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 be positive, f R be an even function, and g R and be an odd function.

(a) Show -ccf(x)dx = 2 0cf(x)dx


We can separate -ccf(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 -ccg(x)dx = 0


We can separate -ccg(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 -ccf(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 0xf(t)e3tdt = 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,,xN = 2} where xj - xj-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 +x4dx to be less than 1100, then we need 4∕N < 1100, 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]. 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() = w(π) - u(π) = 0 so that asin() = 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,Lvfor 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 λ1λ2 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 λ1λ2 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 3


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 f1 has the following three properties
  1. 01|f(x)|dx 0 with equality only when f = 0.
  2. 01|cf(x)|dx = |c| 01|f(x)|dx = |c|∥f
  3. 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) = uλ, 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) = 1λ, g(π) = πλ, 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 limx→∞ = 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 limx→∞ = 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] 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 ga,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 fga,b where ga,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.