MA3422 (Functional Analysis 2) Tutorial sheet 5
[February 20, 2015]
Name: Solutions
1. Let E be a normed space and α ∈ E ∗ . Show
(a) ker α = {x ∈ E : α(x) = 0} is a closed linear subspace of E
Solution: One way to do this is to say that α : E → K is continuous and ker α = {x ∈
E : α(x) = 0} = α−1 ({0}) is the inverse image of a closed set under a continuous
function — hence closed.
Another way would be to take a sequence (xn )∞
n=1 in ker α which converges (in E)
to some x ∈ E. By continuity of α we must have limn→∞ α(xn ) = α(x). But
α(xn ) = 0 for each n and so α(x) = 0 also, or x ∈ ker α. This is enough to show
that ker α is closed in (the metric space) E.
(b) If α 6= 0, there is a continuous linear bijection E → ker α ⊕ K with a continuous
inverse [Hint: Choose x0 ∈ E with α(x0 ) 6= 0 and define
α(x)
x0 , α(x)
]
T (x) = x −
α(x0 )
Solution: Define T as in the hint. It does make sense since
α(x)
α(x)
α x−
x0 = α(x) −
α(x0 ) = 0
α(x0 )
α(x0 )
(using linearity of α) and so the first component is in ker α (and the second is a scalar
of course).
We can check fairly easily that T must be linear:
α(x + y)
T (x + y) =
x+y−
x0 , α(x + y)
α(x0 )
α(x) + α(y)
=
x+y−
x0 , α(x) + α(y)
α(x0 )
α(x)
=
x−
x0 , α(x)
α(x0 )
α(y)
+ y−
x0 , α(y)
α(x0 )
= T (x) + T (y)
α(λx)
T (λx) =
λx −
x0 , α(λx)
α(x0 )
λα(x)
=
λx −
x0 , λα(x)
α(x0 )
α(x)
= λ x−
x0 , α(x) = λT (x)
α(x0 )
Also T is bounded (let us use the norm k · k1 on E ⊕ F ) since
α(x)
+ |α(x)|
kT (x)k1 = x
−
x
0
α(x0 ) kx0 k
+ kαkkxk
≤ kxk + |α(x)|
|α(x0 )|
kx0 k
≤ kxk + kαkkxk
+ kαkkxk
|α(x0 )|
kx0 k
=
1+
kαk + kαk kxk
|α(x0 )|
To show it is a bijection, one way would be to write down the inverse map and show
that it works. Define
S : ker α ⊕ K → E
by
S(x, λ) = x +
λ
x0
α(x0 )
(for x ∈ ker α, λ ∈ K). We have
α(S(x, λ)) = α(x) +
and so
λ
α(x0 ) = 0 + λ = λ
α(x0 )
T (S(x, λ)) = S(x) −
λ
x0 , λ = (x, λ)
α(x0 )
On the other hand, for x ∈ E, S(T (x)) = x is clear. So T has an inverse S and is a
bijection.
Finally S is easily seen to be linear and bounded since
|λ|
|λ|
kS(x, λ)k ≤ kxk +
kx0 k ≤ 1 +
(kxk + |λk)
α(x0 )
α(x0 )
So T and its inverse s are continuous.
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2. In E = CP [0, 1] = {f ∈ C[0, 1] : f (0) = f (1)} show that there is α ∈ E ∗ with α(f ) = 0
for f (x) ≡ 1 and α(g) = 1 for g(x) ≡ x(x − 1). [Hint: Hahn-Banach.]
Solution: Let M = span{f, g} = {λ1 f + λ2 g : λ1 , λ2 ∈ K} and define α : M → Ik by
α(λ1 f + λ2 g) = λ2
Since f and g are linearly independent (if λ1 f + λ2 g = 0, then λ1 f (0) + λ2 g(0) = 0 ⇒
λ1 1 + λ2 0 = 0 ⇒ λ1 = 1 and then evaluation at x = 1/2 [or any x ∈ (0, 1)] gives
λ2 g(1/2) = 0 ⇒ λ2 = 0).
So α is well defined on M and clearly linear. Since M is finite dimensional, α must be
bounded and there is a constant C > 0 so that |α(h)| ≤ Ckhk holds for all h ∈ M .
By the Hahn-Banach theorem there is a linear extendsion β : CP [0, 1] → K of α which
satisfies |β(h)| ≤ Ckhk for all h ∈ CP [0, 1]. But then β ∈ E ∗ and β(f ) = α(f ) = 0,
β(g) = α(g) = 1.
Richard M. Timoney
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