Brandon Behring
Functional Analysis HW 2
Exercise 2.6 The space C[a, b] equipped with the L1 norm || · ||1 defined by
Z
||f ||1 =
b
|f (x)|dx
a
is incomplete. If fn → f with respect to the sup-norm || · ||∞ then fn → f
with respect to || · ||1 . However, the converse does not hold.
Proof: To show that C[a, b] equipped with L1 is incomplete, take any
sequences of continuous functions that converges to the Heaviside unit
step function in the L1 norm. Since step functions are not continuous,
this would prove that our space is incomplete. For example, take the
following functions on C[−1, 1]

if − 1 ≤ x < n1
 0
n
1
(x + n ) if − n1 ≤ x < n1
fn (x) =
 2
1
if n1 ≤ x ≤ 1
and
f (x) =
0
1
if − 1 ≤ x < 0
if 0 ≤ x ≤ 1
We can compute the integra by looking at the areas in the graph to find
that
Z +1
2
||f − fn ||L1 [−1,1] =
|f − fn |dx = → 0.
n
−1
This shows that convergence in the L1 norm does not imply convergence
in the sup-norm on C[−1, 1] as||f − fn ||L1 [−1,1] → 0 but ||f − fn ||∞ = 21
for all n.
To show that if fn → f with respect to the sup-norm then fn → f with
respect to || · ||1 , we use the mean value theorem for integrals (as both f
and fn are continuous) to show that for some c ∈ [a, b] we have
Z
||f − fn ||1
b
|f (x) − fn (x)|dx
=
a
=
≤
|f (c) − fn (c)| · (b − a)
sup |f (y) − fn (y)|(b − a) = ||f − fn ||∞ (b − a).
y∈[a,b]
1
Thus, if fn → f with respect to the sup-norm then fn → f with respect
to L1 norm .
Exercise 2.9 Let w : [0, 1] → R be a nonnegative, continuous function. For
f ∈ C([0, 1]), we define the weighted supremum norm by
||f ||w = sup {w(x)|f (x)|}.
0≤x≤1
if w(x) > 0 for 0 < x < 1, then || · ||w defines norm on C([0, 1]) and if
w(x) > 0 for 0 ≤ x ≤ 1 this norm is equivalent to the usual sup-norm.
However, if w(x) = x, || · ||x is not equivalent to the usual sup-norm.
Solution: We first show that || · ||w defines a norm.
(i) ||f ||w ≥ 0 for all f ∈ C([0, 1]).
Proof: It is clear from
||f ||w = sup {w(x)|f (x)|},
0≤x≤1
that ||f ||w is a supremum of nonnegative numbers as w(x) > 0 so
clearly ||f ||w ≥ 0.
(ii) ||λf ||w = |λ|||f ||w for all f ∈ C([0, 1]) and λ ∈ R.
Proof: A direct computation gives
||λf ||w
sup {w(x) · |λ · f (x)|}
=
0≤x≤1
sup {w(x) · |λ| · |f (x)|}
=
0≤x≤1
=
|λ| sup {w(x) · |f (x)|}
=
|λ|||f ||w .
0≤x≤1
(iii) ||f + g||w ≤ ||f ||w + ||g||w for all f, g ∈ C([0, 1]).
Proof: Preceding as we do in the case for the sup-norm
||f + g||w
=
sup {w(x)|f (x) + g(x)|}
0≤x≤1
≤
sup {w(x)|f (x)| + w(x)|g(x)|}
0≤x≤1
≤
sup {w(x)|f (x)|} + sup {w(y)|g(y)|}
0≤x≤1
0≤y≤1
= ||f ||w + ||g||w .
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(iv) ||f ||w = 0 implies that f (x) = 0 for all x ∈ [0, 1].
Proof: If ||f ||w = sup0≤x≤1 {w(x)|f (x)|} = 0 then we must have
w(x)|f (x)| = 0 for all x ∈ [0, 1]. Since w(x) > 0 for all x ∈ [0, 1],
then f (x) = 0 for all in x ∈ [0, 1]. This final step fails if there exists
x ∈ [0, 1] such that w(x) = 0.
