Let $\bar{B}$ be the closed unit ball in $\mathbb{R}^n$, $C^k\left(\bar{B}\right)$ the Banach space of all real function defined on $\bar{B}$ with continuous derivatives up to order $k$, with norm
$$\Vert f \Vert = \sum_{h\le k} \Vert \partial_{i_1}\dots \partial_{i_h}f\Vert_\infty$$
Are polynomials dense in $C^k\left(\bar{B}\right)$?
Can you give me references about that subject?
Thanks a lot...
Are polynomials dense in $C^k\left(\bar{B}\right)$?
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functional-analysis
1 Answers
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By Stone-Weierstrass, the polynomials are dense in $C(\overline{B})$. Do the rest by induction: if polynomials are dense in $C^{k-1}$, write a function in $C^k$ in terms of integrals of its partial derivatives, and approximate those partial derivatives by polynomials...
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0Sorry for being dense, but isn't $C^k(\bar{B}) \subset C(\bar{B})$ and therefore a subset of the former that is dense in the latter automatically dense in the former? Edit: Oh sorry: $C^k$ is equipped with a different norm than C... – 2011-05-19
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0I'm almost a decade late, but on the off chance that someone reads this, can they explain how this works? Take the case $n=2, k=1$. It seems to me that this answer is suggesting we approximate $\nabla f$ by a polynomial vector field $G$ using the Stone-Weierstrass theorem, and then approximate $f$ by integrating $G$. But in order for this to work, don't we need $G$ to be conservative? Is it obvious how we can guarantee this? Again, sorry for the necromancy. – 2018-06-24
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0That's not what I meant. In the case $n=2$, $k=1$ you can write $f(x,y) = f(0,0) + \int_0^x f_1(s,0)\; ds + \int_0^y f_2(x,t)\; dt$ (where $f_1$ and $f_2$ are the partial derivatives). – 2018-06-24
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0Now is the idea to approximate $f_1$ and $f_2$ by polynomials $p_1$ and $p_2$? Then set $p(x,y) = f(0,0) + \int\limits_0^x f_1(s,0) ds + \int\limits_0^y f_2(x,t) dt$. Now ${\partial p \over \partial x}(x,y) = p_1(x,0) + \int\limits_0^y {\partial p_2 \over \partial x} p_2(x,y)$. Why is this close to $f_1(x,y)$? Thanks for the feedback by the way! – 2018-06-25
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0The integrands in the definition of $p$ should be $p_1$ and $p_2$. – 2018-06-25