142 CHAPTER 5. FUNCTIONS ON NORMED LINEAR SPACES
which converges to 0. Thus
limm→∞
supx/∈[−δ ,δ ]
cm(1− x2)m
= 0 (5.6)
Now let φ n (t) ≡ cm(1− t2
)m. Consider f ∈ C ([−1,1]) and extend to let f (x) = f (1) ifx > 1 and f (x) = f (−1) if x <−1 and define pm (x)≡
∫ 1−1 f (x− t)φ m (t)dt. Then
|pm (x)− f (x)| ≤∫ 1
−1| f (x− t)− f (x)|φ m (t)dt ≤
∫ 1
−1X[−δ ,δ ] (t) | f (x− t)− f (x)|φ m (t)dt +
∫ 1
−1X[−1,1]\[−δ ,δ ] (t) | f (x− t)− f (x)|φ m (t)dt
Choose δ so small that if |x− y| < δ , then | f (x)− f (y)| < ε . Also let M ≥ maxx | f (x)|.Then
|pm (x)− f (x)| ≤ ε
∫ 1
−1φ m (t)dt +2M
∫ 1
−1X[−1,1]\[−δ ,δ ] (t)φ m (t)dt
= ε +2M∫ 1
−1X[−1,1]\[−δ ,δ ] (t)φ m (t)dt
From 5.6, The second term is no larger than 2M∫ 1−1 X[−1,1]\[−δ ,δ ] (t)εdt ≤ 4Mε whenever
m is large enough. Hence, for large enough m, supx∈[−1,1] |pm (x)− f (x)| ≤ (1+4M)ε .Since ε is arbitrary, this shows that the functions pm converge uniformly to f on [−1,1].However, pm is actually a polynomial. To see this, change the variables and obtain
pm (x) =∫ x+1
x−1f (t)φ m (x− t)dt
which will be a polynomial. To see this, note that a typical term is of the form∫ x+1
x−1f (t)a(x− t)k dt,
clearly a polynomial in x. This proves Corollary 5.10.2 in case [a,b] = [−1,1]. In thegeneral case, there is a linear one to one onto map l : [−1,1]→ [a,b].
l (t) =b−a
2(t +1)+a
Then if f ∈C ([a,b]) , f ◦ l ∈C ([−1,1]) . Hence there is a polynomial p such that
maxt∈[−1,1]
| f ◦ l (t)− p(t)|< ε
Then letting t = l−1 (x) = 2(x−a)b−a − 1, for x ∈ [a,b] ,maxx∈[a,b]
∣∣ f (x)− p(l−1 (x)
)∣∣ < ε butx→ p
(l−1 (x)
)is a polynomial. This gives an independent proof of that corollary. ■
The next result is the key to the profound generalization of the Weierstrass theorem dueto Stone in which an interval will be replaced by a compact and later a locally compact setand polynomials will be replaced with elements of an algebra satisfying certain axioms.
Corollary 5.10.3 On the interval [−M,M], there exist polynomials pn, pn (0) = 0, andlimn→∞ ∥pn−|·|∥∞
= 0. recall that ∥ f∥∞≡ supt∈[−M,M] | f (t)|.