MTH631 Lectures 20 23 25 Merged by VUBuddy

18.100A: Complete Lecture Notes

Lecture 25:

Power Series and the Weierstrass Approximation Theorem

Last time, we asked three questions about interchanging limits:

Q1. Hence, we ask three questions about interchanging limits:

Q1. If fn : S → R, fn continuous and fn → f pointwise or uniform, then is f continuous?

2. If fn : [a, b] → R, fn differentiable, and fn → f with f ′ n → g, then is f differentiable and does g = f ′?
3. If fn : [a, b] → R, with fn and f continuous such that fn → f , then does

Z b

a fn = Z b

a f ?

The answer to the above questions are all no, if the convergence is pointwise as seen by the following counterexamples:

1. Let fn(x) = xn on [0, 1] is continuous ∀n. As we noted earlier, fn(x) → f (x) = x ∈ [0, 1)

x = 1 . Notice that

f is not continuous.

2. Let fn(x) = xn+1 n+1 on [0, 1]. Then, fn → 0 pointwise on [0, 1]. However,

f ′ n(x) → g(x) = x ∈ [0, 1)

x = 1

Thus, g(x)̸ = (0)′ = 0 at x = 1.

3. Consider the functions

fn(x) = 4n2x x ∈ 0, 1 2n 4n − 4n2x x ∈  1 2n , 1 n x ∈  1 n , 1

as described in the previous lecture. Then, fn(x) → 0 pointwise on [0,1] as we showed last time. However,

Z 1 fn = 1 2 (base)(height) = 1 2n · 2n = 1̸ → 0 = Z 1 0.

We now prove that the answer to the three questions above is yes if convergence is uniform.

Theorem 2

If fn : S → R is continuous for all n, f : S → R, and fn → f uniformly, then f is continuous.

Proof : Let c ∈ S and let ϵ > 0. Since fn → f uniformly, ∃M ∈ N such that ∀n ≥ M , ∀y ∈ S,

|fn(y) − f (y)| < ϵ 3 . Since fM : S → R is continuous, ∃δ0 > 0 such that ∀|x − c| < δ0,

|fM (x) − fM (c)| < ϵ 3 .

Choose δ = δ0. If |x − c| < δ, then

|f (x) − f (c)| ≤|f (x) + fM (x)| + |fM (c) − f (c)| + |fM (x) − fM (c)| < ϵ 3 + ϵ 3 + ϵ 3 = ϵ.

Theorem 3

If fn : [a, b] → R is continuous for all n, f : [a, b] → R and fn → f uniformly, then

Z b

a fn → Z b

a f.

Proof : Let ϵ > 0. Since fn → f uniform, ∃M0 ∈ N such that ∀n ≥ M0, ∀x ∈ [a, b],

|fn(x) − f (x)| < ϵ b − a .

Then, for all n ≥ M = M0, we have

Z b

a fn − Z b

a f Z b

a |fn − f | < Z b

a ϵ b − a = ϵ.

Remark 4. Notationally, this states that

lim n→∞ Z b

a fn = Z b

a lim n→∞ fn = Z b

a f.

Theorem 5

If fn : [a, b] → R is continuously differentiable, f : [a, b] → R, g : [a, b] → R, and

fn → f pointwise,

f ′ n → g uniformly,

then f is continuously differentiable and g = f ′.

Proof : By the FTC, ∀n∀x ∈ [a, b],

fn(x) − fn(a) = Z x

a f ′ n.

Thus, by the previous two theorems,

f (x) − f (a) = lim n→∞(fn(x) − fn(a))

= lim n→∞ Z x

a f ′ n Z x

a g. Therefore, f (x) = f (a) + R x a g. Thus, by the FTC, f is differentiable and f ′ = ( R x a g)′ = g.

We now return back to our study of power series, answering some questions we asked at the beginning of Lecture 23.

Theorem 6

Let P∞ j=0 aj (x − x0)j be a power series of radius of convergence p ∈ (0, ∞]. Then, ∀r ∈ (0, p), P∞ j=0 aj (x − x0)j

converges uniformly on [x0 − r, x0 + r].

Proof : Let r ∈ [0, p). Then, ∀j ∈ N ∪{0}, ∀x ∈ [x0 − r, x0 + r],

|aj (x − x0)j | ≤|aj |rj =: Mj .

Now,

lim j →∞M 1/j j = lim j→∞ |aj |1/j r = r p p < ∞

p = ∞

since p−1 = limj→∞ |aj |1/j . Since r < p, we have

lim j→∞ M 1/j j < 1 =⇒ X

j=0 Mj converges.

By the Weierstrass M-test, it follows that P∞ j=0 aj (x − x0)j converges uniformly on [x0 − r, x0 + r].

Theorem 7

Let P∞ j=0 aj (x − x0)j be a power series with radius of convergence p ∈ (0, ∞]. Then,

1. ∀c ∈ (x0 − p, x0 + p), P∞ j=0 aj (x − x0)j is differentiable at c and

d dx X

j=0 aj (x − x0)j = X

j=0 jaj (x − x0)j−1.

2. ∀a, b such that x0 − p < a < b < x0 + p,

Z b

a X

j=0 aJ (x − x0)j dx = X

j=0 aj  (b − x0)j+1 j + 1 − (a − x0)j+1 j + 1

Remark 8. Since

lim j→∞((j + 1)|aj+1|)1/j = lim j→∞ ((j + 1)|aj+1|1/(j+1)(j+1)/j = lim k→∞ |ak|1/k = p,

1. implies P aj (x − x0)j is infinitely differentiable and

k!ak =  dk

dxk X aj (x − x0)j x=x0 .

Weierstrass Approximation Theorem

Remark 9. This theorem essentially states: " Every continuous function on [a, b] is almost a polynomial."