linear independence in a dual pair

Question

The following is an excerpt from the Topological Vector Spaces by Schaefer:

I don't see how the underscored sentence work. Suppose $$ \langle x,y_n\rangle=0 $$ for all $x\in F_n$. Why this implies $$ y_n\in\hbox{span}\{y_1,y_2,\cdots,y_n\}? $$ I guess it might have something to do with the direct sum $F=F_n+M_n$. But I don't see the connection.

AnatolyVorobey · Accepted Answer · 2015-10-23T00:48:00.133

0

Consider $y' = y_n - \langle \overline{x}_1,y_n\rangle y_1 - \langle \overline{x}_2,y_n\rangle y_2 - \cdots - \langle \overline{x}_{n-1},y_n\rangle y_{n-1}$.

Can you prove that $y'$ vanishes on all $\overline{x}_i$ and therefore on $M_n$?
Can you prove that $y'$ vanishes on all $x \in F_n$?

Now use the direct sum and conclude what $y'$ must be, and what that implies.

edited Oct 23 '15 at 00:48

answered Oct 22 '15 at 15:43

AnatolyVorobey

2,452

Do you mean $\overline{x}_1$ in your answer? – Oct 22 '15 at 22:53
Yeah, forgot about overlines. EDIT: put them in. – AnatolyVorobey Oct 23 '15 at 00:46
OK. By the direct sum property, (1) and (2) implies that $\langle x,y'\rangle=0$ for all $x\in X$ and thus $y'=0$ (since $F$ and $G$ are in duality). – Oct 24 '15 at 14:32
That's correct! – AnatolyVorobey Oct 25 '15 at 00:11

linear independence in a dual pair

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