Math 508: Advanced Analysis

Homework 9

Lawrence Tyler Rush

<me@tylerlogic.com>

November 14, 2014

http://coursework.tylerlogic.com/courses/upenn/math508/homework09

1

Let c ∈ ℝ be positive, f ∈

be an even function, and g ∈

and be an odd function.

(a) Show ∫ _-c^cf(x)dx = 2 ∫ ₀^cf(x)dx

We can separate ∫ _-c^cf(x)dx as

∫ c ∫ 0 ∫ c f(x)dx = f (x)dx + f(x )dx -c -c 0

Through a change of variable, setting x = h(u) where h(u) = -u, we obtain

∫ c ∫ 0 ′ ∫ c -cf(x)dx = c f(h(u))h(u)du+ 0 f(x)dx ∫ 0 ∫ c = f(- u)(- 1)du + f(x)dx c∫ ∫ 0 0 c = - c f(u)du+ 0 f(x)dx ∫ c ∫ c = f(u)du+ f (x)dx ∫0c ∫0c = f(x)dx+ f (x)dx 0∫ c 0 = 2 f(x )dx 0

since f is even.

(b) Show ∫ _-c^cg(x)dx = 0

We can separate ∫ _-c^cg(x)dx as

∫ c ∫ 0 ∫ c -cg(x)dx = -cg(x)dx + 0 f (x)dx

Through a change of variable, setting x = h(u) where h(u) = -u, we obtain

∫ c ∫ 0 ∫ c g(x)dx = g(h(u))h′(u)du+ g(x)dx -c c 0 ∫ 0 ∫ c = c g(- u)(- 1)du + 0 g(x)dx ∫ 0 ∫ c = - - g(u )du + g(x)dx ∫ c ∫ 0 0 c = c g(u)du+ 0 g(x)dx ∫ c ∫ c = - g(u)du+ g(x)dx ∫0c ∫0c = - g(x)dx+ g(x)dx 0 0 = 0

since g is odd.

(c) Show ∫ _-c^cf(x)g(x)dx = 0

Since f(-x)g(-x) = f(x)(-g(x)) = -f(x)g(x) then f(x)g(x) is an odd function. By the previous part of this problem

∫ c f(x)g(x)dx = 0 - c

2 Find function f and constant c so that ∫ ₀^xf(t)e^3tdt = c + x - cos(x²)

Let’s guess that f(t) = e^-3t(1 + 2tsin(t²)) so that

∫ x 3t ∫ x 2 ∫ x ∫ x 2 ∫ x 2 0 f(t)e dt = 0 (1 + 2tsin(t))dt = 0 dt+ 0 2tsin(t )dt = x+ 0 2tsin(t )dt

(2.1)

Defining g(s) = √-
s we have g(0) = 0 and g(x²) = x so that by putting t = g(s) we have

∫ x ∫ x2 2tsin (t2)dt = 2(g(s)) sin((g(s))2)g′(s)ds 0 ∫02 ( ) = x 2(√s-)sin(s) -√1- ds 0 2 s ∫ x2 = sin(s)ds 0 = - cos(x2)+ cos(0) = 1- cos(x2)

through a change of variable. Substituting this result back into equation 2.1 we obtain

∫ xf(t)e3tdt = 1+ x- cos(x2) 0

so that we see our definition of f works with c = 1.

3

Let f(x) = √ ----4-
9+ x

and define a partition P of [0,2] by P = {0 = x₀,x₁,…,x_N = 2} where x_j - x_j-1 = Δx = 2∕N. Since f is a monotonically increasing function on [0,2], then

N∑ N∑ U(P,f) = f(xi)Δx = -2 f(xi) i=1 N i=1

and

N N L(P,f) = ∑ f(x )Δx = -2 ∑ f(x ) i=1 i-1 N i=1 i-1

leaving us with an error in estimation of

∑N ∑N U (P,f)- L(P,f) = 2- f(xi)- 2- f(xi-1) =-2(f(2)- f(0)) = 2-(5- 3) = 4 N i=1 N i=1 N N N

Hence, if we’d like the error in our computation of ∫ ₀² √------
9 +x4 dx to be less than 1∕100, then we need 4∕N < 1∕100, i.e. N > 400.

4

Let f(s) be a smooth function and c be a constant. Define

∫ x+ct u(x,t) = f (s)ds x-ct

Applying the fundamental theorem of calculus then yields the following four equations

∂xu = f (x + ct)- f(x - ct) ∂tu = cf(x+ ct)+ cf(x - ct) ∂2xu = f′(x+ ct)- f′(x - ct) ∂2tu = c2f′(x + ct) - c2f′(x- ct)

and so we see that

∂2tu = c2f′(x+ ct) - c2f′(x- ct) = c2(f′(x+ ct)- f′(x - ct)) = c2∂2xu

5

(a)

If we set

( ∫ x ∫ x ) u = 1 - cos(cx) f(t)sin(ct)dt+ sin (cx) f (t)cos(ct)dt c 0 0

(5.2)

then

′ ∫ x ∫ x u = sin(cx) f(t)sin(ct)dt +cos(cx) f(t) cos(ct)dt 0 0

and

( ∫ x ∫ x ) u′′ = c cos(cx) f(t)sin (t)dt- sin(cx) f(t)cos(ct)dt + sin2(cx)f(x)+ cos2(cx)f(x) ( ∫0 ∫0 ) x x ( 2 2 ) = c cos(cx) 0 f(t)sin (t)dt- sin(cx) 0 f(t)cos(ct)dt + f(x) sin (cx)+ cos (cx)

Hence u′′ + c²u = f(x).

