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Let $V=C[0,1]$ be a topological vector space with the topology induced by the separating family of seminorms $\{p_t\}_{t\in [0,1]}$, where $p_t(f)=|f(t)|$. Prove that $\Lambda\in (C[0,1],T_p)^*$ iff there are $t_1,\dots, t_n\in [0,1]$ and $\alpha_1,\dots,\alpha_n\in\Bbb{C}$ s.t $\Lambda(f)=\sum_{i=1}^n\alpha_if(t_i).$

My attempt: I know that a linear functional $\Lambda$ is continuous in this topology iff there are $\lambda_1,\dots,\lambda_n\geq 0, t_1,\dots,t_n\in [0,1]$ s.t $|\Lambda(f)|\leq\sum_{i=1}^n\lambda_i P_{t_i}(f)=\sum_{i=1}^n\lambda_i |f(t_i)|$. I thought about taking $\alpha_i\in\Bbb{C}$ s.t $|\alpha_i|=\lambda_i$ and claim $\Lambda(f)=\sum_{i=1}^n\alpha_i f(t_i)$, but I wasn't able to show equality.

Math101
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    A basic lemma in linear algebra says that if $f,f_i, 1 \leq i \leq n$ are linear functionals and $\cap ker (f_i)\subset \ker f$ then $f$ is a linear combination of $f_i's$. – Kavi Rama Murthy Dec 04 '21 at 12:02
  • @KaviRamaMurthy Do you happen to have a proof? wasn't able to find a proof (for the case of more then one functional) that doesn't use advanced tools – Math101 Dec 04 '21 at 12:19
  • See https://math.stackexchange.com/questions/709781/intersection-of-kernels-and-linear-dependence-of-functionals – Kavi Rama Murthy Dec 04 '21 at 12:29
  • And specifically Daniel Fischer's answer in that thread which is elementary. – Henno Brandsma Dec 04 '21 at 12:53
  • https://math.stackexchange.com/questions/386842/intersection-of-kernels-and-linear-dependence-of-linear-maps I believe I found an even more elementary proof (last comment) – Math101 Dec 04 '21 at 12:58

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If $f: V \to \Bbb R$ is a (linear) continuous functional, we know by continuity at $0$ that for some basic open subset of $\mathcal{T}_p$ (presumably the topology generated by the semi-norms $p_t, t \in [0,1]$ and topologists would simply say that $f: C_p([0,1]) \to \Bbb R$ is continuous and linear) then there are $t_i \in [0,1], i=1, \ldots,n$ and $r_i>0$ so that $$f \in \bigcap \pi_{t_i}^{-1}[(-r_i, r_i)] \subseteq (-1,1)$$ (this are the basic sets from the standard subbase; I use that the weak topology induced by the seminorms $p_t$ is the same as the one induced by the evaluations/projections $\pi_t$). From this inclusion it follows quite easily that $\bigcap_{i=1}^n \text{ker}(\pi_{t_i}) \subseteq \text{ker}(f)$ (a point common to all kernels is in the intersection and thus in the kernel of $f$).

By a standard result on functionals (e.g. from here, we get that $f$ must be a linear combination of the $\pi_{t_i}, i = 1, \ldots n$ which is what is claimed.

Henno Brandsma
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