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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