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I would really appreciate it if you could give me some advice on the part (c) of Exercise 13 of first chapter of Walter Rudin's book "Functional Analysis":

Let $C$ be the vector space of all complex continuous functions on $[0, 1]$. Define \begin{equation} d(f,g) = \int_0^1 \frac{\lvert f(x) - g(x) \rvert}{1 + \lvert f(x) - g(x) \rvert} \ dx \ . \end{equation} Let $(C, \sigma)$ be $C$ with the topology induced by this metric. Let $(C, \tau)$ be the topological vector space defined by the semi-norms \begin{equation} P_x(f) = \lvert f(x) \rvert, \qquad (0 \leq x \leq 1), \end{equation}

(a) Prove that every $\tau$-bounded set in $C$ is also $\sigma$-bounded and that the identity map $id: (C, \tau) \rightarrow (C, \sigma)$ therefore carries bounded sets into bounded sets.

(b) Prove that $id: (C, \tau) \rightarrow (C, \sigma)$ is nevertheless not continuous, although it is sequentially continuous (by Lebesgue's dominated convergence theorem). Hence $(C, \tau)$ is not metrizable. Show also directly that $(C, \tau)$ has no countable local base.

(c) Prove that every continuous linear functional on $(C, \tau)$ is of the form \begin{equation} f \rightarrow \sum_{i=1}^n c_i f(x_i) \end{equation} for some choice of $x_1, \ldots, x_n$ in $[0, 1]$ and some $c_i \in \mathbb{C}$.

First part of this question is here and the second part is here. I have no idea solving the third part too.

Thanks in advance.

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Hossein Moradi
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There is a general fact about a TVS $X$ defined by a family of seminorms $\{\rho_{\alpha}\}_{\alpha \in J}$ (which is perhaps proved in the textbook before this):

Theorem: If $\varphi : X\to \mathbb{C}$ is a continuous linear functional, then $\exists C>0$ and $\alpha_1, \alpha_2,\ldots, \alpha_n \in J$ such that $$ |\varphi(x)| \leq C \sum_{j=1}^n \rho_{\alpha_j}(x) $$

Applying this to your present situation tells you that if $\varphi : C \to \mathbb{C}$ is a $\tau$-continuous linear functional, then $\exists C > 0$ and $x_1,x_2,\ldots, x_n \in [0,1]$ such that $$ |\varphi(f)| \leq C \sum_{j=1}^n |f(x_i)| $$ Now consider the linear functionals $\varphi_x : f \mapsto f(x)$ and note that $$ \cap_{j=1}^n \ker(\varphi_{x_i}) \subset \ker(\varphi) $$ and so you get your result.

Prahlad Vaidyanathan
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