Math 516
Professor Lieberman
January 19, 2005
HOMEWORK #1 SOLUTIONS
Chapter 10
7. To show that P is a linear manifold, we let p1 and p2 be in P and let α1 and α2 be real
numbers. Then
k
n
X
X
i
p1 (x) =
ai x , p2 (x) =
bi xi
i=0
i=0
for nonnegative integers k and n. Now set N = max{k, n} and define ai = 0 if i ∈
{k + 1, . . . , N } and bi = 0 if i ∈ {n + 1, . . . , N }. Then
(α1 p1 + α2 p2 )(x) =
N
X
(α1 ai + α2 bi )xi ,
i=0
so α1 p1 + α2 p2 ∈ P .
P is not closed because the sequence of polynomials hpn i defined by
pn (x) =
n
X
xi
i=0
i!
converges uniformly to ex , and this function has infinitely many nonzero derivatives at
x = 0, so it can’t be in P .
In addition to the obvious closed linear manifolds {0} and C[0, 1], we can take
M = {f ∈ C[0, 1] : f (0) = 0}.
Since (αf + βg)(0) = 0 if f and g are in M and α and β are real numbers, M is a linear
manifold. It’s closed because, if hfn i is a sequence in M which converges in C[0, 1] to a
function f , then fn (0) → f (0), so f (0) = 0, which means f ∈ M , so M is closed.
11. First, we show that kxk1 defines a pseudonorm. Let x and y in X, let ε > 0 be given, and
let m1 and m2 be elements of M such that
ε
ε
kx − m1 k ≤ kxk1 + , ky − m2 k ≤ kyk1 + .
2
2
It follows that
kx + yk1 = inf kx + y − mk ≤ kx + y − (m1 + m2 )k
m∈M
≤ kx − m1 k + ky − m2 k ≤ kxk1 + kyk1 + ε.
Since this inequality is true for every ε > 0, we have kx + yk1 ≤ kxk1 + kyk1 . Next, we let
α ∈ R and x ∈ X. If α = 0, then kαxk1 = 0 = |α|kxk1 . If α 6= 0, then take m1 and m2 in
M so that
kαx − m1 k ≤ kαxk1 + ε, kx − m2 k ≤ kxk1 + ε.
Then
kαxk1 ≤ kαx − αm2 k = |α|kx − m2 k ≤ |α|(kxk1 + ε).
1
2
Since ε > 0 is arbitrary, it follows that kαxk1 ≤ |α|kxk1 . In addition,
1
|α|kxk1 ≤ |α|kx − m1 k = kαx − m1 k ≤ kαxk1 + ε,
α
so we also have |α|kxk1 ≤ kαxk1 . The combination of these inequalities is kαxk1 = |α|kxk1 .
Next, we show that ϕ has kernel M̄ . If ϕ(x) = 0, then
inf kx − mk = 0,
m∈M
so, for any δ > 0, there is m ∈ M such that kx − mk ≤ δ. It follows that x ∈ M̄ by the
definition on page 147 of the text, which means that the kernel is a subset of M̄ . Conversely,
if x ∈ M̄ , then, for any δ > 0, there is m ∈ M such that kx − mk < δ, so
inf kx − mk = 0,
m∈M
and therefore ϕ(x) = 0, which means that M̄ is a subset of the kernel. Hence the kernel is
exactly M̄ .
If O is an open set in X, then for any x ∈ O, there is δ > 0 such that ky − xk < δ
implies that y ∈ O. Now suppose x0 ∈ ϕ(O), so x0 = ϕ(x) for some x ∈ O. Suppose
y 0 ∈ X 0 satisfies the inequality kx0 − y 0 k1 ≤ 2δ with δ as above and let y1 ∈ X be such that
ϕ(y1 ) = y 0 . Then there is m ∈ M such that k(x − y1 ) − mk < δ. If we set y = y1 + m, then
ϕ(y) = ϕ(y1 ) + ϕ(m) = ϕ(y1 ) = y 0 and kx − yk < δ. It follows that y ∈ O, so y 0 ∈ ϕ(O),
which means that ϕ(O) is also open.
12. Let hx0n i be an absolutely summable sequence in X/M and, for each n, let x̃n be a point
in X with ϕ(x̃n ) = x0n . Then, for each n, there is a point mn ∈ M such that kx̃n − mn k ≤
kx̃n k1 + 2−n . But
X
kx̃n k1 < ∞
since hx0n i is absolutely summable. If we set xn = x̃n − mn , it follows that
X
X
kxn k ≤
kx̃n k1 + 2−n < ∞,
P
so hxn i is also absolutely summable. Hence x = xn exists. But, since ϕ is continuous, it
follows that
X X
X
x0n =
ϕ(xn ) = ϕ
xn = ϕ(x),
so hx0n i is also summable.
15. (a) Let kxk = 1. Then kxk1 ≤ kxk because 0 ∈ M , and hence kϕk ≤ 1. Conversely,
fix x ∈ X ∼ M . Then kxk1 > 0, so, for any ε > 0, there is m ∈ M such that kxk1 ≥
kx − mk − εkxk1 . Because kxk1 = kx − mk1 , it follows that (1 − ε)kx − mk1 ≥ kx − mk,
which means that kϕk ≥ 1 − ε. But ε > 0 is arbitrary, so kϕk ≥ 1. Combining these two
inequalities yields kϕk = 1.
17. If f is bounded, then it’s continuous. In addition, {0} is closed, so f −1 (0) must be closed,
but this says that the kernel is closed.
If the kernel is closed, then we consider two possibilities: If f = 0, then clearly f is
continuous. If f 6= 0, then there is a point x ∈ X such that f (x) 6= 0. Let y = x/f (x).
Then f (y) = 1. In addition, because the kernel is closed, there is a δ > 0 such that B(y, δ)
does not intersect the kernel. We claim that f (z) > 0 if z ∈ B(y, δ). To this end, define g
on [0, 1] by
g(λ) = f (y + λ(z − y)) = (1 − λ)f (y) + λf (z).
3
Since ky − [y + λ(z − y)]k = kλ(y − z)k < λδ ≤ δ if z 6= y and ky − [y + λ(z − y)]k = 0 < δ
if z = y, we see that g is never zero and that it’s linear and hence continuous. Since
g(1) = f (y) = 1, it follows that f > 0 in B(y, δ). Now, suppose w ∈ B(0, δ). Then
w + y ∈ B(y, δ), so f (w) = f (w + y) − f (y) > −1. Moreover, −w + y ∈ B(y, δ), so
f (w) = f (y) − f (−w + y) < 1. In other words, |f (w)| < 1. It follows that |f (z)| ≤ 1/δ if
kzk ≤ 1 and hence f is bounded.