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Extra info for Applied math. Part 1: Integral

Example text

We shall show that the maximality of γ implies that it is, in fact, not just a partial cluster point but a cluster point of the net (γα ). To see this, suppose that γ ∈ j∈J Xj , with J ⊆ I, so that γ is a cluster point of (γα J)α∈A . We shall show that J = I. By way of contradiction, suppose that J = I and let k ∈ I \ J. Since γ is a cluster Department of Mathematics King’s College, London 3: Product Spaces 31 point of (γα J)α∈A in j∈J Xj , it is the limit of some subnet (γφ(β) J)β∈B , say. Now, (γφ(β) (k))β∈B is a net in the compact space Xk and therefore has a cluster point, ξ ∈ Xk , say.

Setting t = 1 and t = −1, we obtain f (x) + µ ≤ p(x + x1 ) Department of Mathematics for x ∈ M (taking t = 1) King’s College, London 5: Vector Spaces 49 and −f (x) − µ ≤ p(−x − x1 ) for x ∈ M , (taking t = −1). Replacing x by −y and y ∈ M in this last inequality we see that µ must satisfy µ ≤ p(x + x1 ) − f (x) x∈M f (y) − p(y − x1 ) ≤ µ y ∈ M. and The idea of the proof is to show that such a µ exists, and then to work backwards to show that f1 , as defined above, does satisfy the boundedness requirement.

Uk ∈ C such that k λ((z1 , . . , zk , 0)) = ui zi . i=1 Hence, for any (z1 , . . , zk+1 ) ∈ Ck+1 , we have λ((z1 , . . , zk+1 )) = λ((z1 , . . , zk , 0)) + λ((0, . . , 0, zk+1 )) k = ui zi + zk+1 λ((0, . . , 0, 1)) i=1 k+1 = ui zi i=1 Department of Mathematics King’s College, London 5: Vector Spaces 45 where uk+1 = λ((0, . . , 0, 1)) and the result follows. The case of a conjugate linear functional can be proved similarly. Alternatively, it can be deduced from the linear case by noticing that if : Cn → C is conjugate linear, then the map z → λ(z) = (z) is linear and so has the above form, λ(z) = n n i=1 ui zi .

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