1. ## Riesz Map

Prove that the Reisz map $\mathcal{R}:V^x\to V$ defined by $<\mathcal{R}(f),y>=f(y)$ for all y in V, is an isomorphism.

Let $\mu, \lambda\in F$

$\mathcal{R}(\lambda f+\mu g),y>=(\lambda f+\mu g)(y)=$
$\lambda f(y)+\mu g(y)=<\lambda\mathcal{R}(f),y>+<\mu\mathcal{R}(g), y>$

I am not sure how to show it is onto and one to one with the Reisz map.

2. ## Re: Riesz Map

Originally Posted by dwsmith
Prove that the Reisz map $\mathcal{R}:V^x\to V$ defined by $<\mathcal{R}(f),y>=f(y)$ for all y in V, is an isomorphism.

Let $\mu, \lambda\in F$

$\mathcal{R}(\lambda f+\mu g),y>=(\lambda f+\mu g)(y)=$
$\lambda f(y)+\mu g(y)=<\lambda\mathcal{R}(f),y>+<\mu\mathcal{R}(g), y>$

I am not sure how to show it is onto and one to one with the Reisz map.
Is this really what the Riesz map is? Usually the Riesz map is defined by $T:V\to V^\ast$ where $T(v)(x)=\langle v,x\rangle$. Is this what you mean? Regardless, most likely to show bijectivity you just need to show that the map is injective, and since the dual space is of equal dimension to the space, this implies bijectivity.

3. ## Re: Riesz Map

Originally Posted by Drexel28
Is this really what the Riesz map is? Usually the Riesz map is defined by $T:V\to V^\ast$ where $T(v)(x)=\langle v,x\rangle$. Is this what you mean? Regardless, most likely to show bijectivity you just need to show that the map is injective, and since the dual space is of equal dimension to the space, this implies bijectivity.

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 $f(y)=f(x)$ or would it start differently?

4. ## Re: Riesz Map

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).

5. ## Re: Riesz Map

Originally Posted by Deveno
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).
How do I show onto?

6. ## Re: Riesz Map

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.

(amplification:

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).

7. ## Re: Riesz Map

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 $V$ is. Conversely, the map $V^\ast\to V$ can't possibly hope to be a monomorphism if $V$ is infinite dimensional since in that case $\dim V^\ast>\dim V$.

8. ## Re: Riesz Map

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).