Math 205 - Calculus I
Solutions to Homework due April 19
Question 1. In this question, we will consider the Lp metric on the space C([0, 1]) of continuous (real-valued)
functions on [0, 1]. Here, the Lp norm (for 1 ≤ p ≤ ∞) is given by
Z
||f ||p =
1
1/p
|f (x)|p dx
0
and
||f ||∞ = sup{ |f (x)| | x ∈ [0, 1]}.
Let f, g ∈ C([0, 1]) be given by the following:
f (x) = 1/2 − x
g(x) = |1/2 − x| .
(a) Compute f + g and f − g. Write your answers in piecewise form.
(b) Compute ||f ||∞ , ||g||∞ , ||f + g||∞ , and ||f − g||∞ . Does the parallelogram law hold for these vectors
with the L∞ metric?
(c) Compute ||f ||p , ||g||p , ||f + g||p , and ||f − g||p .
(d) For what values of p does the parallelogram law hold for the vectors f and g?
(e) Using the above, what can you conclude about which Lp norms on C[0, 1] are not induced by inner
products?
Solution 1.
(a) Notice that f (x) = 1/2 − x ≥ 0 if and only if x ≤ 1/2. Thus, subtracting, we get
0
if 0 ≤ x ≤ 1/2
f (x) − g(x) =
2x − 1 if 1/2 < x ≤ 1
Similarly,
f (x) + g(x) =
1 − 2x
0
if 0 ≤ x ≤ 1/2
.
if 1/2 < x ≤ 1
(b) Taking the supremum (maximum) of each of these functions, we get the following:
||f ||∞ = 1/2
||g||∞ = 1/2
||f + g||∞ = 2
||f − g||∞ = 2.
Computing the sum of the square of the diagonals, we get 22 + 22 = 4. Computing the sum of the square
of the side lengths, we get 2 · (1/2)2 + 2 · (1/2)2 = 1. Sine these values are not equal, the parallelogram
law fails.
1
(c) Notice that since g(x) = |f (x)|, then considering the equation for || · ||p , we get that ||f ||p = ||g||p . First,
we compute ||f ||pp . Doing so, we get
||g||pp = ||f ||pp =
Z
1
|f (x)|p dx =
0
1/2
Z
(1/2 − x)p dx +
Z
0
1
(x − 1/2)p dx =
1/2
1/2
1
1
−1
=
(1/2 − x)p+1 +
(x − 1/2)p+1 p+1
p+1
0
1/2
1
1
1
1 1
1
1
1
+
=2·
=
.
p+1
p+1
p+1
p+12
p+12
p+12
p + 1 2p
Thus,
||g||p = ||f ||p =
1
(p + 1)−1/p .
2
Computing the remaining norms, we have
||f + g||pp =
1
Z
|f (x) + g(x)|p dx =
1/2
Z
(1 − 2x)p dx =
0
0
1/2
1
1
p+1 =
(1 − 2x) .
−2(p + 1)
2(p
+ 1)
0
Similarly,
||f − g||pp =
Z
1
Z
|f (x) − g(x)|p dx =
0
1
(2x − 1)p dx =
1/2
1
1
p
(2x − 1) 2(p + 1)
1/2
=
1
.
2(p + 1)
Thus, we have that
||f + g||p = ||f − g||p = 2−1/p (p + 1)−1/p .
(d) We check for which values of p the parallelogram laws. It holds if and only if
||f + g||2p + ||f − g||2p = 2||f ||2p + 2||g||2p .
Plugging in our terms from above, we get that
1
1
2−2/p (p + 1)−2/p + 2−2/p (p + 1)−2/p = 2 · (p + 1)−2/p + 2 · (p + 1)−2/p .
4
4
First, we cancel the common (p + 1)−2/p , we get (after simplifying a bit more)
21−2/p = 1 = 20 .
This last equation is true if and only if p = 2.
(e) Using the above, we see that when p 6= 2 (including p = ∞), || · ||p is not induced by an inner product.
Question 2. (à la Julian) Let V be an inner product space and {v1 , v2 , . . . , vm } be an orthogonal collection
of vectors. Show that this collection is linearly independent. Note: This is similar to a theorem we proved
in class, but has weaker hypotheses (and thus more powerful than the class’ theorem). Also, recall that a
collection of vectors is orthogonal if hvi , vj i = 0 if and only if i 6= j.
2
Solution 2.
Assume that for some choice of αi ∈ C we have that
α1 v1 + α2 v2 + · · · + αm vm = 0.
Taking the dot product of this linear combination with vi we get that
0 = h0, vi = hα1 v1 + α2 v2 + · · · + αm vm , vi i =
α1 hv1 , vi i + α2 hv2 , vi i + · · · + αm hvm , vi i.
Since this collection is orthogonal, all these terms are 0 except for the i-th term. Thus,
0 = αi hvi , vi i.
Since this dot product is non-zero, it must be the αi = 0. This is true for every i, thus this collection is
linearly independent.
Question 3. Consider the vector space
C(T) = {f : [−π, π] → C | f is continuous and f (−π) = f (π)},
and the L2 inner product given by
Z
π
hf, gi =
f (t)g(t) dt.
