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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.

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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
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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
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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

2 Answers 2

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.

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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.

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