Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $d\in\mathbb N$ and $$ P_d := \left\{p : K \to \mathbb{R} :\quad p(x) = \sum\limits_{i=1}^d p_i x^i\quad\text{ where }\quad \{p_i\}_{i=1}^d \subset \mathbb{R}\right\}, $$ the set of all polynomials of degree at most $d$, where $K$ is a compact set. We have to show that $P_d$ isn't dense in the space of all continuous functions $C(K,\mathbb{R})$.

Our teacher gave us a hint: Consider a continuous function $f$ which has $d + 1$ zeros and show that all polynomials which approximate $f$ at its zeros with accuracy $\varepsilon > 0$ are uniformly bounded (Lagrange interpolation). Use this observation to show that there are continuous functions $f$ which cannot be approximated by $P_d$ with given accuracy $\varepsilon > 0$.

I can't figure out how to use this hint to solve the problem. Please help me out. Thanks.

share|cite|improve this question
I think that you assume $K\subset \mathbb{R}$. Right? – Norbert Jul 3 '12 at 11:02
The statement as given is definitely wrong, as for $K = \emptyset$, $P_d = C(K, R)$. Also, "dense" in respect to what metric? – penartur Jul 3 '12 at 11:03
Can you (uniformly) approximate $x \mapsto x^2$ with polynomials of degree one? Essentially, your teacher's suggestion is this: a polynomial of degree $d+1$, with $d+1$ distinct roots, cannot be uniformly approximated by polynomials of degree (at most) $d$. – Siminore Jul 3 '12 at 11:03
@penartur I think we can assume $K \neq \emptyset$, since functions defined on the empty set are of little interest :-) – Siminore Jul 3 '12 at 11:04
We can see that $P_d$ is a finite dimensional subspace of a normed space, hence closed. Since it's in general strict (except maybe some trivial cases), it cannot be dense. – Davide Giraudo Jul 3 '12 at 11:10
up vote 1 down vote accepted

Assume $\{p_n\}_n$ is a bounded sequence in $P_d$. We can write $$ p_n(x)=a_n (x-x_{1,n})(x-x_{1,n})\cdots (x-x_{d,n}) $$ for suitable sequences $\{a_n\}$, $\{x_{i,n}\}_n$, $i=1,\ldots ,d$. Since $x \in K$, a compact set, up to subsequences we can assume $$ a_n \to a_\infty, \quad x_{i,n} \to x_{i,\infty} $$ as $n \to +\infty$. But then $\{p_n\}$ converges uniformly on $K$ to the polynomial $$ p_\infty (x) = a_\infty (x-x_{1,\infty})\cdots (x-x_{d,\infty}). $$ In particular, $\deg p_\infty \leq d$.

Remark. This is a rephrasing of Davide Giraudo's comment.

share|cite|improve this answer
How does {pn} converge uniformly to the polynomial p infinity(x)? – johnathan Jul 3 '12 at 12:13
And how does proving that deg p infinity is less or equal to d do the job? – johnathan Jul 3 '12 at 12:14
It is a finite product of uniformly convergent sequences. – Siminore Jul 3 '12 at 12:14
Essentially, this argument shows that you can only approximate polynomials of degree at most $d$. Clearly, in $C(K)$ there are elements that are not polynomials of degree $d$ (for instance there are polynomials of degree $d+1$). Just one remark: $P_d$ is closed under uniform convergence, whatever $K$ may be. Your problem becomes false if $K$ is a set of $d$ points. – Siminore Jul 3 '12 at 12:16

In an infinite dimensional normed space $X$, a finite dimensional subspace $F$ cannot be dense. Indeed, we can show that $F$ is necessarily closed, for example by induction on the dimension. Furthermore, $F$ is necessarily strict.

Apply this to $X:=C(K,\Bbb R)$ and $F:=P_d$, provided that $K$ is infinite, or at least contains strictly more than $d$ points.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.