Riesz representation theorem

Theorem Let $(X,⟨\cdot ,\cdot ⟩)$ be a complete Hilber space, and $L:X\to \mathbb{F}$ a continuous (or bounded) linear map. Then there exists a unique $x_L\in X$ such that

$$L(x)=⟨x_L,x⟩\forall x\in X$$

moreover $\Vert L\Vert =\Vert x_L\Vert$.

Proof The main idea is to pick an element in the orthogonal complement of the Kernel. If $Ker(L)=X$, then the map is trivial (everthing is zero), in this case $x_L=0$. Suppose now $Ker(L)\ne X$. Since $L$ is continuous, then $Ker(L)$ is a closed set (the preimage of $\{0\}$), then there exists an element $y\in Ker(L{)}^{\perp }$, since it’s linear we can scale it such that $\Vert y\Vert =1$.

Consider a vector of the form

$$z=L(x)y-L(y)x$$

we observe that $z\in Ker(L)$, in fact

$$L(z)=L(x)L(y)-L(y)L(x)=0$$

so that

$$0=⟨y,z⟩=⟨y,L(x)y-L(y)x⟩$$ $$=L(x)⟨y,y⟩-L(y)⟨y,x⟩$$ $$=L(x)-L(y)⟨y,x⟩$$

if we set $x_L=L(y)y$ then

$$L(x)=L(y)⟨y,x⟩=⟨x_L,x⟩$$

To show uniqueness, suppose there exists another $x_L^{\prime }$, taking the difference

$$0=⟨x_L,x⟩-⟨x_L^{\prime },x⟩=⟨x_L-x_L^{\prime },x⟩,\forall x\in X$$

but the only element orthogonal to $X$ is the zero vector.

Computing the operator norm

$$\Vert L\Vert =\sup \{|L(x)|:\Vert x\Vert =1\}$$ $$=\sup \{|⟨x_L,x⟩|:\Vert x\Vert =1\}$$

but using Cauchy-Schwarz inequality

$$|⟨x_L,x⟩|\le \Vert x_L\Vert \Vert x\Vert =\Vert x_L\Vert$$

this upper bound is thight, since

$$|L(y)|=|⟨x_L,y⟩|=\Vert x_L\Vert \square$$