Proposition: || · ||w is equivalent to the usual sup-norm.
Proof: We need to show that there exists m > 0 and M > 0 such
that
m||f ||∞ ≤ ||f ||w ≤ M ||f ||∞
for all f ∈ C[0, 1]. Assuming w(x) is continuous (this is not explicitly
stated in the problem, but I believe it was implied), we have from
Theorem 1.68 that w(x) attains its minimum and maximum m and
M for some values x, y ∈ [0, 1]. Let x ∈ [0, 1] be a value where we
achieve a minimum, w(x) = m, and y ∈ [0, 1] be such a value where
it achieves a maximum, w(y) = M . As w(x) > 0 for all x ∈ [0, 1]
clearly m > 0 and M > 0 since these values are obtained by values
in [0, 1]. Thus, for all x ∈ [0, 1] we have that
0 < m < f (x) < M < ∞
and thus
sup {m · |f (x)|} ≤ sup {w(x) · |f (x)|} ≤ sup {M · |f (x)|}
0≤x≤1
0≤x≤1
0≤x≤1
which implies what we wanted
m||f ||∞ ≤ ||f ||w ≤ M ||f ||∞ .
This proof clearly breaks down if w(x) = 0 for any x ∈ [0, 1] because
we can no longer find an m > 0. In fact- it isn’t even true that
|| · ||w is even a norm as we no longer even have that ||f ||w implies
f (x) = 0. For an explicit example, consider fn (x) = (1 − x)n . This
does not converge to a continuous function in the sup norm- it is one
at zero and zero otherwise; thus ||fn (x)||∞ = 1 for all n. However,
||fn (x)||w(x)=x → 0.
Exercise 2.13 Consider the scalar initial value problem
u̇(t)
= |u(t)|α ,
u(0)
=
0.
Then the solution is unique if α ≥ 1 but not if 0 ≤ α < 1.
Proof: We will examine four cases α > 1, α = 1, 0 < α < 1 and α = 0.
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(i) If α = 0, then we have du
dt = 1 and u(0) = 0. The only solution is then
u(t) = t. I believe there is a typo in the problem and that we have
that is non unique for 0 < α < 1.
(ii) If α = 1, then we have du
dt = |u(t)| with u(0) = 0. In the context
of theorem 2.26, we need to show that f (u, t) = |u| is bounded in a
rectangle R around the origin and is a Lipschitz continuous function
of u, uniformly in t. The latter of course being trivial as we have no
explicit t dependence. The Lipschitz condition is also clear from
|f (t, u) − f (t, v)| = | |u| − |v| | ≤ |u − v|.
(See also Example 2.19)
(iii) If α > 1, then we must show that f (u, t) = |u|α is bounded in a
rectangle around the origin and is a Lipschitz continuous function
of u. Let the rectangle R be contained in [−A, A] × [−A, A]. Then
f (u, t) = |u|α < Aα so f (u, t) is bounded. Noting that
−αu(−c)α−1 if c < 0
0
f (u) =
+αucα−1
if c ≥ 0.
and α > 1 gives us |f 0 (u)| ≤ αAα−1 for all u ∈ [−R, R]. The bounded
derivative and the mean value theorem together gives us Lipschitz
continuity as
|f (u, t) − f (v, t)| = f 0 (c)|u − v| ≤ αAα−1 |u − v|
for all (u, t), (v, t) ∈ [−A, A] × [−A, A].
Theorem 2.26 now implies a unique solution. Since u(t) = 0 solves
the ODE- it must be the only solution.
(iv) If 0 < α < 1 we will follow Example 2.23 (α = 12 ) to construct a
general counterexample. Solving the ODE f˙ = f α
Z
Z
f −α df
=
1
f (t)1−α
1−α
f (t)
dt
= t−C
=
1
(1 − α) (t − C) 1−α
suggests we try
u(t) =
0
1
(1 − α) (t − C) 1−α
if t ≤ C
if t > C.
We now have a whole class of functions that solves our ODE as C is
an arbitrary positive number so the solution is not unique.
