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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10. Fréchet <strong>Spaces</strong><br />
Inthischapter wewillconsider completelocallyconvex topologicalvectorspaces—<br />
this class includes, in particular, Banach spaces.<br />
Theorem 10.1 Let (X,T) be a topological vector space with topology determined<br />
by a countable separating family of seminorms. Then (X,T) is metrizable.<br />
Moreover, the metric d can be chosen to be translation invariant in the sense that<br />
d(x+a,y +a) = d(x,y) for any x,y and a in X.<br />
Proof Let {p n : n ∈ N} be a countable separating family of seminorms which<br />
determine the topology T on X. For any x,y in X, put<br />
d(x,y) =<br />
∞�<br />
n=1<br />
1<br />
2 n<br />
p n (x−y)<br />
1+p n (x−y) .<br />
Then d is well-defined and clearly satisfies d(x,x) = 0, d(x,y) = d(y,x) ≥ 0 and<br />
d(x+a,y+a) = d(x,y), x,y,a ∈ X. Now, p n (x−z) ≤ p n (x−y)+p n (y−z) and<br />
the map t ↦→ t/(1+t) is increasing on [0,∞) and so<br />
pn (x−z)<br />
1+p n (x−z) ≤ pn (x−y)<br />
1+p n (x−y) + pn (y −z)<br />
1+p n (y −z)<br />
which implies that d(x,z) ≤ d(x,y) + d(y,z), x,y,z ∈ X. Finally, we see that<br />
d(x,y) = 0 implies that p n (x − y) = 0 for all n, and so x = y since {p n } is a<br />
separating family. Thus d is a translation invariant metric on X.<br />
We must show that d induces the vector topology on X determined by the<br />
family {p n }. To see this, we note that a net (x ν ) in X converges to x with respect<br />
to the vector topology T if and only if (p n (x ν −x)) converges to 0 for each n ∈ N.<br />
This is equivalent to d(x ν ,x) → 0, that is, x ν → x in the metric topology on X<br />
induced by d. It follows that the two topologies have the same closed sets and<br />
therefore the same open sets, that is, they coincide.<br />
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