Approximation of Compactly Supported Continuous Functions by Polynomials

Math 402/502 Final Project: Approximation of compactly supported continuous functions by polynomials

Charles Clauss and Joel Upston May 6, 2015

1 Abstract

In this report we will be following Section 3.8 from Terrence Tao’s Analysis II as he presents the Weier- strass approximation theorem. The goal will be to provide a fully ﬂedged explanation and proof of every piece that Tao leaves as exercises for the reader. The theorem requires three main points, those being an approximation to the identity, convolution, and the relation between any chosen function and a carefully constructed polynomial. In essence, a continuous function on a closed interval that is convolved with a polynomial approximation to the identity will yield a new polynomial function that is uniformly close to the original function. After showing all of the machinery necessary to achieve this result we will also make a few remarks regarding extensions of this theorem, however the reader should note the original hypotheses that require a compactly supported, continuous function.

2 Compactly supported functions

As we build toward the theorem we will need a few diﬀerent pieces. A logical place to start will be to state some important properties of compactly supported functions that we will need.

We would like to approximate continuous functions with polynomials, but are polynomials continuous? As a matter of fact, they are. If we consider the constant functions f(x) = c for all c ∈ R, and the identity function f(x) = x, we can see that:

limx→x0 f(x) = c = f(x0) and limx→x0 f(x) = x0 = f(x0). Therefore both of these functions are continuous everywhere, and additionally, arithmetic of functions pre- serves continuity, so we can build any polynomial out of these two functions, and it will also be continuous.

A function f : R → R is supported on [a, b] iﬀ f(x) = 0 ∀ x∈ / [a, b]. It is compactly supported iﬀ ∃ [a, b] such that f(x) is supported. Finally, if f is continuous and compactly supported on [a, b] then it should be R ∞ R b clear that −∞ f(x) = a f(x).

Given that we are considering functions that are continuous and of the form f :[a, b] → R, we are on a compact set, and it follows immediately that they are uniformly continuous and bounded.

3 Polynomial approximation to the identity

We would like an identity function that we can approximate with polynomials that will serve as a tool for approximation of more complicated functions. Given > 0 and 0 < δ < 1, a function f : R → R is an (, δ) approximation if:

(a) f is supported on [−1, 1], and f(x) ≥ 0 for all −1 ≤ x ≤ 1. R ∞ (b) f is continuous, and −∞ f = 1. (c) f(x) ≤ for all δ ≤ |x| ≤ 1.

1 Given our deﬁnition of the identity, we will see that we will want it to be a polynomial. We can do this by following a few careful steps. First we will compare two functions by diﬀerentiation to show that (1 − x2)n ≥ 1 − nx for x ∈ R, n ∈ Z, 0 ≤ x ≤ 1, n ≥ 1 (1 − x2)n ≥ 1 − nx −2nx(1 − x2)n−1 ≥ −n x(1 − x2)n−1 ≤ 1. This holds for x positive and bounded above by 1.

0 ≤ x ≤ 1 0 ≤ x2 ≤ 1 0 ≥ −x2 ≥ −1 1 ≥ 1 − x2 ≥ 0 1 ≥ (1 − x2)n ≥ 0 x ≥ x(1 − x2)n ≥ 0. A symmetric argument holds for −1 ≤ x ≤ 0 using 1 + nx instead.

Now we want to show that the function c(1 − x2)n can approximate the identity. Therefore we must show that each of the listed properties of an approximation holds.

First, deﬁne the function f to be our indicated function on the interval [−1, 1] and 0 everywhere else. Then f is supported on [−1, 1] by deﬁnition of support, and we just showed previously that it is also positive.

Continuity has already been established for all polynomials, so it remains to show that the integral is unit and that the tails are suﬃciently small.

