Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f:[a,b] \rightarrow \mathbb{R}$ be of class $C^n$. Let $ x_0, ..., x_m$ be different numbers from $[a,b]$.

Does for each $\varepsilon >0$ there exist a polynom $P$ such that $P^{(k)}(x_i)=f^{(k)}(x_i)$ for $i=0,...,m$, $k=0,...,n$ and $sup_{x \in [a,b]} |f(x)-P(x)|< \varepsilon$?

share|improve this question
1  
I have never found an adequate reference for what one would call a "differentiable Stone--Weierstrass theorem", but I would assume that such a thing exists. However, I do happen to know a theorem found in Real Algebraic Manifolds by J. Nash that say: Let $f: U \rightarrow \R^n$ be analytic, let $A\subseteq U$ be compact and bounded by an analytic manifold. Then there exists a sequence of polynomials $\{f_n\}$ so that for any $\alpha$: [ D^\alpha f_n \rightarrow D^\alpha f ] I don't know much about Fourier series, which appear in the proof. Maybe it works with $C^n$ in place of $C^\omega$. –  Malte Mar 13 '12 at 21:27
3  
@Malte: this is true. See mathoverflow.net/questions/85153/…. But I think that the OT wants something different: a uniform approximation of the function and exact value of the function and derivatives at the given points. It is not required that the derivative converges uniformly. –  Valerio Capraro Mar 13 '12 at 21:38
add comment

2 Answers 2

up vote 10 down vote accepted

The problem may be split into two independent and classical ones: the Hermite interpolation, and the Weierstrass approximation.

First, we want a polynomial $p\in \mathbb{R}[x]$ with given derivatives at some given nodes $x _ 0,\dots, x _ m $. This is an instance of the Hermite interpolation problem; yours has exactly one solution $p$ with $ \operatorname{deg}(p) < (m+1)(n+1) $ (the degree one would expect in terms of number of linear conditions). So, given your $f\in C^n$, you can find a polynomial with $p^{(j)}(x _ i)=f^{(j)}(x _ i)$ for all $0 \le i \le m$ and $0 \le j \le n$.

Second, as a consequence,
$$\frac{f(x)-p(x)}{\prod _ {i=0}^ m(x- x _ i)^n}$$ is (extends to) a continuous function on $[a,b]$ that vanishes on the points $x _ i$. By the Stone-Weierstrass approximation theorem there is a polynomial vanishing on the points $x _ i$ as well, whose uniform distance from that function on the interval $[a,b]$ is less than, say, $\epsilon (b-a)^{-n(m+1)}$. In other words, there is a polynomial $q\in \mathbb{R}[x]$ such that

$$\bigg\| \frac{f(x)-p(x)}{\prod _ {i=0}^ m(x- x _ i)^n} - q(x) \prod _ {i=0} ^ m(x- x _ i) \bigg\|_{\infty, [a,b]} < \epsilon (b-a)^{-n(m+1)}\ , $$

therefore the polynomial $P(x):= p(x)+ q(x) \prod _ {i=0} ^ m(x- x _ i) ^{n+1}$ fullfills the requirements, for $P^{(j)}(x _ i)=p^{(j)}(x _ i)=f^{(j)}(x _ i)$ for all $0 \le i \le m$ and $0 \le j \le n$, and $$\|f- P\|_{\infty,[a,b]} < \epsilon\ .$$

btw. Incidentally, some time ago I happen to notice that one can find the solution of the Hermite interpolation problem as an application of the Chinese Remainder Theorem in the ring of polynomials, and wrote here the details.

edit. As to why The set $A$ of all polynomial functions on $[a,b]$ that vanish at given points $x_0,\dots, x_m$ is dense in all continuous functions on $[a,b]$ that vanish in $x_0,\dots, x_r$. One way, a bit abstract but quite immediate is, to see it as a corollary of the Stone-Weierstrass theorem (A separating closed algebra of real valued functions on a compact space $X$ is either $C(X)$ or a maximal ideal $M_x\subset C(X)$, the set of all functions vanishing at $x$). Consider $X=$ the topological quotient of $[a,b]$ obtained identifying all points $x_i$ to a point $\xi$. All functions in $A$ factor through to the quotient map, and define a closed separating algebra of continuous functions on $X$ that vanish on the identified point $\xi$. Thus, this algebra contains all continuous functions on $X$ that vanish on $\xi$, which is the thesis read on the quotient.

