MA3422 (Functional Analysis 2) Tutorial sheet 7
[March 13, 2015]
Name: Solutions
1. Let p be a seminorm on a vector space E over K and let Tp be the topology on E generated
by p. Verify directly that (E, Tp ) is a locally convex topological vector space.
Show also that p : E → R is continuous.
Solution: The convexity part is not an issue since the basic open sets in the natural base
are the open balls Bp (x, r) = {y ∈ E : p(x − y) < r} (with x ∈ E and r > 0) and these
are convex.
We just have to prove that addition and scalar multiplication are continuous. Consider the
addition map, which we might denote A : E × E → E, where A(x, y) = x + y. Take
an open set U ⊆ E in the topology Tp and then we aim to show that the inverse image
A−1 (U ) = {(x, y) : A(x, y) ∈ U } = {(x, y) : x + y ∈ U } is open. Let (x0 , y0 ) ∈ A−1 (U )
and choose r > 0 so that
Bp (x0 + y0 , r) = {z ∈ E : p(x0 + y0 − z) < r} ⊆ U.
We can then see that if (x, y) ∈ Bp (x0 , r/2) × Bp (y0 , r/2), then
p(x0 + y0 − (x + y)) = p((x0 − x) + (y0 − y)) ≤ p(x0 − x) + p(y0 − y) <
r r
+ =r
2 2
and so x + y ∈ Bp (x0 + y0 , r) ⊆ U ⇒ (x, y) ∈ A−1 (U ). So we have
(x0 , y0 ) ∈ Bp (x0 , r/2) × Bp (y0 , r/2) ⊆ A−1 (U )
and (x0 , y0 ) is an interior point of A−1 (U ) (because the product Bp (x0 , r/2) × Bp (y0 , r/2)
is a basic open set for the product topology on E × E). So A−1 (U ) is open.
To do someting similar for scalar multiplication, let M : K × E → E be the scalar multiplication map, thet is M (λ, x) = λx. Take U ⊆ E open and (λ0 , x0 ) ∈ M −1 (U ) =
{(λ, x) ∈ K × E : λx ∈ U }. Choose r > 0 with Bp (λ0 x0 , r) ⊆ U . If |λ − λ0 | <
min(r/(2(p(x0 ) + 1)), |λ0 | + 1) and p(x − x0 ) < r/(4|λ0 | + 2) then
p(λx − λ0 x0 ) ≤ p(λx − λx0 ) + p(λx0 − λ0 x0 )
= |λ|p(x − x0 ) + |λ − λ0 |p(x0 )
r
rp(x0 )
≤ (|λ0 | + |λ − λ0 |)
+
4(|λ0 | + 2) 2(p(x0 ) + 1)
(2|λ0 | + 1)r r
≤
+
4|λ0 | + 2
2
= r
In short we have two positive quantities δ1 > 0 and δ2 > 0 so that |λ − λ0 | < δ1 and
p(x − x0 ) < δ2 implies λx ∈ Bp (λ0 x0 , r) ⊆ U . So
(λ0 , x0 ) ∈ {λ : |λ0 − λ| < δ1 } × Bp (x0 , δ2 ) ⊆ M −1 (U )
and thus (λ0 , x0 ) is an interior point of M −1 (U ). As this it true for all points (λ0 , x0 ) ∈
M −1 (U ), we have that M −1 (U ) is open (in K × E).
So addition and scalar mutiplication are continuous as required.
There remains the statement that p is continuous. It is a consequence of the triangle inequality that
|p(x) − p(y)| ≤ p(x − y)
(Idea is that p(x) = p((x − y) + y) ≤ p(x − y) + p(y) ⇒ p(x) − p(y) ≤ p(x − y). But that
is also true with x and y exchanged so that p(y) − p(x) ≤ p(y − x) = p(x − y). Putting
these together we get |p(x) − p(y)| ≤ p(x − y).) So p(x − y) < ε ⇒ |p(x) − p(y)| < ε
and p is therefore continuus.
2. Let E and F be normed spaces and T : E → F a linear transformation.
We say that T is weakly bounded if α ◦ T : E → K is bounded for each α ∈ F ∗ .
Show that if T is weakly bounded and E is a Banach space, then T is bounded.
[Hint: uniform boundedness theorem applied to the family of linear operators Tα = α ◦
T : E → K (with α ∈ F ∗ and kαk ≤ 1) and question 2 on sheet 6.]
Solution: The operators Tα = α ◦ T : E → K (for α ∈ F ∗ and kαk ≤ 1) are clearly linear
(composition of linear transformations) and we are assuming they are each bounded. We
also have pointwise boundedness because
|Tα (x)| = |α(T x)| ≤ kαkkT xk ≤ kT xk
holds for all α ∈ F ∗ with kαk ≤ 1.
By the uniform boundedness theorem (since E is complete)
kTα k < ∞
sup
α∈F ∗ ,kαk≤1
and that amounts to
sup
sup
|Tα (x)| < ∞
α∈F ∗ ,kαk≤1 x∈E,kxk≤1
or
sup{|α(T x)| : x ∈ E, α ∈ F ∗ , kxk ≤ 1, kαk ≤ 1} < ∞
From the previous sheet, we know that this implies that T is bounded.
Richard M. Timoney
2