Extending lemma 3.9 - Rudin's functional analysis to infinite dimension I was wondering if there's an infinite dimensional extension somewhere of the following.

Lemma 3.9 Suppose $\Lambda_1,\ldots,\Lambda_n$ and $\Lambda$ are linear functionals on a vector space $X$. Let
$$
N=\bigcap_{i=1}^n \left\{ x : \Lambda_ix =0\right\}
$$
the following three properties are equivalent


a) There are scalars $\alpha_1,\ldots,\alpha_n$ such that
$$
\Lambda = \sum_{i=1}^n \alpha_i \Lambda_i
$$


b) There's a $\gamma<\infty$ such that
$$
\left| \Lambda x \right| \leq \gamma \max_{1\leq i \leq n} \left| \Lambda_i x \right| 
$$


c) $\Lambda x = 0$ for every $x \in N$

The proof of the theorem relies at some point on the Riesz representation theorem in finite dimensions to determine the existance of the $\left\{ \alpha \right\}_{i=1}^n$, I wonder if this can somewhat be replaced with the usage of the Riesz representation theorems using measures or the dual of $L^p$.
Spefically in the lemma $\Lambda = \sum_{i=1}^n \alpha_i \Lambda_i$, I wonder if this can be extended to representations like $\Lambda = \sum_{i=1}^\infty \alpha_i \Lambda_i$, or even $\Lambda = \int_Q \Lambda_i d\mu$ (the latter maybe using some vector valued integration).
Any paper or reference would be useful. Thank you
 A: Yes, this lemma has versions for infinite sequences of linear functionals $\Lambda_i$ on a vector space, at least the equivalence of a) and b):

Asuume that $\Lambda_i(x)\to 0$ and, for some constant $c\ge 0$, $$|\Lambda(x)|\le c\sup\{|\Lambda_i(x)|:i\in\mathbb N\}$$ for all $x\in X$. Then there are scalars $\alpha_i$ with $\sum_i|\alpha_i|<\infty$ and $\Lambda(x)=\sum_i\alpha_i \Lambda_i(x)$ for all $x\in X$.

The proof is a combination of Hahn-Banach and a representation theorem: The set $$L=\{(\Lambda_i(x))_{i\in\mathbb N}: x\in X\}$$ is contained in $c_0$ by the first assumption and it is a subspace by the linearity of all $\Lambda_i$. Moreover, the linear functional $\varphi:L\to \mathbb K$ defined by $(\Lambda_i(x))_{i\in\mathbb N}\mapsto \Lambda(x)$ is well-defined, linear and continuous by the second assumption and the linearity of $\Lambda$. By Hahn-Banach, there is a continuous linear extension $\phi:c_0\to \mathbb K$ which is represented as $\phi((z_i)_{i\in\mathbb N})=\sum_i\alpha_iz_i$ for a sequence $\alpha\in\ell^1$. For $x\in X$ we thus get $$\Lambda(x)=\varphi((\Lambda_i(x))_{i\in\mathbb N})=\phi((\Lambda_i(x))_{i\in\mathbb N})=\sum_i\alpha_i\Lambda_i(x).$$
Replacing the simple representaion theorem for the dual of $c_0$ by other ones (e.g., for the Hilbert space $\ell^2$) you can easily get other versions of this theorem, also for families of functionals indexed, say, by $t\in K$ for a compact set $K$. Then you have to require the continuity of all $t\mapsto \Lambda_t(x)$ and $|\Lambda(x)|\le c \sup\{|\Lambda_t(x)|:t\in K\}$ to get a Borel measure $\mu$ on $K$ such that $\Lambda(x)=\int_K\Lambda_t(x)d\mu(t)$ for all $x\in X$.
I don't know whether such theorems are explicitely stated in the literature. But I consider this as an important principle exemplifying the slogan domination implies representation.
Edit. Condition (c) of course follows from (a) and (b) also for infinte sequences of $\Lambda_i$ but, in general, it is not sufficient: The existence of a Hamel basis on $X=c_0$ implies that there are discontinuous linear functionals $\Lambda:c_0\to\mathbb K$ and for $\Lambda_i(x)=x_i$ one has $N=\{x\in X: \Lambda_i(x)=0$ for all $i\in\mathbb N\}=\{0\}$.
