Basic Analysis – Gently Done Topological Vector Spaces
Basic Analysis – Gently Done Topological Vector Spaces
Basic Analysis – Gently Done Topological Vector Spaces
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120 <strong>Basic</strong> <strong>Analysis</strong><br />
because X = �∞ k=1kV2 , since V2 is a neighbourhood of 0 and so is absorbing. By<br />
Baire’s category theorem, k0T(V2 ) has non-empty interior for some k0 ∈ N. But<br />
y ↦→ k0y is a homeomorphism of Y onto Y so that k0T(V2 ) = k0T(V2 ) and we<br />
deduce that T(V2 ) has non-empty interior, as claimed.<br />
In particular, this means that there is an open set U contained in T(V2 ).<br />
Putting W = U −U, we see that W is an open neighbourhood of 0 with<br />
W ⊆ T(V 2 )−T(V 2 ) ⊆ T(V 1 )<br />
which completes the first part of the argument.<br />
Next we shall show that T(V 1 ) ⊆ T(V). To see this, we construct a sequence<br />
(y n ) n∈N , with y n ∈ T(V n ), such that y n+1 − y n ∈ T(V n ). The y n are defined<br />
recursively. First fix y 1 to be any point of T(V 1 ). Suppose that n ≥ 1 and that y n<br />
has been chosen in T(V n ). Arguing as above, but now with V n+1 instead of V 1 , we<br />
see that T(V n+1 ) is a neighbourhood of 0. Hence y n −T(V n+1 ) is a neighbourhood<br />
of y n , and, since y n belongs to the closure of T(V n ), we have<br />
� yn −T(V n+1 ) � ∩T(V n ) �= ∅.<br />
Hence there is x n ∈ V n such that T(x n ) ∈ y n −T(V n+1 ). Set y n+1 = y n −T(x n ).<br />
Then y n+1 ∈ T(V n+1 ) which completes the construction of the sequence (y n ).<br />
Let s n = x 1 +x 2 +···+x n , for n ∈ N. Then for n > m,<br />
d(s n ,s m ) = d(s n −s m ,0)<br />
= d(x n +···+x m+1 ,0)<br />
≤ d(xn +···+x m+1 ,xm+1 )+d(x m+1 ,0)<br />
< d(xn +···+x m+2 ,0)+ 1<br />
2m+1 since xm+1 ∈ Vm+1 ,<br />
< 1 1<br />
+···+<br />
2n 2m+1. Hence (s n ) is a Cauchy sequence in X and therefore converges (because X is<br />
complete) to some x ∈ X, with d(x,0) < r (since d(x,0) ≤ d(x,x 1 )+d(x 1 ,0) <<br />
d(x,x 1 ) + r/2 and d(s n ,x 1 ) = d(x n + ··· + x 2 ,0) ≤ r/2 for all n > 1, so that<br />
d(x,x 1 ) = limd(s n ,x 1 ) ≤ r/2). It follows that x ∈ V 0 ⊆ V. Furthermore,<br />
T(s n ) =<br />
n�<br />
T(xj ) =<br />
j=1<br />
n�<br />
j=1<br />
(y j −y j+1 ) = y 1 −y n+1 .<br />
Now, y n ∈ T(V n ) and so there is some v n ∈ V n such that d(y n ,T(v n )) < 1<br />
n and<br />
hence<br />
d(y n ,0) ≤ d(y n ,T(v n ))+d(T(v n ),0) < 1<br />
n +d(T(v n ),0).<br />
Department of Mathematics King’s College, London