Let ${P}$ denote the set of all polynomial with variable $x\in [0,1]$, I need to know what is the closure of ${P}$ in $C[0,1]$?
Well, Stone-Weierstrass theorem says: If $f\in C[0,1]$ then there exists a sequence of polynomials $p_n(x)$ which converges uniformly to $f$. So can I say that $closure{P}=C[0,1]$?
Closure of the set of all polynomial with variable $x\in [0,1]$
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real-analysis
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0If you mean closure with respect to the supremum norm, that's exactly the statement of Stone-Weierstrass theorem. – 2012-07-19
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0yes with supnorm – 2012-07-19
2 Answers
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Yes. The Stone-Weierstrass approximation theorem tells you that $P$ is dense in $C[0,1]$ with respect to $\|\cdot\|_\infty$ and a set $D$ is dense in a set $S$ if the closure of $D$ equals $S$. (by definition)
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0Talking about the definition of being dense. Can't we say that $[0,1)$ is dense in $[0,1)$? – 2012-07-19
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0@Ilya Yes of course, the entire space is always dense in itself. – 2012-07-19
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0@HenningMakholm Well in the given case the whole space is complete with respect to $\|\cdot\|_\infty$. But the definitions I've seen of "dense set" don't put any requirement on the space, in particular, they don't require the space to be complete. – 2012-07-19
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0@MattN: Sorry -- had gotten myself very confused there. – 2012-07-19
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0@HenningMakholm Np : ) I'm glad I'm not the only one to whom this happens. – 2012-07-19
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I thought Stone-Weierstrass theorem says $C[0,1]$ is the uniform closure of $P$? A sequence in $P$ may converge to a discontinuous function on $[0,1]$.
With respect to the norm defined the in answer above we can say it is the closure of $P$.