R ∞ Consider −∞ f since the function is zero outside the interval [−1, 1], we can say immediately that R ∞ R 1 −∞ f = −1 f

R 1 2 N Examining the integral −1 c(1 − x ) for some finite N, we can use binomial theorem to expand our polynomial, and exchange sum and integral due to integration on a compact set. N R 1 2 N R 1 P N n 2n −1 c(1 − x ) dx = c −1 n (−1) x dx n=0 N R 1 2 N P N n R 1 2n −1 c(1 − x ) = c n (−1) −1 x dx n=0 N R 1 2 N P N n 2n+1 1 −1 c(1 − x ) = c n (−1) x /(2n + 1) |−1 n=0 N R 1 2 N P N n −1 c(1 − x ) = 2c n (−1) /(2n + 1). n=0 N P N n −1 Thus the sum is finite and we can define c(N) := (2 ∗ n (−1) /(2n + 1)) such that the integral is 1. n=0 Notice that the shape of these functions becomes very narrow as N becomes large, so the constant c depends on N, and must also become very large to compensate.

1 1 N 2 Lastly, for any < c we can choose δ = (1 − ( c ) ) , and if we look again at the derivative of this functional form, we can see that from [−1, 0] it is increasing and from [0, 1] it is decreasing. Therefore for δ < |x| < 1 we can conclusively say that f(x) < .

4 Convolutions

We will also need a few important properties of convolution in order to construct needed polynomials. Deﬁnition of convolution is given by, Z ∞ f ∗ g(x) = (f(y)g(x − y))dy. −∞ Then we can show properties of convolutions given by the Proposition 3.8.11.

2 Proposition 1. Let f : R → R, g : R → R, h : R → R be continuous and compactly supported functions then the following properties hold,

1. The convolution of f ∗ g(x) is also a continuous and compactly supported function. 2. The convolution is commutative f ∗ g(x) = g ∗ f(x).

3. The convolution is a linear operator such that f ∗ (g + h)(x) = f ∗ g(x) + f ∗ h(x) and let c ∈ R then f ∗ (cg)(x) = c(f ∗ g(x)).

Proof of 1. First we will show that the convolution of f ∗g(x) is continuous. First let us use that g : R → R is continuous. Therefore ∀0 > 0 ∃δg such that for every x, y ∈ R that satisfy |x − y| < δg, then |g(x) − g(y)| < .Therefore ∀ > 0, we let = then ∃δ such that ∀x, y ∈ , |x − y| < δ , so that 0 0 M(b − a) g R g

|g(x) − g(y)| < 0

. Then we can use the fact that |f(x)| < M for all x ∈ [a, b], since it is continuous on a compact set so it is uniformly bounded. Therefore we use these facts and triangle inequality to see that ∀ > 0 ∃δ = δg such that for every x, y ∈ R that satisfy |x − y| < δ ,

Z ∞ Z ∞ |f ∗ g(x) − f ∗ g(x0)| = | f(y)g(x − y)dy − f(y)g(x0 − y)dy| −∞ −∞ Z ∞ Z ∞ = | f(y)(g(x − y) − g(x0 − y))dy| ≤ |f(y)||(g(x − y) − g(x0 − y))|dy −∞ −∞ Z ∞ Z b ≤ |f(y)|dy = |f(y)|dy ≤ M(b − a) = . M(b − a) −∞ M(b − a) a M(b − a) Now to show that f ∗ g(x) is compactly supported we just need to look at the deﬁnition. So WLOG let a < b and c < d and let f be compactly supported on [a, b] and let g be compactly supported on [c, d]. Then since f is compactly supported on [a, b],

Z ∞ Z b f ∗ g(x) = f(y)g(x − y)dy = f(y)g(x − y)dy. −∞ a Therefore since g is also compactly supported on [c, d] if x − y 6∈ [c, d] then g(x − y) = 0. So then since y ∈ [a, b] so if x 6∈ [a+c, b+d] then g(x−y) = 0 therefore f ∗g(x) is compactly supported on [a+c, b+d]. Z ∞ Proof of 2. By deﬁnition, f ∗ g(x) = f(y)g(x − y)dy. So then, −∞

Z ∞ f ∗ g(x) = f(y)g(x − y)dy −∞ du We let u = x − y so y = u − x and then = −1. Therefore dy Z ∞ Z −∞ Z ∞ Z ∞ f ∗ g(x) = f(y)g(x − y)dy = − f(u − x)g(u)du = f(u − x)g(u)du = g(y)f(y − x)dy. −∞ ∞ −∞ −∞

Proof of 3.

