Math 432 - Real Analysis II Solutions to Homework Due May 1

Math 432 - Real Analysis II Solutions to Homework due May 1

Recall that a function f is called even if f(−x) = f(x) and called odd if f(−x) = −f(x) for all x. We saw that these classes of functions had a particularly nice orthogonality relationship. More precisely, we considered the inner product space

C([−1, 1]) = {f :[−1, 1] → R | f is continuous} and its inner product Z 1 hf, gi = f(t)g(t) dt. −1 Furthermore, we considered the susbpace of even functions and odd functions given by

E = {f ∈ C([−1, 1]) | f(−x) = f(x)}

O = {f ∈ C([−1, 1]) | f(−x) = −f(x)}. In class, we proved that E⊥ = O.

(a) Show that if f ∈ E ∩ O, then f is equal to the zero function (i.e., the zero vector in this vector space). (b) Use (a) to show that C([−1, 1]) = E ⊕ O. To do this, show that you can write any f ∈ C([−1, 1]) uniquely as f = fe + fo, where fe ∈ E and fo ∈ O. Note that you cannot use the decomposition theorem from class since that requires your subspace to be ﬁnite dimensional, which this is not.

(c) Show that if a function f is odd, then f(0) = 0. (d) Show that the derivative of an even function is odd and the derivative of an odd function is even.

n n−1 (e) Let p(x) = anx + an−1x + ··· + a1x + a0 be a polynomial. Show p is odd if and only if ak = 0 for all even k. As a hint, take a derivative and evaluate the derivative at 0; continue in this pattern.

(f) Similar to above, show that the polynomial p is even if and only if ak = 0 for all odd k. (g) Consider the polynomial p(x) = −5x7 − 7x6 + x4 − 2x2 + x − 1. Decompose p as an even function and an odd function. Use the above to verify that your your components are indeed even and odd. (h) Decompose f(x) = ex as an even plus odd function. The even and odd function components have special names; what are they?

Solution 1.

(a) If f ∈ E ∩ O, then f(x) = f(−x) = −f(x) for all x. Since f(x) = −f(x) for all x, then f(x) = 0 for all x. Thus, f is the zero constant function.

(b) In class, we learned that we can write any function on [−1, 1] as f = fe + fo where

f(x) + f(−x) f(x) − f(−x) f (x) = and f (x) = . e 2 o 2

Note that in particular fe ∈ E and fo ∈ O. Assume that we have another expression of f as an even plus and an odd function. Thus, f = ge + go for ge ∈ E and go ∈ O. Then, consider fe − ge ∈ E. Notice that fe − ge = (f − fo) − (f − go) = go − fo ∈ O. Thus, fe − ge ∈ E ∩ O. By above, we have that fe − ge = 0, the constant 0 function. Thus, fe = ge. A similar computation shows that fo = g0. Thus, there is a unique way to decompose f as an even plus an odd function. So, C([−1, 1]) = E ⊕ O.

1 (c) If f is odd, then f(−x) = −f(x). When x = 0, we get that f(0) = −f(0). The only way for this to be true is if f(0) = 0. (d) If f is even, then f(−x) = f(x). Diﬀerentiating, we get that −f 0(−x) = f 0(x), and thus f 0(−x) = −f 0(x). Thus, the derivative of an even function is odd. Now, assume that g is odd. Then g(−x) = −g(x). Diﬀerentiating, we get that −g0(−x) = −g0(x) and thus g0(−x) = g0(x). Thus, g0 is even.

(e) If ak = 0 for all even k, then computing p(−x), we get a constant −1 term that can be factored out. Thus, p(−x) = −p(x). Conversely, assume that p is odd. Then, plugging in 0, we get that p(0) = a0. By the above, we must have that p(0) = 0, thus, a0 = 0. Taking two derivatives, we get that

00 n−2 n−3 p (x) = n(n − 1)anx + (n − 1)(n − 2)an−1x + ··· + 2a2.

Since this is two derivatives of an odd function, it is itself odd. Thus, evaluating at 0, we get that 00 p (0) = 2a2 = 0. Thus, a2 = 0. Continuing in this way, we get that ak = 0 for all even k.

(f) If ak = 0 for all odd k, then replacing x by −x will not change the polynomial. Thus, p(−x) = p(x). 0 n−1 Conversely, assume that p is even. Then, its derivative p (x) = nanx + ··· + a1 is an odd function. 0 Thus, evaluating at 0 should yield 0. Thus, we get that 0 = p (0) = a1. Taking two more derivatives, we get another odd function and, after evaluating at 0, we must have that a3 = 0. Continuing in this way, we get that ak = 0 for all odd k.

