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There is a problem about which I have no ideas at all:

For each $t\in [0,1]$, we define the functional $\delta_t$ on $C [0,1]$ as $\delta_t(f)=f(t)$. Let $I (f) = \displaystyle\int\limits_0^1f (t) dt$. Is there a bounded linear functional $F$ on $C [0,1]^\ast$ such that $F (\delta_t) = 1$ for any $t\in[0,1]$ and such that $F (I) = 2$?

Maybe someone can give some hints? Maybe we need to prove that the set $\{\delta_t\}$ is dense in $C [0,1]^\ast$ and then the answer would be no?

thing
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    Hint: let's denote $L = Span{\delta_t}$. You have exactly two possibilities: either $I \in \overline{L}$ and such $F$ doesn't exist or $I \notin \overline{L}$ and such $F$ exists due to Hahn-Banach theorem – Matsmir Dec 29 '19 at 13:17
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    Hint for details regarding Matsmir's hint: Let $G\in C^{**}$ be defined by $G(\phi)=\phi(1)$. So $G(\delta_1)=G(I)=1$. Because such a $G$ exists, your question is equivalent to asking whether there exists $F$ with $F(\delta_t)=0$ for all $t$ and $F(I)\ne0$. – David C. Ullrich Dec 29 '19 at 13:24
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    Regarding :Maybe we need to prove that ${\delta_t}$ is dense...": Of course you meant the span of that set. No, that span is dense in the weak* topology, but not in the norm topology, which is what would be relevant. – David C. Ullrich Dec 29 '19 at 13:26
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    @thing $I$ belongs to $\overline{Span{\delta}}$ iff $I$ can be approximated by linear functionals of the form $\sum_{i = 1}^n \alpha_i \delta_{t_i}$ in norm topology. But you can always find a function $f \in C[0,1]$ that is $0$ on $t_i$ and $I(f)$ is arbitrary close to $||f||$. – Matsmir Dec 29 '19 at 15:43
  • thanks, i completely understand. – thing Dec 30 '19 at 06:02

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