Proof of the Riesz representation theorem in finite dimensions

Proof

We build $y$ from an orthonormal basis, and then prove uniqueness with $x = y - y'$.

1. Let $V$ be a finite-dimensional inner product space over $\mathbb{K} = \R$ or $\C$, with inner product $\ip{\cdot}{\cdot}$ linear in the first argument. Let $f \colon V \to \mathbb{K}$ be a linear form.
2. For all $u, w \in V$ and $a \in \mathbb{K}$, conjugate symmetry and linearity in the first argument give $\ip{u}{a w} = \overline{a} \ip{u}{w}$. In the same way, $\ip{u}{\cdot}$ is additive.
3. Let $e_1, \dots, e_n$ be an orthonormal basis of $V$. Set $y = \sum_{i=1}^{n} \overline{f(e_i)}\, e_i$.
4. Let $x \in V$. By the formula for coordinates, $x = \sum_{i=1}^{n} \ip{x}{e_i} e_i$.
5. Since $f$ is linear, step 4 gives $f(x) = \sum_{i=1}^{n} \ip{x}{e_i} f(e_i)$.
6. By step 2, $\ip{x}{y} = \sum_{i=1}^{n} \overline{\overline{f(e_i)}} \ip{x}{e_i} = \sum_{i=1}^{n} f(e_i) \ip{x}{e_i}$.
7. Steps 5 and 6 give $f(x) = \ip{x}{y}$ for all $x \in V$. This proves existence.
8. For uniqueness, let $y, y' \in V$ with $\ip{x}{y} = f(x) = \ip{x}{y'}$ for all $x \in V$.
9. By step 2, $\ip{x}{y - y'} = \ip{x}{y} - \ip{x}{y'} = 0$ for all $x \in V$.
10. Take $x = y - y'$. Then $\ip{y - y'}{y - y'} = 0$, and positivity of the inner product gives $y - y' = 0$. So $y = y'$.