Abstract Nonsense

Crushing one theorem at a time

Interesting Theorem Regarding Polynomials

Point of post: This post proves that the uniform limit of polynomials of a fixed degree is a polynomial of degree less than or equal to that fixed degree.


Surely most of us are aware of the Stone-Weierstrass theorem (see here for a more generalized version for compact Hausdorff spaces) which states that for any [a,b]\subseteq\mathbb{R} one has that the polynomials \mathbb{P}[a,b] are dense in the space C[a,b] of continuous functions with the infinity norm. Another interpretation of this is that every continuous function f can be arbitrarily and uniformly approximated by a sequence of polynomials \{p_n\}_{n\in\mathbb{N}} is the sense that \|p_n-f\|_{\infty}\to 0. One may begin to wonder “what kinds of polynomials do I need to approximate these functions?” in the sense that could I approximate f with polynomials of only even degree? What about odd degree? What bout prime degree? This seems like a valid question since, either the Stone-Weierstrass theorem is done by constructing the polynomials (see the first link and the notion of the Bernstein polynomials) or it’s proven in full generality (as in the second link). The first way leaves one wondering if there isn’t a ‘better’ choice of polynomials to approximate f, I mean, it’s conceivable that using Bernstein polynomials while a valid way to prove the theorem is overly complicated. The second proof gives the reader absolutely zero idea what the polynomials used to approximate the functions look like. So with this in mind, one may ask an an example of such a thought experiment (although a naive one) whether continuous functions can be approximate by polynomials which all have a fixed degree N. Well, in this post we will show that not only can you not approximate all the continuous functions with such sequences of polynomials but you can really only approximate other polynomials of degree less than or equal to N! In other words, we’ll prove that the space of all polynomials of degree less than or equal to N is closed in C[a,b] with the infinity norm. In fact, we’ll prove something MUCH stronger

Theorem: Let N\in\mathbb{N} be fixed and define the sequence \left\{p_n\right\}_{n\in\mathbb{N}} of polynomials by\displaystyle p_n(x)=\sum_{k=0}^{N}\alpha_{n,k}x^k for some real numbers \displaystyle \alpha_{n,0},\cdots,\alpha_{n,N}. Suppose that there exists N+1 points \{x_0,\cdots,x_N\} such that p_n(x_k) converges for k=0,\cdots,N. Then, \left\{p_n\right\}_{n\in\mathbb{N}} converges uniformly to a polynomial of degree less than or equal to N+1 on every compact subspace of \mathbb{R}.

Proof: We first show that if K\subseteq\mathbb{R} is compact then the convergence of the coefficient sequences implies the uniform convergence of \{p_n\}_{n\in\mathbb{N}} to the polynomials whose coefficients are the limits of the coefficient sequences. Indeed, suppose that K\subseteq\mathbb{R} is compact. Then, if \alpha_{n,k}\to\alpha_k then \displaystyle p_n(x)\xrightarrow{\text{unif.}}\sum_{k=0}^{N}\alpha_k x^k. To see this merely let M=\max\{\|x\|_{\infty},\cdots,\|x^N\|_{\infty}\}. Then, given \varepsilon>0 choosing N large enough so that \displaystyle \left|\alpha_{k,n}-\alpha_k\right|<\frac{\varepsilon}{MN} implies that

\displaystyle \begin{aligned}\left\|\sum_{k=0}^{N}\alpha_{k,n}x^k-\sum_{k=0}^{N}\alpha_k x^k\right\|_{\infty} &\leqslant \sum_{k=0}^{N}\left\|\left(\alpha_{k,n}-\alpha_k\right)x^k\right\|_{\infty}\\ &=\sum_{k=0}^{N}\left|\alpha_{n,k}-\alpha_k\right|\left\|x^k\right\|_{\infty}\\ &< \frac{\varepsilon}{MN}\sum_{k=0}^{N}\left\|x^k\right\|_{\infty}\\ &\leqslant \varepsilon\end{aligned}


from where it follows by the arbitrariness of \varepsilon that \displaystyle p_n(x)\xrightarrow{\text{unif.}}\sum_{k=0}^{N}\alpha_{k}x^k as desired. We now show that since p_n(x_k) converges for each k=0,\cdots,n then \alpha_{k,n}\to\alpha_k. Indeed, consider that this fact is equivalent to saying that

\displaystyle \lim_{n\to\infty}\begin{pmatrix}\alpha_{n,0}+\cdots+\alpha{n,N}x_0^N\\ \vdots\\ \alpha_{n,0}+\cdots+\alpha_{n,N}x_N^N\end{pmatrix}=\begin{pmatrix}p_{\infty}(x_0)\\ \vdots\\ p_{\infty}(x_N)\end{pmatrix}


where clearly \displaystyle p_{\infty}(x_k)=\lim_{n\to\infty}p_n(x_k). Or, if we define \bold{x}_n=\left(\alpha_{n,0},\cdots,\alpha_{n,N}\right)^{\top}, y=\left(p_{\infty}(x_0),\cdots,p_{\infty}(x_N)\right)^{\top}, and V=\left[x_i^j\right]_{i,j=0}^{N} then this can be rephrased as \displaystyle \lim_{n\to\infty}V\bold{x}_n=\bold{y}. But, since V is invertible (it’s a Vandermonde matrix) and since V^{-1}:\mathbb{R}^{N+1}\to\mathbb{R}^{N+1} (being a linear operator on a finite dimensional vector space) is continuous we may conclude that

\displaystyle V^{-1}(\bold{y})=V^{-1}\left(\lim_{n\to\infty}V\bold{x}_n\right)=\lim_{n\to\infty}V^{-1}V\bold{x}_n=\lim_{n\to\infty}\bold{x}_n


and thus \displaystyle \lim_{n\to\infty}\bold{x}_n converges and thus each sequence converges. The fact then that \{p_n\}_{n\in\mathbb{N}} converges uniformly to a polynomial of degree less than or equal to N then follows from the first part of the proof. \blacksquare

Remark: A simpler proof for the second part would be to recall that all norms on finite dimensional vector spaces induce the same topology. In particular note that \mathbb{P}_N[\{x_0,\cdots,x_N\}] with the infinity norm is a finite dimensional vector space in which \{p_n\}_{n\in\mathbb{N}} converges. That said, \displaystyle \left\|\sum_{k=0}^{n}\beta_k x^k\right\|=\sum_{k=0}^{N}|\beta_k| is a norm on \mathbb{P}_N[\{x_0,\cdots,x_N\}] and thus by prior comment (since both norms induce the same topology) it follows that \{p_n\}_{n\in\mathbb{N}} converges in this norm. But, this clearly implies that each \{\alpha_{n,k}\}_{n\in\mathbb{N}} converges.


1. Apostol, Tom M. Mathematical Analysis. Reading, MA: Addison-Wesley Pub., 1974. Print.


February 9, 2011 - Posted by | Analysis, Fun Problems | , ,

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