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Can you help me please with this question?

Prove that there exists a $\mu$-Borel regular measure on $[0,1]$ so that to all polynomial $p$ of degree $\leq n$ $(n\in\mathbb{N})$ one has:

$$\int_{[0,1]}p(t) ~d\mu(t)=\sum_{k=1}^{n}p^{(k)}\left ( \frac{k}{n} \right )$$

Thanks a lot!

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What are conditions on a linear functional to insure that it is integration against some measure? Are those conditions satisfied for the linear functional on your right-hand side? – GEdgar Jul 22 '12 at 0:16
up vote 2 down vote accepted

Consider $C[0,1]$, the space of continuous function on $[0,1]$ with the $\sup$ norm. Let $\Pi_n$ be the finite dimensional subspace consisting of polynomials of degree $n$ or less. Then $L:\Pi_n \to \mathbb{C}$ defined by the right hand side above is a continuous linear functional on this subspace.

Can you extend this functional to a continuous linear functional on $C[0,1]$?

What does the Riesz representation theorem tell you then?

Here is an explanation of why $L$ is continuous:

Suppose $a \in \mathbb{C}^{n+1}$. Then define the polynomial $ p(t) = \sum_{k=0}^n a_k t^k$ (with $a = (a_0,\cdots, a_n)$). Differentiating gives $p'(t) = \sum_{k=0}^{n-1} (k+1)a_{k+1} t^k$, which can be represented by the point $(a_1,\cdots,n a_n, 0)$. So, we can define $D:\mathbb{C}^{n+1} \to \mathbb{C}^{n+1}$ to be the continuous linear operator $D(a) = (a_1,\cdots,n a_n, 0)$. Similarly, if we select $t^*\in [0,1]$, then $p(t^*) = \sum_{k=0}^n a_k (t^*)^k$, which is trivially a linear function of $a$. So let $E_t :\mathbb{C}^{n+1} \to \mathbb{C}$ be the continuous linear functional $E_t(a) = \sum_{k=0}^n a_k t^k$. Then it is clear that the map $\Lambda : \mathbb{C}^{n+1} \to \mathbb{C}$ defined by $\Lambda(a) = \sum_{k=1}^n E_{\frac{k}{n}} D^k (a)$ is also a continuous linear functional.

It remains to show continuity of the map $\pi: \Pi_n \to \mathbb{C}^{n+1}$, where $\pi(p)$ is the unique element of $\mathbb{C}^{n+1}$ such that $p(t) = \sum_{k=0}^n \pi(p)_k t^k$. If $\pi$ is continuous, then it follows that $L = \Lambda \circ \pi$ is continuous. This can be done explicitly using the Vandermonde matrix, or more simply by noting that the polynomials $v_k(t) = t^k$ form a basis for the finite dimensional $\Pi_n$, and define $\pi(v_k) = e_k$. Then $\pi$ can be defined on $\Pi_n$ by extension. Finally, note that $n(p) = \|\pi(p)\|$ defines a norm on $\Pi_n$, and since norms on a finite dimensional space are equivalent, it follows that $n(p) \leq K \|p \|_{\infty}$, where $K>0$ and $\|\cdot \|_{\infty}$ is the $\sup$ norm on $C[0,1]$. Hence the map $\pi$ is continuous.

Note: The extension of the functional $L$ to $C[0,1]$ is only guaranteed to agree with the right hand side above on $\Pi_n$. In particular, for $f \in C[0,1] \setminus \Pi_n$, the function $f$ may not even be differentiable.

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You could add an argument to the fact that L is indeed continuous, this is the heart of the proof and not obvious from what is written now. – Did Jul 22 '12 at 8:03
@did: It seems to me that the continuity of $L$ is trivial (it is defined on a finite dimensional space) and the existence of a continuous extension is guaranteed by a certain well-known theorem... am I missing something? – Nate Eldredge Jul 22 '12 at 14:43
@did: The Hahn Banach theorem guarantees this. For example, Rudin's "Real & Complex Analysis", Theorem 5.16. – copper.hat Jul 22 '12 at 16:20
@did: I can write down an explicit formula for $L$ defined on $C[0,1]$ that matches the right hand side above on $\Pi_n$. If $f \in C[0,1]$, then only the values $f(\frac{k}{n})$ for $k=0,\cdots, n$ appear in the formula. The HB theorem is not needed, although it saves a few lines in the proof. – copper.hat Jul 22 '12 at 17:05
copper: You are right. I first thought that you proposed to define L on the whole space C[0,1] by the sum on the RHS of the equation in the question... But I see now that this is definitely not what you wrote and I fail to see what led me to err like this since your first post is quite clear on the extension step (perhaps I began to imagine things because, to me, this is a finite dimensional problem, which C[0,1] is not the natural context of). Anyway, sorry about the noise. – Did Jul 22 '12 at 20:24


  • Show that there exists some coefficients $(\lambda_i)_{1\leqslant i\leqslant n}$ such that $\sum\limits_{k=1}^{n}p^{(k)}\left ( \frac{k}{n} \right )=\sum\limits_{i=1}^{n}\lambda_ia_i$ for every polynomial $p(x)=\sum\limits_{i=0}^{n}a_ix^i$.
  • Show that, for every $1\leqslant i\leqslant n$, there exists a measure $\mu_i$ such that $\int\limits_0^1t^i\,\mathrm d\mu_i(t)=1$ and $\int\limits_0^1t^k\,\mathrm d\mu_i(t)=0$ for every $0\leqslant k\leqslant n$ such that $k\ne i$.
    Sub-hint: One can find $\mathrm d\mu_i(t)=p_i(t)\,\mathrm dt$ for some polynomial $p_i$ of degree at most $n$.
  • Consider the measure $\mu=\sum\limits_{i=1}^{n}\lambda_i\mu_i$.
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