(g) Using the equation for pe and po, we get that

(−5x7 − 7x6 + x4 − 2x2 + x − 1) + (5x7 − 7x6 + x4 − 2x2 − x − 1 p (x) = = −7x6 + x4 − 2x2 − 1; e 2 (−5x7 − 7x6 + x4 − 2x2 + x − 1) − (5x7 − 7x6 + x4 − 2x2 − x − 1) p (x) = = −5x7 − x. o 2

(h) Using the equation for fe and fo mentioned above, we get that

ex + e−x ex − e−x f (x) = and f (x) = . e 2 o 2

The function fe is also known as cosh x and fo is known as sinh x.

Question 2. Consider the ﬁnite dimensional vector space Pn, the space of polynomials with real coeﬃcients of degree at most n, along with its inner product given by

Z 1 hf, gi = f(t)g(t) dt. −1

(a) A common basis for P1 is {1, x}. Use the Gram-Schmidt Process to turn this into an orthonormal basis for P1. 2 (b) A common basis for P2 is {1, x, x }. Use the Gram-Schmidt Process to turn this into an orthonormal basis for P2. ⊥ (c) Let U be the subspace of P1 of constant functions. Find U . (d) Compute the orthogonal complement of the U ⊥ you computed in (c). Verify that (U ⊥)⊥ = U.

Solution 2.

2 (a) First, we normalize the vector 1. Since Z 1 h1, 1i = 1 dx = 2, −1 1 then normalizing the vector, we get that e1 = √ . To compute the next vector, we ﬁrst compute the 2 dot product of e1 with x. Doing so, we get that

Z 1 2 1 x x he1, xi = √ d = √ = 0. −1 2 2 2 −1 Thus, u2 = x − he1, xie1 = x. To normalize this, we ﬁrst compute

Z 1 3 1 2 x 2 hx, xi = x dx = = . −1 3 −1 3 Thus, r3 e = x. 2 2 Thus, an orthonormal basis for P1 is ( ) 1 r3 √ , x . 2 2

(b) We can use our previous work to ﬁnd the Gram-Schmidt process for P2. First, we compute the various components of u3: Z 1 3 1 2 1 1 1 2 1 2 x 1 x , √ √ = x , 1 = x dx = = . 2 2 2 2 −1 6 −1 3 * + ! r3 r3 Z 1 r3 r3 x2, x x = x3 dx x = 0. 2 2 −1 2 2 Thus, 2 2 2 u3 = x − hx , e2ie2 − hx , e1ie1 = 1 x2 − . 3 Normalizing, we compute that Z 1 Z 1 2 1 hx2 − 1/3, x2 − 1/3i = (x2 − 1/3)2 dx = x4 − x2 + dx = −1 −1 3 9

5 1 x 2 3 1 2 4 2 18 20 10 8 − x + x = − + = − + = . 5 9 9 −1 5 9 9 45 45 45 45 Thus, r 8 ||x2 − 1/3|| = . 45 So, normalizing, we get that r45 e = (x2 − 1/3). 3 8 Thus, the orthonormal basis obtained via the Gram-Schmidt process is ( ) 1 r3 r45 √ , x, (x2 − 1/3) . 2 2 8

3 Question 3. In this question, we will apply orthogonal projections to the space C([−1, 1]) of continuous functions on [−1, 1]. We will project onto the ﬁnite-dimensional subspaces P1 and P2 investigated in Question 2. In what follows, using a computer program that can perform numerical integration (like Grapher, Sage, or Wolfram Alpha) is recommended, as well as using decimal expansions.

(a) Consider the function f(x) = ex ∈ C([−1, 1]). Compute the orthogonal projection of this function onto x the subspace P1. In other words, compute PP1 (e ). Your orthonormal basis for P1 from Question 2a should be handy in this computation.

x x (b) Compute ||e − PP1 (e )||.

(c) Compare this to how “good” of an approximation the ﬁrst-degree Taylor polynomial, T1(x) = 1 + x is for ex by comparing the norm computation from (b) to ||ex − (1 + x)||.

x (d) Repeat steps (a) - (c) but with the subspace P2. That is, compute PP2 (e ) and compare the values of the norms x x x 2 ||e − PP1 (e )|| and ||e − (1 + x + x /2)||.

Solution 3.

(a) To compute the orthogonal projection of ex, we ﬁrst compute the various inner products of ex with the basis elements: 1 Z 1 ex ex, √ = √ dx ≈ 1.662 2 −1 2 * + r3 Z 1 r3 ex, x = xex dx ≈ .9011 2 −1 2 x Thus, the projection of e onto P1 is 1 r3 P (ex) = 1.662√ + .9011 · x = P1 2 2 1.1752 + 1.1036x.

(b) Computing the distance via the norm, we get s Z 1 √ x x x x 2 ||PP1 (e ) − e )|| = (PP1 (e ) − e ) dx ≈ .0527 = .2295. −1

(c) The distance between ex to 1 + x is given by s Z 1 √ ||ex − (1 + x))|| = (ex − 1 − x)2 dx ≈ .1212 = .3481. −1 Thus, the projection of ex onto the space of linear functions is better than the ﬁrst-degree Taylor polynomial for ex.

