Elements of Dual of Infinite_Dim. Space have Codimension One

[Bacle]

Hi:
In the case of a finite-dimensional normed space V, it is relatively-straightforward to
show that the kernel of any element of V* has 1 .

( Assume DimV=n):

We take a linear map L:V-->F ; F the base field. We choose a basis to represent L,
then we consider F as a vector space over itself; F is then 1-dimensional over
itself, then ,( by rank-nullity) , L has rank 1 , so it must have nullity n-1.

I don't see, though, how, in the case of V infinite-dimensional, how to show that, given
L in V* , that the kernel of L is a maximal (strict) subspace; I don't know if we still
say that KerV* has codimension 1.

Any Ideas?

Thanks.

[micromass]

Let W be a linear subspace of V which strictly contains ker(L). Our goal is to show that W=V.

Take w in W such that L(w)=1. This exists, since W contains the kernel strictly.
Now, take v in V, and let a=L(v). Then v-aw is an element of the kernel, and is thus contained in W. Thus v=aw+(v-aw) is an element of W. This shows that V=W.

[Bacle]

Sorry, I don't get it; you showed that the kernel is the whole space. Then the map
should be the zero map, right?. Also: how did you use the fact that V is infinite-dimensional?

[Bacle]

Never mind, sorry, I jumped the gun; I got it. Thanks.