Z ∞ Z ∞ f ∗ c(g + h)(x) = f(y)(cg(x − y) + ch(x − y)dy = c f(y)(g(x − y) − h(x − y))dy = −∞ −∞ Z ∞ Z ∞ c f(y)g(x − y)dy + c f(y)h(x − y)dy = c(f ∗ g(x)) + c(f ∗ h(x)) −∞ −∞

3 5 Building the Weierstrass approximation theorem

We will show that we can create polynomials that are very close to our original function on the interval [0, 1]. After which we will show that we can do this for any interval [a, b].

Lemma 1. Let f : R → R be a continuous function supported on [0, 1] and let g : R → R be a continuous function supported on [-1,1] which is a polynomial on [-1,1]. Then f ∗ g(x) is a polynomial on [0,1]. Proof. Since g is a polynomial on [-1,1] then ∃n such that

n X j g(x) = cjx ∀x ∈ [−1, 1]. j=0

Then since f is compactly supported on [0,1],

Z ∞ Z 1 f ∗ g(x) = f(y)g(x − y)dy = f(y)g(x − y)dy. −∞ 0 So for x ∈ [0, 1] then x−y ∈ [−1, 1] therefore g(x−y) is a polynomial which we can substitute in the integral. Thus using the binomial formula,

j X j! (x − a)j = (b − a)j−m(x − b)m, m!(j − m)! m=0 which holds for a,b real numbers and for any real number x, therefore let a=y and b=0. Thus we see that

n n j X X X j! c (x − y)j = c (−y)j−m(x)m. j j m!(j − m)! j=0 j=0 m=0 Then by rearranging the sum since the sums are ﬁnite and of ﬁnite terms and by re indexing we obtain,

n n n X X X j! c (x − y)j = c (−y)j−m(x)m j j m!(j − m)! j=0 m=0 j=m n n−m X X (j + m)! = c (−y)j(x)m. j+m m!j! m=0 j=0 Therefore if we now substitute inside the integral to get, Z 1 f ∗ g(x) = f(y)g(x − y)dy 0 n n−m Z 1 X X (j + m)! = f(y) c (−y)j(x)mdy j+m m!j! 0 m=0 j=0 n n−m X Z 1 X (j + m)! = xm f(y) c (−y)jdy. j+m m!j! m=0 0 j=0

Then we deﬁne Cm as, n−m Z 1 X (j + m)! C := f(y) c (−y)jdy. m j+m m!j! 0 j=0

Therefore Cm is a constant and so n X m f ∗ g(x) = Cmx , m=0 which is a polynomial for x ∈ [0, 1]. Next we will need to show that a convolution with a bounded continuous function and an approximation to the identity are close to the original continuous function.

Proof. First we need to show that we can bound g, the (, δ) approximation to the identity, by

Z δ 1 − 2 ≤ g(y)dy ≤ 1. −δ Then we know that g is supported on [1,1] and g(y) ≥ 0 for all −1 ≤ x ≤ 1, so by deﬁnition of compactly supported and laws of integration we can see that, Z ∞ Z 1 1 = g(y)dy = g(y)dy ≥ 0. −∞ −1 Moreover we can split the domain of the integral and see that,

Z δ Z δ Z 1 g(y)dy ≥ 0, g(y)dy ≥ 0, and g(y)dy ≥ 0. −1 −δ δ

So,

Z −δ Z δ Z 1 g(y)dy + g(y)dy + g(y)dy = 1 −1 −δ δ Z δ Z −δ Z 1 g(y)dy = 1 − g(y)dy − g(y)dy −δ 1 δ ≤ 1.

