# Is the set of polynomials of degree less than or equal to $n$ closed?

This question is in relation to the space $C(I)$, $I = [a, b]$. Define $P_n =\{ a_0+\dots+a_nx^n \mid a_i \in \mathbb{R}\}$ (any or all $a_i$ could be zero); clearly $P_n \subset C(I)$. The norm I'm using is $\lVert f\rVert_I = \sup_I |f(x)|$. Is $P_n$ closed under $\lVert\cdot\rVert_I$?

I am almost sure the answer is "yes", but I can't seem to prove it. My first instinct was to biject $P_n$ to $\mathbb{R}^{n+1}$, using coefficients as coordinates, and prove that sequences of degree-$n$ polynomials converge to degree-$n$ polynomials, but I can't prove that the metric $\lVert\cdot\rVert_I$ is equivalent to the standard metric on $\mathbb{R}^{n+1}$. Next, intuitively given a function $f \in C(I), \notin P_n$ I should be able to find some $\epsilon > 0$ such that there are no low-degree polynomials "nearby", then that function was not a limit point of $P_n$ so $P_n$ is closed. Again I have no idea how to prove this. What should I do?

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A general result is that a finite-dimensional subspace of a normed space is closed. Maybe here a quite simple argument can be given. –  Davide Giraudo Nov 12 '12 at 23:04
@ChrisEagle fixed. –  Julien Clancy Nov 13 '12 at 0:27

Generally, any finite dimensional subspace of a topological vector space is closed. However, in this case one can use properties of polynomials to show this directly.

Choose distinct points $x_0,...,x_n \in I$. Let $V$ be the $(n+1)\times (n+1)$ Vandermonde matrix $V=\pmatrix{1 &x_0 & \cdots & x_0^n\\ \vdots & \vdots & & \vdots \\1 &x_n & \cdots & x_n^n }$. A tedious calculation shows that $\det V \neq 0$ iff the $x_k$ are distinct. If $p$ is a polynomial of the form $p(x) = a_0+a_1x+...+a_nx^n$, then we see that (taking the coefficients as a element $a \in \mathbb{R}^{n+1}$) $\pmatrix{p(x_0) \\ \vdots \\ p(x_n)} = V a$, or equivalently $a = V^{-1} \pmatrix{p(x_0) \\ \vdots \\ p(x_n)}$.

Now define the operator $N: C(I) \to C(I)$ as follows: Let $f \in C(I)$, and define $a(f) = V^{-1} \pmatrix{f(x_0) \\ \vdots \\ f(x_n)}$, and let $N(f)(x) = f(x) - \sum_{k=0}^n [a(f)]_k x^k$. Noting that pointwise evaluation is continuous with respect to the $\sup$ norm, we see that $N$ is continuous, and furthermore, $N(f) = 0$ iff $f$ is a polynomial of degree $n$ or less. Hence $P_n = N^{-1}\{0\}$, the inverse image of a closed set, hence $P_n$ is closed.

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Very clever. Coincidentally we calculated the Vandermonde determinant recently in class. –  Julien Clancy Nov 13 '12 at 0:02

It suffices to show that any sequence $(p_k)_{k\in\mathbb{N}}\subset P_n$ that converges with respect to $\|\cdot\|_I$ converges to a polynomial $p(x)\in P_n$. So, let $(p_k)$ be a sequence; for each $k$,

$$p_k(x)=a_0^{(k)}+a_1^{(k)}x+\ldots+a_n^{(k)}x^n$$

Claim: $p_k(x)$ converges (with respect to your norm) to the polynomial $p(x)=a_0+\ldots+a_nx^n$ where

$$a_i=\lim_{k\rightarrow\infty}a_i^{(k)}$$

I'll let you flush out the details.

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In the details you flush out, there is the fact that the sequences $\{a_i^{(k)}\}_i$ are convergent. –  Davide Giraudo Nov 12 '12 at 23:05
@DavideGiraudo This is where I got stuck on my first attempt. Can you give me a hint as to how to do this? –  Julien Clancy Nov 13 '12 at 0:00