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Let $C^k([0,1])$ be the space of such complex-valued functions on $[0,1]$ that are continuously differentiable at least $k$ times ($k\in\mathbb N$). It is well known that $C^k([0,1])$ is a Banach space with the norm $$f\mapsto\|f\|\equiv\sum_{j=0}^k\|f^{(j)}\|_{\text{uniform}},$$ where the “zeroth” derivative is the function itself.

Now suppose that $I:C^k([0,1])\to\mathbb C$ is a bounded, linear functional. By the Riesz representation theorem, there exist a unique regular complex Borel measure $\mu$ on $[0,1]$ and unique constants $(c_0,\ldots,c_{k-1})\in\mathbb C^k$ such that $$I(f)=\int f^{(k)}\,\mathrm d\mu+\sum_{j=0}^{k-1}c_jf^{(j)}(0)\quad\forall f\in C^k([0,1]).\tag{*}$$ See also this thread.


I want to prove the existence and uniqueness of a slightly more general form of representation. Namely, assume that $(S_0,\ldots,S_{k-1})$ are given linear functionals on $C^k([0,1])$ and that $(S_0,\ldots,S_{k-1})$ separate polynomials of degree at most $k-1$. That is, if \begin{align*} f(x)\equiv&\,\sum_{j=0}^{k-1} p_jx^j\quad\forall x\in[0,1]\\ g(x)\equiv&\,\sum_{j=0}^{k-1}q_j x^j\quad\forall x\in[0,1],\end{align*} where $(p_0,\ldots,p_{k-1},q_0,\ldots,q_{k-1})$ are complex coefficients, and $f$ is not identically equal to $g$, then there exists some $j\in\{0,\ldots,k-1\}$ such that $S_j(f)\neq S_j(g)$.


Conjecture: There exists a unique regular complex Borel measure $\nu$ on $[0,1]$ and unique constants $(b_0,\ldots,b_{k-1})\in\mathbb C^k$ such that$$I(f)=\int f^{(k)}\,\mathrm d\nu+\sum_{j=0}^{k-1}b_jS_j(f)\quad\forall f\in C^k([0,1]).$$


I guess the proof of this conjecture is structurally similar to that of $(*)$; but I feel stuck. In particular, I wonder if the correctness of the conjecture hinges on whether the functionals $(S_0,\ldots,S_{k-1})$ are assumed, in addition, to be bounded on $C^k([0,1])$. Any hint or suggestion would be greatly appreciated.

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Your answer is correct: you need the boundedness of $S_j$. Assuming that, here is a shorter proof. Let $P\subset C^k[0,1]$ be the space of polynomials of degree at most $k-1$. The functionals of the form $\int f^{(k)}\,d\nu$ vanish on $P$ and therefore define functionals on the quotient space $C^k/P$. Also, every functional on $C^k/P$ is of this form, by the RRT.

Recall that the dual of quotient space is naturally identified with a subspace of the dual. Namely, it's the subspace $P^\perp\subset (C^k)^*$, the annihilator of $P$. It remains to prove that $P^\perp \cup \{S_j\}$ span $(C^k)^*$. To this end, note that for any $f\in (C^k)^*$ the restriction of $f$ to $P$ is a linear combination of $S_j$. Subtracting this linear combination from $f$ we get an element of $P^\perp$.

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  • $\begingroup$ Thank you for your answer. Your approach is interesting and elegant. Shorter as it is, however, I feel more comfortable with the approach I outlined, as it is more elementary—it basically collapses to linear algebra 101 once we assume the $S$'s are bounded, using the representation $(*)$—even though the details, admittedly, are somewhat tedious and less elegant than yours, indeed. $\endgroup$ – triple_sec Jul 21 '14 at 2:14
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I think I got it now. First of all, the functionals $(S_0,\ldots,S_{k-1})$ must be bounded (with respect to the $C^k$-norm), otherwise the conjecture may not hold. Indeed, if $k=1$, then one can construct an unbounded linear functional $S:C^1([0,1])\to\mathbb C$ (using Zorn's lemma and the fact that $C^1([0,1])$ is infinite-dimensional) that separates the zero-degree polynomials (i.e., the constant functions). In turn, the functional $I(f)\equiv f(0)$ is linear and bounded on $C^1([0,1])$. If the claimed representation existed for $S$, then its coefficient must satisfy $b\neq 0$. (Otherwise $0=I(x)=I(x+1)=1$, since $x$ and $x+1$ have the same derivative, which is impossible). But then the unbounded functional $S$ could be represented as $1/b$ times the difference between two bounded linear functionals, $I$ and $f\mapsto\int f'\,\mathrm d\nu$, which is impossible.

Once one adds the extra assumption that the $S$'s must be bounded, the result is logically quite straightforward (though algebraically tedious, so I will just sketch the proof). For each $j\in\{0,\ldots,k-1\}$, $S_j$ can be represented in the form $(*)$ (this is where the boundedness of the $S$'s is crucial). Given that the $S$'s are assumed to separate low-order polynomials, it follows that the complex matrix$$\left[\begin{array}{cccc}\dfrac{S_0(1)}{0!}&\dfrac{S_1(1)}{0!}&\cdots&\dfrac{S_{k-1}(1)}{0!}\\\dfrac{S_0(x)}{1!}&\dfrac{S_1(x)}{1!}&\cdots&\dfrac{S_{k-1}(x)}{1!}\\\vdots&\vdots&\ddots&\vdots\\\dfrac{S_0(x^{k-1})}{(k-1)!}&\dfrac{S_1(x^{k-1})}{(k-1)!}&\cdots&\dfrac{S_{k-1}(x^{k-1})}{(k-1)!}\end{array}\right]$$ is invertible. Based on this observiation, the coefficients of $I$ and the $S$'s according to the representation $(*)$ yield, after some linear algebra, a unique representation of $I$ as a function of the $S$'s according to the conjecture.

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