Look at (2) on page 222 under the Riesz Representation Theorem. The map you wrote is (1) but only (2) is being called the Reisz map.
How do I show this is one to one?
Would I start if or would it start differently?
well, you would suppose first that R(f) = R(g) (these are the riesz vectors for f and g).
therefore <R(f),y> = <R(g),y> for all vectors y. but this means that f(y) = g(y) for all y in V, so f = g.
EDIT: drexel28's map is the inverse riesz map. since we have an isomorphism, we're really talking about the same thing: a 1-1 correspondence (bijection), it's a "dual" thing (and that is a million dollar pun).
you don't like alex's suggestion (using the fact that dim V = dim V*)?
very well, for any v in V, we need to find some f in V* such that R(f) = v.
how about f = <v,_>? then R(f) is defined by <R(f),y> = f(y) = <v,y>, and since <R(f),y> = <v,y> for every y,
we conclude R(f) = v.
if <R(f),y> = <v,y>, then <R(f) - v,y> = 0 for all y in V. but R(f) - v is some vector in V,
so <R(f) - v,R(f) - v> = 0, so by positive definiteness of the inner product, R(f) - v = 0, so R(f) = v).
I make only one remark. The reason why, to me at least, it makes more sense to define the Riesz map as I did is that it produces a monomorphism regardless of what kind of vector space is. Conversely, the map can't possibly hope to be a monomorphism if is infinite dimensional since in that case .
well, you and Mr. Roman better have a l'il talk then. i mean, sheesh calling the wrong map the Riesz map!
i agree with you, it makes more "sense" to think of V as the natural starting place (domain), and V* as the "derived" thing (image),
although his way of defining things does make the name "Riesz vector" seem more meaningful.
(myself, i find myself wondering how we really know which vectors are really vectors, and which are co-vectors. what if the math elves switched them while we were sleeping? i guess that shows why you should keep an infinite-dimensional vector space on hand, so you can embed your finite-dimensional space in it, and see if its dual is bigger).