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So here's the question I'm trying to answer:

Suppose $p_n(x) = \sum_{k=1}^N a_k^{(n)} x^k$ is a sequence of polynomials such that $p_n \to f$ uniformly over $[0,1]$ for some function $f:[0,1] \to \mathbb{R}$. Prove that $f$ must itself be an $N^\text{th}$ degree polynomial.

I've already shown that if each $a_k^{(n)} \to a_k$, then $p_n(x) \to p(x) = \sum_{k=1}^N a_k x^k$ uniformly (earlier part of the problem). I'm thinking that there's some way to show that if $p_n \to f$, then $a_k^{(n)}$ converges for each $k$. This certainly works for $k = 0$, since we can guarantee that the sequence $a_k^{(n)} = p_n(0)$ is Cauchy. I've gotten stuck in trying to extend this to other coefficients; I'm thinking there's some trick involving subtracting the $a_0^{(n)}$ off and dividing by $x$, maybe some fancy induction along those lines.

Other potentially helpful thoughts: we can guarantee that $f$ is continuous since it is the uniform limit of continuous functions. Remember also that we have a compact domain, so that all of these functions are bounded and achieve their max/min.

Any comments, hints, or answers are very much appreciated.

Michael Hardy
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Ben Grossmann
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    One lazy way is to exploit the one-one correspondence between a polynomial $f$ and the tuple of its values $(f(x_0), \ldots, f(x_{N}))$, where $(x_i)_{i=0}^N$ are $N+1$ distinct points, chosen arbitrarily. It is also useful to keep in mind that when all of $f(x_i)$ are very small, all the other values $f(x)$ are also small. – Dan Shved Feb 14 '14 at 05:38
  • @DanShved I'm not sure I understand what you're going for here. For one, I'm not given that $f$ is a polynomial; that's what I'm trying to prove (or part of it, anyway). – Ben Grossmann Feb 14 '14 at 05:50
  • Well, by subtracting a suitable polynomial from each of your functions, you can safely assume that $f(x_i)=0$ for each $i=0,\ldots,N$. So $p_n(x_i)$ will get small when $n$ is large. Then you can use what I said in my comment above to prove that $f(x)=0$ for all $x$. It will follow that the original $f$ (before we adjusted it by subtracting a polynomial) was a polynomial. – Dan Shved Feb 14 '14 at 08:46
  • Or you can develop further your own idea with the coefficients and combine it with mine just a little bit. If $p(x) = a_0 + a_1 x + \ldots + a_N x^N$, then, as you've noticed, $a_0 = p(0)$, so if $(p_n)$ converges uniformly, then $(a_0^{(n)})$ also converges. To show the same for $a_k$ when $k>0$, you can use the fact that $a_k$ is a linear combination of values $p(x_0), \ldots, p(x_N)$ with some fixed coefficients. – Dan Shved Feb 14 '14 at 08:55
  • @DanShved I understand now. Somehow, I was making this more complicated than it needed to be. Thank you. – Ben Grossmann Feb 14 '14 at 17:03
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    Alternatively, you can use that the space of polynomials of degree $\leq N$ is finite-dimensional, hence closed in $C[0,1]$. – Luiz Cordeiro Feb 14 '14 at 21:58
  • @LuizCordeiro is correct -- this is probably the quickest way about it. In general though the polynomial may be of a degree strictly smaller than $N$. For example, the sequence $p_n(t)=\frac{1}{n}t^N$ converges uniformly to $0$ on $[0,1]$. – Zack Cramer Apr 23 '16 at 18:33

2 Answers2

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A proof that for my criterion is elegant is using facts of functional analysis:

Let $P_N$ the set of all polynomials of degrre $N$, this is a vector space and isometrically isomorphic to $\mathbb{R}^{N+1}$ via $$\begin{array}{rcl}\Phi: P_{N} &\rightarrow & \mathbb{R}^{N+1}\\ a_0+a_1 x +\cdots+a_N x^N & \mapsto & (a_0,\ldots,a_N).\end{array}$$ Given any norm $\left\|\:\:\right\|$ in $\mathbb{R}^{N+1}$ we define the norm in $P_{N}$ for any $p(x)=a_0+a_1 x+\cdots+a_N x^N$ as $$\left\|p\right\|:=\left\|(a_0,\ldots,a_N)\right\|.$$ Since $\mathbb{R}^{N+1}$ is complete, then $P_N$ is complete with this norm, therefore $(P_N,\left\|\:\:\right\|)$ is a Banach space, but we know that all norms in a finite-dimensional space are equivalent, therefore, $P_N$ is Banach space with any norm, in particular, with the norm $$\left\|p \right\|_{\sup}:=\sup_{x\in[0,1]}\left|a_0+a_1 x+\cdots+a_N x^N \right|.$$

From the real analysis courses we know that if $\left\{p_n\right\}$ converges uniformly then it does so in the norm $\left\|\:\right\|_{\sup}$, therefore, $\left\{p_n\right\}$ is a Cauchy sequence in the Banach space $(P_{N}, \left\|\:\right\|_{\sup})$, which allows us to conclude that $f\in P_N$.

Michael Hardy
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Diego Fonseca
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You may start by sampling the points $0/n, 1/n, ..., (n-1)/n$, and make a vector $$\left( \begin{array}{c} p_m(0) - f(0)\\p_m\left(1/N\right) - f(1/N)\\ \vdots \\ p_m\left(N/N\right) - f( N/N) \end{array} \right) = \left(\begin{array}{cccc} 1& 0& \cdots& 0\\ 1& (1/N)& \cdots & (1/N)^N\\ \vdots& \vdots& \ddots& \vdots\\ 1& N/N& \cdots& ( N/N )^N \end{array}\right)\left( \begin{array}{c}a_0^{(m)}\\a_1^{(m)}\\ \vdots\\ a_N^{(m)} \end{array}\right) - \left( \begin{array}{c}f(0)\\f(1/N)\\ \vdots\\ f(N/N) \end{array}\right)$$

Let us denote this matrix by $V$, the polynomial vector $\vec{p}_m$, the vector with $f$ as $\vec{f}$, and the coefficient vector $\vec{a}_m$. The matrix is called a Vandermonde matrix, and it is invertible. Thus $$\vec{a}_m = V^{-1} (\vec{p}_m - \vec{f}) + V^{-1}\vec{f}$$ and we know that $\vec{p}_m - \vec{f} \to \vec{0}$ as $m\to \infty$. This means that $\vec{a}_m \to V^{-1} \vec{f}$, since every linear operator on a finite dimensional vector space is continuous.

Now we know that the coefficients converge to some vector $\vec{a} = (a_0, a_1, ..., a_N)^T$, and we can demonstrate that $p_m(x) \to p(x) = a_N x^N + \cdots + a_0$.

Joel
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    I'll review how I actually solved the problem, but I'm fairly certain it was similar to this. Presumably, since $N$ is the degree of the polynomial, we should have $N+1$ points, though. Thanks for the input! – Ben Grossmann Jun 30 '14 at 15:52
  • Absolutely. There is always something I miss. – Joel Jun 30 '14 at 16:58