−π
Consider the vectors given by
inx ∞
e
e−2ix e−ix 1
eix e2ix
√
= ..., √ , √ , √ , √ , √ ,... .
2π n=−∞
2π
2π
2π
2π
2π
In what follows, we will prove that this collection of vectors in C(T) is an orthonormal set.
(a) Using Euler’s formula (eiθ = cos θ + i sin θ) and properties of sin and cos to show the following:
(i) If n ∈ Z is even, then einπ = 1.
(ii) If n ∈ Z is odd, then einπ = −1.
(iii) einx = e−inx
(b) Show that this collection of vectors forms an orthonormal set.
Solution 3.
(a)
(i) Using Euler’s formula, we have that
ei(nπ) = cos(nπ) + i sin(nπ).
Since n is even, then cos(nπ) = 1 and sin(nπ) = 0. Thus, einπ = 1.
(ii) If n is odd, then cos(nπ) = −1 and sin(nπ) = 0. Thus, einπ = −1.
(iii) Using Euler’s formula, we have that
einπ = cos θ + i sin θ = cos θ − i sin θ.
Since cos is an even function and sin is an odd function, we have that cos(θ) = cos(−θ) and
− sin(θ) = sin(−θ). Thus, continuing our above computation, we have this is equal to
cos(−θ) + i sin(−θ) = e−iθ .
3
(b) First, we show that each vector has length 1. Taking the dot product of the n-th term with itself, we
have that
inx inx Z π inx
e
e
einx
e
√ ,√
√ · √ dx =
=
2π
2π
2π
2π
−π
Z π
Z π
1
1
einx e−inx dx =
1 dx = 1.
2π −π
2π −π
Next, we will show that distinct vectors are orthogonal by showing that their dot product is 0. Computing
the dot product of the n and m terms, we get that
inx imx Z π
e
e
1
√ ,√
einx · eimx dx =
=
2π −π
2π
2π
π
Z π
1 ei(n−m)x 1
i(n−m)x
e
dx =
=
2π −π
2π (n − m) −π
1
ei(n−m)π − e−i(n−m)π .
2π(n − m)
If n − m is even, then both of the terms in the parentheses are 1 and the dot product is zero. Similarly,
if n − m is odd, then both terms are −1 and the dot product is zero. Thus, these vectors are orthogonal.
So, this collection of vectors is an orthonormal set.
In class, we showed that Fourier Analysis (and some easy facts about orthonormal bases) imply that any
f ∈ C(T) can be written as
f=
∞ X
n=−∞
einx
f, √
2π
∞
einx
1 X
√
=
hf, einx ieinx .
2π n=−∞
2π
After some a little bit of complex algebra, we were able to write it in a form that was more familiar to many
of us:
∞
a0 X
+
an cos(nx) + bn sin(nx)
f=
2
n=1
where the harmonics an and bn are given by
an =
1
1
hf, cos(nx)i and bn = hf, sin(nx)i.
π
π
Question 4. For the following, we will consider the following sawtooth function f ∈ C(T) given by f (x) = |x|.
This function is called a sawtooth function because, when this function is extended by periodicity to R, it
looks like the teeth of a saw.
(a) Show that |x| sin(nx) is an odd function. Use this to quickly prove that bn = 0 for all n.
(b) Compute an . Your answer will be in terms of n.
(c) Using this, write out the Fourier expansion for f (x). Graph the Fourier sum (i.e., Fourier series where
the sum only contains finitely many terms) and show that this graph does look roughly like a sawtooth.
Solution 4.
(a) Notice that plugging
in −x for x, we get | − x| sin(−nx) = −|x| sin(nx). Thus, this function is odd. So,
Rπ
since an = π1 −π f (x) sin(nx) dx, and integrating an odd function over a symmetric interval gives zero,
we get that an = 0 for all n.
4
(b) Since an = π1
simplified to
Rπ
−π
|x| cos(nx) dx, we can use the fact that this function is even and thus this integral is
an =
2
π
When n = 0, we get that
a0 =
2
π
Z
π
Z
x cos(nx) dx.
0
π
x dx =
0
2 π2
·
= π.
π 2
For n > 0, we use integration by parts to obtain that
π Z π 1
2 x
sin(nx) −
sin(nx) dx =
an =
π n
0
0 n
π 2
2
1
= 2 ((−1)n − 1) .
cos(nx)
π n2
n
π
0
First, notice that if n is even, then (−1)n − 1 = 0. Thus, an = 0 for n even and positive. When n is
odd, then (−1)n − 1 = −2. Thus, for odd terms, we get that
an =
−4
.
n2 π
(c) Using (b), the Fourier series for f can be written as
∞
a0 X
π −4
4
4
+
an cos(nx) = +
cos(x) −
cos(3x) −
cos(5x) + · · ·
2
2
π
9π
25π
n=1
(d) Graphing the first 11 terms of the Fourier series, we get the following graph, which looks very similar to
the graph of f when extended to all R by periodicity.
3
2
1
-5
-4
-3
-2
-1
0
-1
-2
-3
5
1
2
3
4
5