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I am asked to show that every cont. function from a manifold M to $\mathbb{R}$ can be approximated by smooth functions.

Try: let f be a map from M to the reals(R). Let ${s_{i}, U_{i}}$ be our atlas. Then $f(p)$ can be rewritten as $fs_{i}^{-1}$ at $s_{i}(p)$. Now, since $f$ is cont. and $s_{i}$ is a chart, their composition is cont. Now, we have $f$ represented by all these cont. maps according to the appropriate charts. But notice that $fs_{i}^{-1}$ is a map from the reals to the reals, and hence we invoke the Wierstrass approximation theorem and say there exists a polynomial $P(x)_{i}$, such that the supremum of the distance between $P(x)_{i}$ and $fs_{i}^{-1}$ is less than arbitrary $\epsilon_{i}.$

Hence, a candidate for the desired function is $\{P(x)_{i} : x = s_{i}(p)\}$. I hope this makes, its the collection of all these approximating polynomials, in accordance with the charts. Now the issue is showing that this map is smooth. I am assuming the coordinate transformations are supposed to come in here, but i am at a loss.

Thanks in advance.

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  • $\begingroup$ In what sense do you need to approximate? (e.g. locally uniformly?) Approximating on a collection of charts is a good idea; but you will need to choose a subcollection of charts and use a partition of unity. Also, you said you have a function on the reals, when in fact it is on $\mathbb R^n$ - does your version of the Weierstrass theorem apply to this case? $\endgroup$ – Anthony Carapetis Nov 4 '13 at 0:42
  • $\begingroup$ doesn't necessarily need to be locally uniform. The only requirement is that its smooth. $\endgroup$ – masszz Nov 4 '13 at 0:45
  • $\begingroup$ oh yes, this version of Weierstrass theorem does apply. Could you elaborate on the need for the bump functions and taking a sub collection of charts please $\endgroup$ – masszz Nov 4 '13 at 0:46
  • $\begingroup$ but you need to define what you mean by "approximate" - what disqualifies the zero function from "approximating" your continuous function? Usually an approximation by smooth functions means you can find smooth functions that converge to your target function - you need to specify the norm/mode of convergence. $\endgroup$ – Anthony Carapetis Nov 4 '13 at 0:47
  • $\begingroup$ supremum|f(x)-P(x)| < $\epsilon$ for all x on the manifold. $\endgroup$ – masszz Nov 4 '13 at 0:50
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Let $(U_i, s_i)$ be an atlas and $\eta_i$ a smooth partition of unity subordinate to $U_i$: this means

  • each $\eta_i$ is zero outside $U_i$
  • each $x\in M$ has a neighbourhood on which all but finitely many $\eta_i$ are zero
  • $\sum_i \eta_i(x) = 1$ for every $x$.

Now, on each $U_i$ we can approximate in the chart as you suggested - let $p_i$ be a smooth function on $s_i(U_i)$ approximating $f \circ s_i^{-1}$ to within $\epsilon$. We can use $\eta_i$ to patch all of these local approximations together into a global approximation: define $$ \tilde f (x) = \sum_i p_i (s_i(x))\eta_i(x).$$

This is smooth because it is locally the finite sum of smooth functions - if you fix an $x\in M$ and look at $\tilde f$ on the nice neighbourhood provided by the partition of unity, then there are only finitely many non-zero terms in the sum, and each is the product of a smooth function $p_i \circ s_i$ with a smooth function $\eta_i$.

As to the approximation, we have $$\begin{align} |f(x) - \tilde f(x)| &= \left|\sum_i \eta_i(x)\left(f(x) - p_i(s_i(x))\right)\right|\\ &\le \sum_i \eta_i(x) \left| f(x) - p_i(s_i(x))\right| \end{align}$$ where we used $\sum_i \eta_i = 1$ to move $f$ inside the sum. Now, the uniform approximation $|f \circ s_i^{-1} - p_i| < \epsilon$ on the image of the chart $s_i$ means that $|f(x) - p_i(s_i(x))| < \epsilon$ for every $x \in U_i$, and thus we have the bound $$|f(x) - \tilde f(x)| < \sum_i \eta_i(x) \epsilon = \epsilon.$$

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  • $\begingroup$ @sergey: Note that the "subcollection of charts" I referred to before is in a sense implicitly contained in the local finiteness condition on the $\eta_i$ - if you use a complete atlas then a vast majority of the $\eta_i$ will in fact be zero. $\endgroup$ – Anthony Carapetis Nov 4 '13 at 3:09

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