Then to ﬁnd the lower bound,

Z 1 Z −δ Z δ Z 1 1 = | g(y)dy| = | g(y)dy + g(y)dy + g(y)dy| −1 −1 −δ δ Z −δ Z δ Z 1 ≤ | g(y)dy| + | g(y)dy| + | g(y)dy| −1 −δ δ Z −δ Z δ Z 1 ≤ |g(y)|dy + | g(y)dy| + |g(y)|dy. −1 −δ δ

Z δ However since g(y)dy ≥ 0 the absolute values are not necessary. Therefore, −δ

Z δ Z δ 1 ≤ (−δ + 1) + (1 − δ) + g(y)dy ≤ 2 + g(y)dy −δ −δ Z δ 1 − 2 ≤ g(y)dy. −δ Next we need to show that, |f ∗ g(x) − f(x)| ≤ (1 + 4M). (1) Z 1 So now we want to rewrite f(x) as an integral since g(y)dy = 1 therefore, −1

Z 1 Z −δ Z δ Z 1 f(x) = f(x) g(y)dy = f(x)g(y)dy + f(x)g(y)dy + f(x)g(y)dy. (2) −1 −1 −δ δ

5 Using this we see that,

Z 1 Z −δ Z δ Z 1 ! |f ∗ g(x) − f(x)| = | f(y)g(x − y)dy − f(x)g(y)dy + f(x)g(y)dy + f(x)g(y)dy | −1 −1 −δ δ Z 1 Z −δ Z δ Z 1 = | f(x − y)g(y)dy − f(x)g(y)dy − f(x)g(y)dy − f(x)g(y)dy| −1 −1 −δ δ Z −δ Z δ Z 1 = | (f(x − y) − f(x))g(y)dy + (f(x − y) − f(x))g(y)dy + (f(x − y) − f(x))g(y)dy| −1 −δ δ Z −δ Z δ Z 1 ≤ | (f(x − y) − f(x))g(y)dy| + | (f(x − y) − f(x))g(y)dy| + | (f(x − y) − f(x))g(y)dy| −1 −δ δ Z −δ Z δ Z 1 ≤ |f(x − y) − f(x)||g(y)|dy + |(f(x − y) − f(x))||g(y)|dy + |(f(x − y) − f(x))||g(y)|dy. −1 −δ δ

Z −δ Z δ Z 1 |f ∗ g(x) − f(x)| ≤ |f(x − y) − f(x)||g(y)|dy + |g(y)|dy + |(f(x − y) − f(x))||g(y)|dy −1 −δ δ Z −δ Z δ Z 1 = |f(x − y) − f(x)||g(y)|dy + |g(y)|dy + |(f(x − y) − f(x))||g(y)|dy −1 −δ δ since in that integral |y| < δ. Then next we know that |f(x)| ≤ M so by using triangle inequality of |f(x − y) − f(x)| ≤ |f(x − y)| + |f(x)| we obtain,

Z −δ Z δ Z 1 |f ∗ g(x) − f(x)| ≤ (|f(x − y)| + |f(x)|)|g(y)|dy + |g(y)|dy + (|(f(x − y)| + |f(x)|)|g(y)|dy −1 −δ δ Z −δ Z δ Z 1 ≤ (2M)|g(y)|dy + |g(y)|dy + (2M)|g(y)|dy −1 −δ δ Z −δ Z δ Z 1 = 2M |g(y)|dy + |g(y)|dy + 2M |g(y)|dy. −1 −δ δ

Corollary 1. Let f : R → R be a continuous function supported on [0, 1]. Then for every > 0, there exists a function P : R → R which is a polynomial on [0, 1] and such that |P (x) − f(x)| ≤ for all x ∈ [0, 1]. Proof. Since Lemma 2 works for any , δ approximation to the identity let g := g( , δ). Then by 1 + 4M Lemma 2, |f ∗ g(x) − f(x)| < . Then ∃q polynomial on [0,1] such that g = q on [0,1] by Theorem 1. Therefore,

|f ∗ q(x) − f(x)| < .