Note that the same construction holds in general, and provides a characterization of all closed algebras $A$ of continuous functions on a compact space $X$: identifying all points that are not distinguished by the functions of $A$ (that is, under the equivalence relation $x R_A y$ iff $f(x)=f(y)$ for all $f\in A$) one gets a Hausdorff compact quotient space (whether or not $X$ is Hausdorff), and the quotient map $\pi: X\to X/{R _ A}$ induces an isometric isomorphism of algebras $f\mapsto f\circ \pi$ of either $C(X/{R_A})$ or a maximal ideal of it onto $A$; conversely, any Hausdorff quotient of $X$ produces a closed sub-algebra of $C(X)$ this way.

Another way to see it is as a corollary of the classic Weierstrass theorem: Consider $P$ as in your comment below, then add a perturbation $L$ that makes $P+L$ vanish on the points $\{x_i\}$; this has been clearly explained in Ilya Bogdanov's answer. Here you don't have derivatives and $L$ is just a Lagrange interpolation polynomial, which is small in the uniform norm because it is small on the points $\{x_i\}$.

share|improve this answer
    
Very thanks. I have one question. You used the following fact: if a continuous function $g:[a,b]\rightarrow R$ vanishes at points $x_0,...,x_m$ then for every $\delta >0$ there exists a polynomial $Q$ which vanishes at $x_0,...,x_m$ such that $\|g(x)-Q(x)\|_{sup} <\delta$. It is easy for $m=0$, because by Weierstrass threorem we find polynomial $P$ such that $\|f-P\| <\frac{\delta}{2}$ and polynomial $Q:=P-P(x_0)$ will be good. How to prove its for $m\geq 1$? –  arc Mar 14 '12 at 13:59
    
I've edit and added a few lines on this point. –  Pietro Majer Mar 14 '12 at 17:02
add comment

For the sake of simplicity, let us assume that $[a,b]=[0,1]$. $C^k$ always means $C^k[0,1]$. We will even approximate $f$ in $C^n$-norm satisfying your additional condition.

  1. As was mentioned in the comments, you can easily approximate $f$ together with all its derivatives up to $n$th uniformly by a polynomial. In fact, it is enough to approximate $f^{(n)}$ with an adequate accuracy: if $||f'-P'||_C<\varepsilon$ and $f(0)=P(0),$ then $||f-P||_C<\varepsilon.$

  2. Now take the polynomials $Q_{ik}(x)$ such that $Q_{ik}^{(d)}(x_j)=0$ for all $d=0,\dots,n$ and $j=0,\dots,m$ except that $Q_{ik}^{(k)}(x_i)=1$. Such polynomials are easy to construct: for instance, one may take $$ Q_{ik}(x)=c_{ik}(x-x_i)^k\prod_{j\neq i}\left((x-x_i)^{n+1}-(x_j-x_i)^{n+1}\right)^{n+1}\;\; $$ for a suitable constant $c_{ik}.$ Let $M=\max_{i,k}||Q_{ik}||_{C^n}.$ Then, let the approximation in the previous paragraph be $\delta$-accurate with $\delta=\varepsilon/(2M(m+1)(n+1)).$ To correct the values of the polynomial and its derivatives at $x_i,$ it is enough to add the polynomials $Q_{ik}$ multiplied by the coefficients with absolute values $\leq\delta,$ hence the total error will be not more that $\delta+(m+1)(n+1)M\delta<\varepsilon.$

share|improve this answer
    
nice ! –  Pietro Majer Mar 14 '12 at 5:49
add comment

Your Answer

 
discard

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.