Analysis Problem of the Day 54

Today’s problem appeared as Problem 6 on Stanford’s Spring 2022 Analysis Qual:

Problem 6. a) Show that there exists a signed Radon measure $\nu$ on $[0,1]$ such that $\int_0^1 p d\nu = p'(0)$ for all polynomials $p$ of degree at most 100.

b) Is there such a signed measure $\mu$ such that the above identity holds for all polynomials? Prove or disprove.

Solution: a) By Riesz-Markov-Kakutani, every signed Radon measure $\nu$ is a bounded linear functional on $C([0,1]),$ and we are given the action of this linear functional on the subspace of polynomials of degree at most 100. Thus, the existence of such a measure $\nu$ boils down to extending $\nu$ to all of $C(K),$ which can be done via the Hahn-Banach theorem. The only condition one needs to verify is that there is a seminorm on $C(K)$ that bounds $\nu$ on the given subspace. Let $p(x) = \sum_{n=0}^{100} a_n x^n$ be a polynomial of degree at most 100, and define $J_n p$ to be the $n$ -th antiderivative of $p,$ i.e. $J_0 p = p$ and $(J_n p)(x) = \int_0^x (J_{n-1}p)(t)dt.$ Then, given the values of $(J_0 p)(1) (J_1 p)(1),...,(J_{100} p)(1),$ one may uniquely determine the coefficients of the polynomial $p.$ Indeed, $(J_{n}p)(0)=0$ for $1 \leq n \leq 100,$ so $J_{100}p$ is a polynomial of degree at most 200 with no $x^n$ terms for $0 \leq n \leq 99.$ Let $(J_{100} p)(x) = x^{100} q(x)$ for $q$ a polynomial of degree at most 100. Then, $(J_{100}p)(1)=q(1),(J_{99})p(1)=100 q(1)+q'(1),...$ It follows that one may uniquely solve for $q(1),...,q^{(100)}(1)$ from the values of $(J_n p)(1), 0 \leq p \leq 100.$ But since $q$ is a polynomial of degree at most 100, one therefore obtains a unique solution for the coefficients of $q,$ and therefore the coefficients of $p.$

Now, let $a=[a_0,...,a_{100}]$ and $J=[J_0 p(1),...,J_{99} p(1)].$ It follows from the above discussion that there exists an invertible matrix $A$ s.t. $Aa=J,$ and it follows that $|p'(0)|=|a_1| \leq \|a\|_1 =\|A^{-1}J\|_1.$ Notice that the latter is a seminorm as a function of $p$ and can be in fact defined for an arbitrary element $f \in C([0,1]).$ Thus, by Hahn-Banach, $\nu$ is bounded by a seminorm on all of $C([0,1])$ and therefore extends to a bounded linear functional on $C([0,1]),$ i.e. a signed Radon measure.

b) We claim that such a measure $\mu$ does not exist. Proceed by contradiction. Indeed, by integration by parts, it would follow that $\int_0^1 p'(x) x d\mu = p(1)-p(0)-p'(0).$ Defining $\nu = x \mu,$ it would follow that $\int_0^1 g(x) d\nu = \int_0^1 g(x) dm - g(0)$ for all polynomials, and by Weierstrass approximation, for all $g \in C([0,1]),$ where $m$ is the Lebesgue measure on $[0,1].$ But this implies that $\nu = m - \delta_0,$ where $\delta_0$ is the Dirac delta measure at $0.$ But then, $\int \chi_{\{0\}} d\nu = \int 0 d\mu = 0 = -\delta_0(0)=-1,$ which is a contradiction. Thus, such a measure $\mu$ cannot exist.

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