Riesz Representation Theorem proof in Bjork

Question

I have a question regarding the proof of Riesz representation Theorem in the appendix of Bjork textbook. Here is the statement of the theorem: Assume that $F:H\to R$ is a bounded linear functional. Then there exists a unique $g\in H$ such that $F(f)=(f,g)$ for any $f\in H$. Here is the proof: Define $M$ by $M = Ker[F]$. The $M$ is a closed subspace and we can decompose $H$ as $H= M + \hat{M}$. From the definition of $M$ it is clear that $F$ restricted to $\hat{M}$ is linear with trivial kernel. It is thus a vector space isomorphism between $\hat{M}$ and $\mathbb{R}$ so $\hat{M}$ has to be one dimensional and we can write $\hat{M}=\mathbb{R}g$ for some $g\in\hat{M}$. At this stage of the proof I have my question: why $F$ restricted to $\hat{M}$ is an isomorphism? Just injectivity is clear. Many thanks in advance

Answer 1

Suppose that $\hat{M}$ is not the trivial space (otherwise $F$ maps every element of $H$ to $0$ in which case $g=0$). Then there exists an $h\in\hat{M}$ such that $\alpha = F(h) \neq 0$. Since $\hat{M}$ is a subspace it follows that $(\beta/\alpha)h \in \hat{M}$ for all $\beta \in \mathbb{R}$. Thus, by the linearity of $F$ we have $F((\beta/\alpha) h) = \beta$ for all $\beta \in\mathbb{R}$, that is, $F$ restricted to $\hat{M}$ is onto.

Answer 2

For $F\ne 0$ fix $x_0$ with $F(x_0)\ne 0$.Let $S$ be the set of scalars (the real or the complex numbers).Let $A=\{s x_0 | \in S\}$.Let $B=\{y | F(y)=0\}$.Then $A+B=H$ Proof: For any $x\in H$ ,let $U(x)=x_0 F(x)/F(x_0)$ and $V(x)=x-U(x)$. We have $U(x)\in A$ and $V(x)\in B$ and $x=U(x)+V(x)$....... Observe that $x\in A \implies U(x)=x$ and that $F$ restricted to $A$ is an isomorphism.

Answer 3

There is an easier observation. If $x \perp \mathcal{N}(F)$ is normalized so that $F(x)=1$, then $F(y-F(y)x)=0$ gives an orthogonal decomposition $$ y = (y-F(y)x)+F(y)x. $$ Therefore, $$ (y,x) = (F(y)x,x)=F(y)(x,x),\;\;\; y \in H. $$