Question 4. For a ﬁnite dimensional inner produt space V , the Riesz Representation Theorem says that, given any linear function ϕ : V → F, there exists a vector x ∈ V such that ϕ(v) = hv, xi.

For each of the following ϕ, show that it is indeed a linear functional and ﬁnd the corresponding x ∈ V such that ϕ(v) = hv, xi.

4 0 (a) ϕ : P1 → R given by ϕ(p) = p (0). That is, ϕ(p) is obtained by taking a derivative and evaluating at 0.

(b) ϕ : P2 → R given by ϕ(p) = p(0). 0 (c) ϕ : P2 → R given by ϕ(p) = p (0).

Solution 4. n q o √ (a) Consider the orthonormal basis √1 , 3 x . for . Applying ϕ to each of these, we get that ϕ(1/ 2) = 2 2 P1 q 3 q 3 0 and ϕ 2 x = 2 . Therefore, the vector that corresponds to this functional is given by ! √ √ r3 r3 3 ϕ(1/ 2) · 1/ 2 + ϕ x · x = x. 2 2 2

n q q o √ √ (b) For the orthonormal basis √1 , 3 x, 45 (x2 − 1/3) for , we apply ϕ and get ϕ(1/ 2) = 1/ 2, 2 2 8 P2 q 3 q 45 2 q 5 ϕ 2 x = 0, and ϕ 8 (x − 1/3) = − 8 . Thus, the vector that represents the linear functional is ! ! √ √ r3 r3 r45 r45 15 ϕ(1/ 2) · 1/ 2 + ϕ x · x + ϕ (x2 − 1/3) · (x2 − 1/3) = 1/2 − (x2 − 1/3). 2 2 8 8 8

n q q o √ (c) For the orthonormal basis √1 , 3 x, 45 (x2 − 1/3) for , we apply ϕ and get ϕ(1/ 2) = 0, 2 2 8 P2 q 3 q 3 q 45 2 ϕ 2 x = 2 , and ϕ 8 (x − 1/3) = 0. Therefore, the vector that represents the linear functional is ! ! √ √ r3 r3 r45 3 ϕ(1/ 2) · 1/ 2 + ϕ x · x + ϕ (x2 − 1/3) · x2 = x. 2 2 8 2

Question 5. Let V be an n-dimensional inner product space and ﬁx a vector x ∈ V . We saw that we can ∗ deﬁne a linear functional ϕx ∈ V by ϕx(v) = hv, xi. Let U = span{x}. Notice that U is a 1-dimensional subspace if x 6= 0. Furthermore, recall that the kernel of a linear transformation ϕ is the set of all vectors that are sent to 0:

ker(ϕ) = {v ∈ V | ϕ(v) = 0}, which forms a subspace.

⊥ (a) Show that ker(ϕx) = U . (b) Use the Riesz Representation Theorem, the Orthogonal Decomposition Theorem, and (a) to show that the kernel of a linear functional is always a subspace of dimension n − 1 or n. When is it n-dimensional?

(c) Let W be any (n − 1)-dimensional subspace of V . Use (a) to show that W is the kernel of some linear functional. Be sure to deﬁne what that linear functional is.

Solution 5.

5 ⊥ ⊥ (a) We will show that ker(ϕx) = U . First, we will show that ker(ϕx) ⊂ U . To do so, let v ∈ ker(ϕx) and let αx ∈ U. Then, since v ∈ ker(ϕx), we have that 0 = ϕx(v) = hv, xi. Thus, hv, αxi = αhv, xi = α0 = 0. Thus, v ∈ U ⊥. ⊥ ⊥ Next, we will show that U ⊂ ker(ϕx). Let v ∈ U . Then, v is orthogonal to any element of U. In particular, since x ∈ U, hv, xi = 0. Thus,

ϕx(v) = hx, vi = 0.

Thus, v ∈ ker(ϕx). (b) Let ϕ ∈ V ∗ be any linear functional of V . By the Riesz Representation Theorem, there exists a unique x ∈ V such that ϕ(v) = hv, xi. Let U = span{x}. By (a), ker(φ) = U ⊥. By the decomposition Theorem, we have that V = U ⊕ U ⊥ = U ⊕ ker(ϕ). If x 6= 0, then U is 1-dimensional and thus ker(ϕ) is n − 1-dimensional. If x = 0, then U is 0-dimensional and ker(ϕ) is n-dimensional. Thus, ker(ϕ) has dimension n if and only if x = 0 if and only if ϕ is the zero functional. (c) Let W be an (n − 1) dimensional vector space and let W ⊥ be its orthogonal complement. By the Decomposition Theorem, V = W ⊕ W ⊥ and thus W ⊥ is 1-dimensional. Let x be any non-zero vector in ⊥ ⊥ W . Then, W = span{x}. Consider the linear functional ϕx given by ϕx(v) = hv, xi. By (a), we have ⊥ ⊥ that ker(ϕx) = (W ) = W.