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Exercise 3.6 The following integral equation for f : [−a, a] → R arise in a
model of the motion of gas particles on a line:
Z
1
1 a
f (y)dy f or − a ≤ x ≤ a
f (x) = 1 +
π −a 1 + (x − y)2
This equation has a nonnegative unique bounded, continuous solution for
every a such that 0 < a < ∞.
Proof: This is a Fredholm integral equation of the second kind for for
f : [−a, a] → R with g(x) = 1 and
k(x, y) =
1
1
.
π 1 + (x − y)2
By Theorem 3.3 we must show that
Z
sup {
x∈[−a,a]
+a
|k(x, y)|dy} < 1.
−a
By direct calculation we have
Z
+a
|k(x, y)|dy
=
−a
=
=
1
π
Z
+a
−a
1
dy
1 + (y − x)2
Z
1 a−x
1
dy
π −a−x 1 + u2
1
1
tan−1 (a − x) − tan−1 (−a − x) =
tan−1 (a − x) + tan−1 (a + x) .
π
π
Additionally, because the range of tan−1 (x) is
{y ∈ R | −
π
π
< y < },
2
2
we have
tan−1 (a − x) + tan−1 (a + x) < π.
sup
x∈[−a,a]
Together, we have what we need
Z +a
sup {
|k(x, y)|dy} < 1.
x∈[−a,a]
−a
To see that the solution is always positive, we examine the Neumann series
with g = 1
∞
X
f = (1 − K)−1 =
K n.
n=0
5
As this is an infinite sum of positive values, it must be positive.
If a is not finite, then if
1
f (x) = 1 +
π
Z
∞
−∞
1
f (y)dy
1 + (x − y)2
is a solution then
1
f (x+C) = 1+
π
Z
∞
1
1
2 f (y)dy = 1+ π
1 + ([x + C] − y)
−∞
Z
∞
−∞
1
f (u+C)dy
1 + (x − u)2
is also a solution.
Exercise 3.7 There is a unique solution of the following nonlinear boundary
value problem when the constant λ is sufficiently small
−u00 + λ sin u = f (x),
u(0)
=
0,
u(1)
=
0.
Here, f : [0, 1] → R is a given continuous function.
Proof: Let us follow the reasoning set out between Proposition 3.4 and
Theorem 3.5 in the text. However, instead of replacing f by −qv + f , we
now replace f with −λ sin u + f to arrive at
Z
v(x) = −
1
Z
1
λ sin(v(y))g(x, y)dy +
0
g(x, y)f (y)dy
0
which has the form (I − K)v = h where
Z
Kv(x)
=
−λ
1
sin(v(y))g(x, y)dy
0
Z
h(x)
=
1
g(x, y)f (y)dy.
0
We are now able to utilize Theorem 3.3, the Contraction Theorem applied
to Integral Equations, if we can show that
Z 1
sup {λ
| sin(v(y))g(x, y)|dy} < 1
x∈[0,1]
0
6
for sufficiently small λ.
Let
1
Z
|g(x, y)|dy}
A = sup {
x∈[0,1]
0
(I believe the text claims that A = 8, but I can’t verify this- either way it
is clear that it is a finite number as )
then if λ < 1/A and noting that | sin(x)| ≤ 1 for all x ∈ R we have
Z
1
x∈[0,1]
1
Z
| sin(v(y))g(x, y)|dy}
sup {λ
x∈[0,1]
0
|g(x, y)|dy}
≤ λ sup {λ
0
≤ λA < 1.
Let v0 = 0, then
1
Z
v1 (x) = −
Z
λ sin(0)g(x, y)dy +
0
1
1
Z
g(x, y)f (y)dy =
g(x, y)f (y)dy
0
0
which could be in theory calculated if we knew f (y). Next, we have
Z
v2 (x) = −
1
Z
λ sin
0
1
Z
g(x, z)f (z)dz g(x, y)dy +
0
1
g(x, y)f (y)dy
0
Here- we see the non-linear integral equation becomes quite ugly to calculate by iteration- even though we do know vn will converge uniformly to
our solution.
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