Graphical version of the proof of Riesz’ Theorem I am trying to sketch a graphical version of proof of Riesz' representation Theorem:
Let H be a Hilbert space. For every $L \in H^*$ there exists a unique $u_L \in H$ such that: 
\begin{align}
1. & \quad L x = (u_L, x) \text{ for every } x \in H,   \\
2. & \quad \| L \| = \| u_L\|. 
\end{align}
Does anybody have any suggestion, how to proceed? 
 A: The first geometric fact is that the null space of a non-zero linear functional $F$ on a real or complex linear space $X$ is a hyper plane, meaning that it is one dimension shy of being the entire space. Indeed, if $F(x)\ne 0$, you can normalize so that $F(x)=1$, and note that $F(y-F(y)x)=0$, which means every $y\in X$ can be uniquely written as $y = n +\alpha x$, where $n \in \mathcal{N}(F)$ and $\alpha$ is a scalar:
$$
                 y = (y-F(y)x)+F(y)x.
$$
Therefore, if $F$ is a non-zero linear functional, and if $G$ is another linear functional with $\mathcal{N}(F)\subseteq\mathcal{N}(G)$, then either $G=0$ or the null spaces must be equal, which forces $F$ and $G$ to be scalar multiples of each other. This follows from the above decomposition because $F(y-F(y)x)=0$ implies $G(y-F(y)x)=0$ for all $y$, which gives $G(y)=G(x)F(y)$ or $G=\alpha F$ where $\alpha=F(x)$ is a scalar constant.
Finally, in the case of a Hilbert space, if $F$ is a non-zero continuous linear functional, then $\mathcal{N}(F)\ne X$ is a closed subspace. So you can find a unit vector $y \perp \mathcal{N}(F)$, and note that $G(x)=(x,y)$ has the same null space as $F$. Therefore $F$ and $G$ must be constant multiples of each other.
