=−∞

akeikθ . Show that

limk→∞

kmak = limk→∞

kma−k = 0

for every m ∈ N.

2. The proof of Theorem 12.6.2 used the inequality (a+b+ c)2 ≤ 4(a2 +b2 + c2

)whenever a,b and c are nonnegative numbers. In fact the 4 can be replaced with3. Show this is true.

3. Let f be a continuous function defined on [−π,π]. Show there exists a polynomialp such that ||p− f || < ε where ∥g∥ ≡ sup{|g(x)| : x ∈ [−π,π]} . Extend this result