Riesz Representation Theorem

Theorem.

For any finite-dimensional inner product space $V$, there exists a bijective map $\alpha$ from $V$ to its dual $V^{\prime }$ such that

$$V\stackrel{\alpha }{\cong }{V}^{\prime },v\mapsto ⟨\cdot ,v⟩,$$

i.e. for any linear functional $\phi \in V^{\prime }$, there exists unique $v\in V$ such that for any $u\in V$, $\phi (u)=⟨u,v⟩$.

Proof.

To show the injectivity of $\alpha$, suppose $\alpha (v)=\phi =0$ such that for all $u\in V$, $\phi (u)=⟨u,v⟩=0$. This implies that $⟨v,v⟩=0$, which in turn implies that $v=0$ by the definition of the inner product.

To show the surjectivity of $\alpha$, suppose $\phi \in V^{\prime }$ is some linear functional. If $\phi =0$, then clearly $\alpha (0)=\phi =0$. Thus, assume $\phi \ne 0$. In the finite-dimensional case, we are guaranteed that ${null(\phi )}^{\perp }\ne 0$ if $\phi \ne 0$. So, let $w$ be a nonzero unit vector in $null(\phi {)}^{\perp }$. We proceed by conjuring a vector $v=\overline{\phi (w)}w$ which we will use to show that for all $u\in V$, $\phi (u)=⟨u,v⟩$ and thus $\phi =\alpha (v)$. Observe that $v\in null(\phi {)}^{\perp }$ since $v$ is a scalar multiple of $w$. Suppose $u$ is any vector in $V$. We decompose $u$ as such:

$$u=(u-\frac{\phi (u)}{\phi (v)}v)+\frac{\phi (u)}{\phi (v)}v.$$

A few key observations:

$$\phi (v)=\phi (\overline{\phi (w)}w)=\overline{\phi (w)}\phi (w)=|\phi (w){|}^2$$

|| v ||^2 = \langle v, v \rangle = \langle \overline{\varphi(w)}w, \overline{\varphi(w)}w \rangle = \overline{\varphi(w)}\varphi(w) \langle w, w \rangle = \lvert\varphi(w)\rvert

\implies \varphi(v) = || v ||

$$and$$

\varphi \bigg(u-\frac{\varphi(u)}{\varphi(v)}v \bigg) = \varphi(u)-\frac{\varphi(u)}{\varphi(v)}\varphi(v) = 0

\implies \langle u-\frac{\varphi(u)}{\varphi(v)}v, v\rangle = 0

where we used the fact that $v \in null(\varphi)^\perp$ and is thus orthogonal to any vector in the null space of $\varphi$. Then,

\langle u, v \rangle = \langle u-\frac{\varphi(u)}{\varphi(v)}v, v \rangle + \langle \frac{\varphi(u)}{\varphi(v)}v, v \rangle = 0+\frac{\varphi(u)}{||v||^2}\langle v,v \rangle = \varphi(u),

and therefore, $\varphi = \alpha(v)$. $\square$ --- As demonstrated in the linked resource, this proof can easily be extended to show the theorem holds for special infinite-dimensional inner product spaces called Hilbert Spaces. It seems that this theorem serves as a bridge from linear algebra and algebra in general to analysis and measure theory. Resources: [Hilbert Spaces and The Riesz Representation Theorem](https://math.uchicago.edu/~may/REU2021/REUPapers/Adler.pdf)