(b)

Let 0 < c < 1. Let u(x) and w(x) be solutions to the differential equation w′′ + c²w = f(x) and be zero on the boundary points of [0,π]. Then define v(x) = w(x) - u(x). We would then have

′′ 2 ′′ ′′ 2 ′′ 2 ′′ 2 v + c v = w - u +c (w + u) = (w +c w )- (u + cu ) = f(x)- f(x) = 0

(5.3)

Thus the v must have the form

v = asin(cx)+ bcos(cx)

for some a and b as it is the solution of the homogenous equation in 5.3. Hence v(0) = asin(0) + bcos(0) = b = w(0) -u(0) = 0 so that b is 0. Furthermore v(π) = asin(cπ) = w(π) - u(π) = 0 so that asin(cπ) = 0. Since 0 < c < 1, then a = 0. Thus v(x) = 0, which implies w and u are the same. Hence there is only one unique solution when 0 < c < 1.

(c)

Let c = 1, in which case we have u′′ + u = f. Thus with two applications of integration by parts we get

∫ π ∫ π ′′ ∫ π 0 f (x)sinx dx = 0 u sin x dx + 0 usinx dx ∫ π ∫ π = u′(π)sinπ - u′(0)sin0- u′cosdx+ u sin x dx ∫ π ∫ π 0 0 = - u′cosdx+ u sinx dx (0 0 ∫ ) ∫ = - u (π)sinπ - u(0)sin0- π u(- sinx)dx + π usinx dx 0 0 ( ∫ π ) ∫ π = - - u (- sin x)dx + usin x dx = 0 0 0

(d)

Applying are result from part (a) to the situation when c = 1, the previous part implies equation 5.2 becomes

1 ( ∫ x ∫ x ) u = c - cos(cx) f (t)sin(ct)dt+ sin(cx) f(t)cos(ct)dt ∫ x 0 ∫ x 0 = - cos(x) f (t)sin(t)dt+ sin(x) f(t)cos(t)dt ∫ 0 0 = sin(x) xf (t)cos(t)dt 0

which is the unique solution for u.

6

Let L be the differential operator defined by Lw = -w′′ + c(x)w on the interval J = [a,b], where c(x) is some continuous function. Define the inner product by

∫ b ⟨f,g⟩ = f(x)g(x)dx a

(a) Show ⟨Lu,v⟩ = ⟨u,Lv⟩ for u and v that are both zero on the boundary of J.

We first layout a helpful equation obtained via two applications of integration by parts:

∫ b ∫ b u′′vdx = u′(b)v(b)- u′(a)v(a)- u′v′dx a ∫ ba = u′(b)(0) - u ′(a)(0)- u′v′dx a ∫ b ′′ = - uv dx (a ∫ b ) = - u(b)v′(b)- u(a)v′(a) - uv′′dx a ( ∫ b ) = - (0)v′(b)- (0)v′(a) - uv ′′dx a ∫ b ′′ = uv dx a

taking into account that the boundary points for u and v are zero, i.e. u(a) = v(a) = u(b) = v(b) = 0. With the above equation, our job is simple:

∫ b ⟨Lu,v⟩ = (Lu )vdx ∫ab = (- u′′ + c(x)u)vdx a ∫ b ′′ ∫ b = - a u vdx+ a c(x)uvdx ∫ b ∫ b = - uv′′dx+ c(x)uvdx ∫ a a∫ b ′′ b = a u(- v )dx+ a u (c(x)v)dx ∫ b = u(- v′′ +c(x)v)dx ∫ab = u(Lv)dx a = ⟨u,Lv⟩

(b) If Lu = λ₁u and Lv = λ₂v for λ₁≠λ₂ then ⟨u,v⟩ = 0

In considering (λ₁ - λ₂)⟨u,v⟩, we use the previous part of this problem:

(λ1 - λ2)⟨u,v⟩ = λ1⟨u,v⟩- λ2⟨u,v⟩ = ⟨λ1u,v⟩- ⟨u,λ2v⟩ = ⟨Lu, v⟩- ⟨u,Lv ⟩ = ⟨u,Lv ⟩- ⟨u,Lv ⟩ = 0

Since λ₁≠λ₂ we must therefore have ⟨u,v⟩.