However by Lemma 1, f ∗ q(x) is a polynomial on [0,1]. So we let P (x) = f ∗ q(x), and we obtain

|P (x) − f(x)| < .

6 Lemma 3. Let f : [0, 1] → R be a continuous function which equals 0 on the boundary of [0, 1], i.e., f(0)=f(1)=0. Let F : R → R be the function deﬁned by setting F (x) := f(x) for x ∈ [0, 1] and F (x) := 0 for x 6∈ [0, 1]. Then F is also continuous. Proof. Since [0,1] is a interval where f is already continuous and the F is trivially continuous outside of this interval, then the only place we have to check is at x = 0 and x = 1. However this is trivially as well since F (1) = F (0) = 0 and f is continuous on the whole interval [0,1].

Corollary 2. Let f : [0, 1] → R be a continuous function supported on [0, 1] such that f(0) = f(1) = 0. The for every > 0 there exists a polynomial P : [0, 1] → R such that |P (x) − f(x)| ≤ for all x ∈ [0, 1]. Proof. Let us deﬁne F : R → R as, ( f(x), if x ∈ [0, 1] F (x) := . (3) 0, if x 6∈ [0, 1]

Then by previous Corollary, ∃ a function P 0 : R → R which is a polynomial P on [0,1] such that |P 0(x) − F (x)| < for all x ∈ [0, 1]. However when x ∈ [0, 1] then |P (x) − f(x)| < .

Corollary 3. Let f : [0, 1] → R be a continuous function supported on [0, 1]. The for every > 0 there exists a polynomial P : [0, 1] → R such that |P (x) − f(x)| ≤ for all x ∈ [0, 1]. Proof. Let F : [0, 1] → R deﬁned as, F (x) := f(x) − f(0) − x(f(1) − f(0)). (4) Since f(1) − f(0) is some constant and a constant times x is continuous and then adding f(x) which is also continuous makes F (x) continuous on [0,1]. Then by also realizing that F (1) = F (0) = 0 we can apply the previous Corollary so that, ∃ a polynomial Q : [0, 1] → R such that |Q(x) − F (x)| < for all x ∈ [0, 1]. However Q(x) − F (x) = Q(x) + f(0) + x(f(1) − f(0)) − f(x). Therefore let P (x) := Q(x) + f(0) + x(f(1) − f(0)) which is also a polynomial and satisﬁes the condition.

Theorem 1. If [a,b] is an interval, f :[a, b] → R is a continuous function, and > 0, then there exists a polynomial P on [a, b] such that |P (x) − f(x)| ≤ for all x ∈ [a, b]. Proof. Let G : [0, 1] → R be the function, G(x) := f(a + (b − a)x) for all x ∈ [0, 1]. (5) Then G(x) is continuous on [0,1] since it is a composition of continuous functions. Therefore if we apply the previous Corollary, ∃ Q : [0, 1] → R where Q is a polynomial such that |Q(x) − g(x)| < for all x ∈ [0, 1] (6) y − a Then if we let y = a + (b − a)x, so x = . Thus we set P (y) = Q((y − a)/(b − a)), so P (y) is also a b − a polynomial since it is a composition of polynomials. Therefore we obtain, |Q((y − a)/(b − a)) − g((y − a)/(b − a))| < ∀y ∈ [a, b] |P (y) − f(y)| < ∀y ∈ [a, b]

6 Conclusion

In conclusion, we see that this theorem does have some restrictions, for example it has to be on a compact set in R. Otherwise we could ﬁnd a polynomial that approximates ex on the Real line, which is impossible since the exponential function grows faster than any polynomial. However although we did the proof in R we can bring it to higher dimensions Rn, where the image is in R. Even more we can generalize to an arbitrary metric space through the Stone-Weierstrass Theorem.