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Mathematics (2,835)
MAT454H1 (1)
Lecture

# Locally convex spaces

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Department
Mathematics
Course
MAT454H1
Professor
Andresdel Junco
Semester
Winter

Description
Locally convex spaces Suppose X is a real vector space and U is a convex subset containing 0. U is called absorbing if rU = X. Given such a U de▯ne p = p by r>0 U p(x) = inffr ▯ 0 : x 2 rUg. p is called the gauge of U. Proposition 1. p is a positive Minkowksi functional. If U = ▯U it is a semi-norm. Proof: That p(ax) = ap(x) for a > 0 is clear. For the sub-additivity, given x and y in X we may write x = (p(x) + ▯)u and y = (p(y) + ▯)v with u;v 2 U. Thus x + y = (p(x) + ▯ + p(y) + ▯)w, where 1 w = (p(x) + ▯)u + (p(y) + ▯)v) 2 U: p(x) + ▯ + p(y) + ▯ Thus p(x+y) ▯ p(x)+▯+p(y)+▯. Since ▯ > 0 is arbitrary we get subaddivity. Finally if U = ▯U we get p(x) = p(▯x) so p is a semi-norm. Recall that a topological vector space (TVS) is a vector space over K (equals R or C) with a Hausdor▯ topolgy such that the functions (▯;x) 7! ▯x from K ▯X to X and (x;y) 7! x+y from X ▯X to X are continuous. If X ▯1 is a TVS and U is a convex symmetric neighborhood of 0 then n x 2 U for all large n, by the continuity of scalar multiplication, hence x 2 nU). Thus pU is de▯ned. Note that for all x 2 U we have (1 + ▯)x 2 U for all small enough ▯ (why?) and this implies that pU < 1 on U. Evidently p U ▯ 1 on c U so we have U = fx : p(x) < 1g, that is U is the open unit ball for the seminorm p U Recall that a TVS X is called a locally convex space (LCS) if there is a neighborhood base U at 0 consisting of open convex sets. Explicitly, this means that every neighborhood of 0 contains some element of U. We can always assume that such a neighborhood base consists of symmetric sets, by replacing each U 2 U by U \ ▯U, which is again an open convex neighbor- hood of 0. Theorem 1. A real TV S is an LCS i▯ its topology is generated by a family of semi-norms. Proof: We have already discussed (or see Folland) the \if" of \i▯". For the converse suppose U is a neighborhood base at 0 consisting of symmetric convex sets. Consider the semi-norms pU;U 2 U. I claim they generate the toplogy of X. For this it is enough to show that a ne▯ x converges to 0 i▯ p (x ) ! 0. If x ! 0 and U 2 U then x is eventually in ▯U so p (x ) < ▯ U ▯ ▯ ▯ U ▯ eventually. On the other hand if U (▯ ) ! 0 for all U 2 U then eventually pU(x ▯ < 1, that is ▯ 2 U. 1 2 Suppose X and Y are TVS and T : X ! Y is linear. Then T is continuous if and only if T is continuous at 0. We have seen this is true when X and Y are normed and we leave the proof here as an exercise. If X is a TVS we will denote by X the space of continuous linear functionals on X. Theorem 2. Supppose X is a real TVS, U is an open convex subset of X and x 2= U. Then there are f 2 X and a 2 R such that f(U) ▯ (▯1;a) and f(x) ▯ a. Proof: Wolog 0 2 U. Let p = p , a positive Minkowski functional. U De▯ne f on Rx by f(tx) = tp(x). Then f ▯ p on Rx (f(tx) = p(tx) for t > 0, f(tx) is negative for t < 0) so f has an extension, also denoted f, to X with f ▯ p. Then f ▯ p < 1 on U since U is open and p(x) ▯ 1. Now we check that f is continuous. Let N = U \ ▯U. Then f < 1 on N so f > ▯1 on N. Thus jfj < 1 on N so jfj < ▯ on ▯N, giving continuity of f at 0. Remark: If X is a complex TVS then the real- and the complex-valued continuous linear functionals on X are in one-one correspondence as in the case of normed vector spaces: if f is real-linear then F(x) = f(x)▯if(ix) is complex-linear, every complex F is of this form and one is continuous i▯ the other is. (Recall also that if X is a normed vector space then kFk = kfk.) As a consequence the weak toplogy on X, that is the topology generated by X , is the same whether it is considered as a real or complex space TVS. Theorem 3. Suppose X is a real LCS, C is a closed convex subset and ▯ x 2= C. Then there is an f 2 X and a 2 R such that f ▯ a on C and f(x) > a. In particular any closed convex subset C of a real or complex X ▯ is weakly closed, i.e. closed in the topology generated by X . Proof: Wolog x = 0. Let U be a symmeSric convex neighborhood of 0 such that U \ C is empty. Then U + C = fU + c : c 2 Cg is open, convex and does not contain 0. By the previous theorem there is f 2 X such that f < 0 on U +C and of course f(0) = 0. f is not identically 0 on U since for each x there is an n such that1x 2 U. Thus we may choose u 2 U such that n f(u) > 0. Then for any c 2 C f(u + c) < 0 so f(c) < ▯f(u) < 0 = f(0). The claim about closed convex subsets follows immediately in case X is real. The complex case follows by the remarks preceding the proof of Theorem 3. We can apply Theorem 3 in the complex case as well: the conclusion is that there is f 2 X whose real part
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