(c) If Lu = 0 and Lv = f(x), show ⟨u,f⟩ = 0

If Lu = 0 and Lv = f, then we have the following

⟨u,f ⟩ = ⟨u,Lv⟩ = ⟨Lu,v⟩ = ⟨0,v⟩ = 0

7 Compute the arclength of X(t) = (cos t, sin t,t) in ℝ³

We have that

∘ --------------- |X ′(t)| = |(- sint,cost,1)| = sin2t + cos2t+ 1 = √2

so that

∫ ∫ 4π|X ′(t)|dt = 4π √2dt = √2-(4π- 0) = 4π√2 0 0

8

9

(a)

This is virtually problem 5 of the previous homework.

(b)

The value ∥f∥₁ has the following three properties

∫ ₀¹|f(x)|dx ≥ 0 with equality only when f = 0.
∫ ₀¹|cf(x)|dx = |c|∫ ₀¹|f(x)|dx = |c|∥f∥
∫ ₀¹|f(x) + g(x)|dx ≤∫ ₀¹|f(x)| + |g(x)|dx = ∥f∥ + ∥g∥

which make it a norm on C([0,1]).

(c)

10 Compute lim _λ→∞∫ ₀¹| sin(λx)|dx

Since the graph of |sin(λx)| for an arbitrary λ is just a sequence of concave “humps”, we can find the area under a single “hump” and then multiply that times the fraction of these “humps” that are between 0 and 1. One such “hump” is the left-most one in [0,1]. It’s area is that of the area under |sin(λx)| on the interval [0, π
λ

]. Furthermore there are λ
π

of these “humps” over the interval [0,1]. Hence we have

∫ 1 λ ∫ πλ 1 ∫ πλ lim |sin(λx)|dx = lim -- |sin(λx)|dx = --lim λ |sin(λx)|dx λ→ ∞ 0 λ→∞ π 0 π λ→∞ 0

However, on the interval [0, π
λ ], sin(λx) is positive and so from the above equation we get

∫ 1 1 ∫ πλ λli→m∞ 0 |sin(λx )|dx = πλl→im∞ λ 0 sin(λx)dx

Now if we set x = g(u) where g(u) = , by a change of variable, we get

∫ ∫ ( ) ∫ lim 1 |sin(λx )|dx = 1-lim λ πsin(λ (u)) 1- du =-1 lim π sin(u)du λ→ ∞ 0 πλ→ ∞ 0 λ λ π λ→ ∞ 0

since g′(u) = , g(π) = , and g(0) = 0. Hence we are left with

∫ 1 ∫ π lim |sin(λx)|dx = 1- lim sin(u)du = 1- lim (- cos(π)+ cos(0)) = 1-lim 0 = 0 λ→∞ 0 π λ→∞ 0 π λ→∞ πλ→ ∞

11

Define g(x) = f(x) - c. Then lim_x→∞ = 0 and furthermore that

1∫ T 1 ∫ T T- 0 f(x)dx = c+ T- 0 g(x)dx

Without loss of generality we may assume that c = 0. Then, since f is coninuous and lim_x→∞ = 0, f is bounded. Thus there is an M such that |f(x)| < M. Let ε > 0 be given. Also becuase f is continuous, there exists a t such that for x with 0 < t < x we have |f(x) - 0| < ε∕2. Hence we have the following sequence of equations

| | ||1∫ T || 1 (∫ t ∫ T ) ||T- f(x)dx || ≤ T- |f(x )|dx+ |f(x)|dx 0 ( 0 t ) 1 ∫ t ∫ T ≤ T- M dx + ε∕2dx 0 t = M-t+ (T---t)ε∕2 T T

Thus for T such that Mt
T--

< ε∕2, the right-hand side of the above equation becomes

T - t ε∕2 +ε∕2----- < ε T

which implies

∫ -1 T Tli→m∞ T 0 f(x)dx = 0

as desired.

12

(a)

Let f : [0,1] → ℝ be a continuous function such that

∫ 1 f(x )g(x) = 0 0

for all continuous functions g(x). In particular if g(x) = f(x) we have

∫ 1 (f(x))2 = 0 0

Since f(x)² ≥ 0 we must then have f(x)² = 0, and so therefore f(x) = 0.

(b)

This is not true. By problem 6(b) of homework 6, the function

{ 0 x ≤ 0 h (x) = ( x∘----1)x L + x otherwise

approaches L as x approaches ∞. Hence we have that the function g_a,b(x) defined by h(x-a)h(b-x) is zero everywhere except for (a,b) where a,b > 0. Furthermore g(x) ∈ C¹

So, assume for later contradiction that f(x)≠0 is such that ∫ ₀¹f(x)g(x) = 0 for all g(x) in C¹. Then there must be a point x₀ where f is positive. Then there exists a,b ∈ [0,1] with a < b such that x₀ ∈ (a,b) and f is positive on all of (a,b). However, since this is the case, then fg_a,b where g_a,b is as defined above will be positive on (a,b) and zero everywhere else, i.e.

∫ 1 f(x )ga,b(x)dx = 0 0

a contradiction.