The following lemma applies Hahn-Banach theorem in a tricky way.

Suppose ^[1] $\Lambda_{1}, \ldots, \Lambda_{n}$ and $\Lambda$ are linear functionals on a vector space $X .$ We define $N:=\{ x: \Lambda_{1} x=\cdots=\Lambda_{n} x=0 \}$. Then the following three properties are then equivalent:

1. There are scalars $\alpha_{1}, \ldots, a_{n}$ such that
   $$\Lambda = \alpha_1 \Lambda_1 + \cdots + \alpha_n \Lambda_{n};$$
2. There exists $\gamma < \infty$ such that
   $$\Lambda x \leq \gamma \max _{1 \leq i \leq n} \left| \Lambda_{i} x\right|, \quad \forall x \in X;$$
3. $\Lambda x=0$ for every $x \in N$.

It is clear that (1) implies (2) and that (2) implies (3). Assume (3) holds. Let $\Phi$ be the scalar field. Define $\pi: X \rightarrow \Phi^{n}$ by

$$\pi(x)=\left(\Lambda_{1} x, \ldots, \Lambda_{n} x\right).$$

If $\pi(x)=\pi\left(x^{\prime}\right)$, then (3) implies $\Lambda x=\Lambda x^{\prime}.$ Hence $f(\pi(x))=\Lambda x$ defines a linear functional $f$ on $\pi(X).$ Extend $f$ to a linear $F$ on $\Phi^n$. This means that there exist $\alpha_1,\ldots,\alpha_n$ such that

$$F\left(u_{1}, \ldots, u_{n}\right)=\alpha_{1} u_{1}+\cdots+\alpha_{n} u_{n}.$$

Thus

$$\Lambda x=F(\pi(x))=F\left(\Lambda_{1} x, \ldots, \Lambda_{n} x\right) =\sum_{i=1}^{n} \alpha_{i} \Lambda_{i} x,$$

which is (1).

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1. Walter Rudin. Analyse fonctionnelle. French. Trans. English by Adbdellatif Abouhazim.1995. Chap. 3, p. 63. ↩︎