Getting the unique element in the Riesz-Frechet Theorem.

Question

I have this thorem in my book, H', denotes the dual space, that is the set of bounded linear operators from X to the field over X.

The way they got the unique element seems very interesting. Does it follow from the proof of the theorem, that if z is any element in H, and if $f(z)\ne 0$ , then $\frac{z}{f(z)\|z\|^2}$, must be unique?

Is there a simpler way of proving this? That is, lets just say that you only get the info that you have a Hilbert space, you have a bounded linear operator $f: H \rightarrow \mathbb{F}$, then it holds that if $f(z_1)\ne 0$, and $f(z_2)\ne 0$, then $\frac{z_1}{f(z_1)\|z_1\|^2}=\frac{z_2}{f(z_2)\|z_2\|^2}$? Is there a simple proof of this?

Answer 1

You forgot another constraint that $z$ has to follow which is $z \in (\ker f)^\perp$.

Then it is true that if $z_1$ and $z_2$ are such that $f(z_1)\neq 0$ and $f(z_2) \neq 0$, $\frac{z_1}{f(z_1)\|z_1\|^2}=\frac{z_2}{f(z_2)\|z_2\|^2}$ and $z_1,z_2 \in (\ker f)^\perp$ you have $z_1=z_2$.

Answer 2

Notice that on the existence part there is this statement: "Therefore there exist $z\in Ker(T)^\perp$ such that $f(z)=1$." So your claim which is "...that if $z$ is any element in $\mathcal H$, and if $f(z)\ne 0$ , then $\frac{z}{f(z)||z||^2}$, must be unique?" is not true.

It would be better if you note how the proof is done.

Given $f\in\mathcal H'$ we have to find a unique $y\in\mathcal H$ with some properties. We start with giving the existence of a vector $z\in\mathcal H$ then we construct $y$ from $z$. Which gives the existence of $y$ (given in main theorem).

Now for uniqueness we use a very fundamental concept of inner-product space which is "the vector which is perpendicular to every vector, is the null vector." In math notation $\langle a,b\rangle=0,\ \forall a\in V(\text{ the inner-product space})\iff b=0(\text{ null vector})$

Now in the uniqueness we consider $y,w\in\mathcal H$ satisfying the same property then $\langle x,y\rangle=\langle x,w\rangle,\ \forall x\in \mathcal H$ then $\langle x,y-w\rangle =0,\ \forall x\in\mathcal H$. So by the same concept as above we have $y-w$ is the null vector and hence the uniqueness is established.

Note: If you are wondering how $\langle a,b\rangle=0\ \forall a\in V\implies b=0$ then just see that when $a=b$ we have $\langle a,b\rangle=\langle b,b\rangle=||b||^2=0\implies b=0$. due to the norm axioms.