How can you prove the linearity of a functional using a signed Borel measure?

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Here is this week's POTW:

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Let $T : C^1([0,1]) \to \Bbb R$ be a linear functional such that $|T(f)| \le A\|f\| + B\|f'\|$ for all $f \in C^1[0,1]$, where $A$ and $B$ are constants and $\|\cdot\|$ is the supremum norm. Prove that there is a signed Borel measure $\mu$ on $[0,1]$ and a constant $C$ such that $$T(f) = Cf(0) + \int f'\, d\mu$$ for all $f \in C^1([0,1])$.

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I apologize to the MHB members for the following misprint to the problem: the equation originally written $T(f) = Cf(0) + \int f\, d\mu$ is supposed to be $T(f) = Cf(0) + \int f'\, d\mu$. I've corrected the misprint, and I'll give another week before posting a solution.

 

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No answered this week's problem. You can find my solution below.

Spoiler

The mapping $\Lambda : C([0,1]) \to \Bbb R$ given by $\Lambda(g) : x\mapsto T(h)(x)$, where $h(x) = \int_0^x g(t)\, dt$, is a linear functional on $C([0,1])$ such that $\|\Lambda\| \le c_1 + c_2$. Hence, by the Riesz representation theorem, there exists a finite signed Borel measure $\mu$ on $[0,1]$ such that $\Lambda(g) = \int g\, d\mu$ for all $g \in C([0,1])$. Given $f \in C^1([0,1])$, $f(x) = f(0) + \int_0^x f'(t)\, dt$, hence $$T(f) = T(1)f(0) + \Lambda(f') = Cf(0) + \int f'\, d\mu,$$ where $C = T(1)$.