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If $f \in\operatorname{Lip}_K[a, b]$, show that $f$ can be uniformly approximated by polynomials in $\operatorname{Lip}_K[ a, b]$.

Context: $f \in \operatorname{Lip}_K[a,b]$ then it is Lipschitz with constant $K$. The text I am currently using is Real Analysis by Carothers. We have developed Stone-Weierstrass and have seen that $\operatorname{Lip}[a,b]=\cup\operatorname{Lip}_K[a,b]$ for $K\in\mathbb{N}$ is a subalgebra $C[a,b]$.

I was wondering if we could somehow adapt the proof of Weierstrass theorem or maybe I am missing some fact about polynomials on a interval that relates to Lipschitz.

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What do you mean by "uniformly in $LipK[a,b]$"? –  Davide Giraudo Nov 19 '12 at 9:08
    
So that P_n $\in$ LipK[a,b] converge uniformly to f on the interval [a,b]. Where P_n is the sequence of polynomials –  aawaldrop Nov 19 '12 at 17:46

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You have not mentioned which proof of Weirestrass Approximation Theorem you have studied. Let us look at the Weirestrass Approximation Theorem through Bernstein Polynomials.

Let $f\in C[0,1]$. For each $n\in\mathbb{N}$, let us define $B_n(f):[0,1]\rightarrow\mathbb{R}$ by $$ B_n(f)(x):=\sum_{k=0}^{n}\binom{n}{k}f\left(\frac{k}{n}\right)x^k(1-x)^{n-k},\text{ for all }x\in [0,1]. $$ Then, it can be showed that $B_n(f)$ converges uniformly to $f$ in $[0,1]$ and $B_n(f)\in \operatorname*{LipK}[0,1]$.

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Thanks this was exactly what I had been looking for. We did see a treatment with the Bernstein Polynomials but I did not consider that they would all also have the same Lipschitz constant of the function they were approximating. –  aawaldrop Nov 19 '12 at 17:59
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Also to see an elementary proof that $$ B_n \in LipK[0,1] $$ see "Lipschitz constants for the Bernstein polynomials of a Lipschitz continuous functions" by Brown, Elliot, and Paget in the Journal of Approximation Theory 49, 196-199 (1987). –  aawaldrop Nov 19 '12 at 19:19
    
@saugata: I understand that a Bernstein polynomial is a polynomial so you have shown that a sequence of polynomials converges uniformly to $f$, but how do you know that $f$ is Lipschitz? (Is it something about $C[0,1]$ is continuous on compact, which means all functions in the space are uniformly continuous? I know Lipschitz $\implies$ uniform continuity, but I don't see how you are getting the Lipschitz condition for $f$.) –  Sujaan Kunalan Dec 3 at 8:29

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