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Let $$C^{\infty}_b(\mathbb{R}^{n}) = \{f:\mathbb{R}^{n} \to \mathbb{R} \mid f\text{ is smooth and } \Vert f \Vert_{\infty} < \infty\},$$ $$C_{b}(\mathbb{R}^{n}) = \{f:\mathbb{R}^{n} \to \mathbb{R}\mid f\text{ is continuous and bounded}\}.$$

I want to prove that $C^{\infty}_b(\mathbb{R}^{n})$ is dense in $C_{b}(\mathbb{R}^{n})$.

The proofs that I know are totally constructive, I mean: the proofs use explicit function defined by parts.

I want to know if is possible to prove this result using generic functions (such a generic step function) or, at least, a strong theorem.

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    $\begingroup$ In what metric? $\endgroup$ – zhw. Feb 26 at 0:45
  • $\begingroup$ @zhw. $\Vert f \Vert_{\infty} = \sup\{|f(x)|\mid x \in \mathbb{R}^{n}\}$ $\endgroup$ – Lucas Corrêa Feb 26 at 0:52
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    $\begingroup$ The standard approach I know is to use partitions of unity by functions that are infinitely many times differentiable. $\endgroup$ – Will M. Feb 26 at 2:31
  • $\begingroup$ So $C_b$ consists of continuous bounded functions? $\endgroup$ – zhw. Feb 26 at 18:42
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    $\begingroup$ You should edit the question to make all the hypotheses clear. For example, $C^\infty(\mathbb R^n )$ has a well known meaning different from what you have; maybe write that as $C^\infty _b(\mathbb R^n )$ $\endgroup$ – zhw. Feb 28 at 17:19
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There's a stronger result here: Suppose $f$ is continuous on $\mathbb R^n.$ Let $\epsilon>0.$ Then there exists $g\in C^\infty(\mathbb R^n )$ such that $|f(x)-g(x)|<\epsilon$ for all $x\in \mathbb R^n.$

Proof for $n=1:$ Assume first that $f(k)=0$ for all $k\in \mathbb Z.$ Let $I_k=[k,k+1].$ Then by Weierstrass, there is a polynomial $p_k$ such that $|f-p_k|<\epsilon/2$ on $I_k.$ Because $f=0$ at the end points, there is $\delta_k>0$ such that $|f|+|p_k|<\epsilon$ on $[k,k+\delta_k] \cup [(k+1)-\delta_k,k+1].$

Now find $g_k\in C^\infty(\mathbb R)$ such that i) the support of $g_k$ is $I_k$, ii) $0\le g_k\le 1$ everywhere, and $g_k=1$ on $[k+\delta_k,(k+1)-\delta_k].$ Verify that $|f-g_kp_k|<\epsilon$ on $I_k.$

The nice thing about $g_kp_k$ is that all derivatives of it equal $0$ at the endpoints of $I_k.$ Thus the $g_kp_k$ paste nicely together to form a function $g=\sum_{k\in \mathbb Z}g_kp_k$ in $C^\infty(\mathbb R).$ We then have $|f-g|<\epsilon$ everywhere. This is the desired conclusion for such an $f.$

To get the full result, for each $k$ define a bump function $b_k\in C^\infty(\mathbb R )$ whose support is a small interval containing $k,$ with $b_k(k) = f(k).$ The function $b=\sum_{k\in \mathbb Z}b_k$ is then in $C^\infty(\mathbb R ),$ and $f-b=0$ at all integers. We can then apply the above to $f-b$ and this gives the result for $f.$

Added later If $n>1$ we can do something similar. For $k=1,2,\dots$ let $A_k$ be the annular region $\{k-1\le |x|\le k\},$ and let $S_k$ be the sphere $\{|x|= k\}.$ At first we assume $|f|<\epsilon$ on each $S_k.$ Then the ideas for the $n=1$ proof work pretty much the same, with the $A_k$ replacing the $I_k$ and the $S_k$ replacing the endpoints. Also we would use Stone-Weierstrass rather than just Weierstrass.

To get to the $|f|<\epsilon$ assumption, we choose "annular" bump functions $b_k$ supported very close to $S_k,$ with $b_k=1$ on $S_k.$ By SW, there are polynomials $p_k$ such that $|f-p_k|<\epsilon$ on $S_k.$ We then have $|f-\sum_k b_kp_k|<\epsilon$ on each $S_j$ as desired.

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  • $\begingroup$ That is a nice answer! I could not see, but probably is simple the same proof works for $n > 1$? $\endgroup$ – Lucas Corrêa Mar 2 at 22:12
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    $\begingroup$ I added a bit on the $n>1$ case. $\endgroup$ – zhw. Mar 3 at 16:52
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If $f$ is uniformly continuous then let $\phi \in C^\infty_c, \int \phi=1,\phi_k(x) =k^n \phi(kx)$ you'll have $f \ast \phi_k \to f$ uniformly ($*$ for convolution).

If $f$ is only continuous then split it in $f=\sum_{m\in \mathbb{Z}^n} f_m$ where $ f_m = f \prod_{j=1}^n (1-|x_j-m_j|) 1_{x_j-m_j\in [-1,1]}$ and look at $\sum_m f_m \ast \phi_{e_{m,k}}$ where $e_{m,k}$ is large enough such that $ \sup_{|x-y| < 1/e_{m,k}} |f_m(x)-f_m(y)| < 1/k$

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  • $\begingroup$ I don't understood one point. Probably is easy, but I cannot justify why $f_{m} \ast \phi_{e_{m,k}}$ is $C^{\infty}$ $\endgroup$ – Lucas Corrêa Mar 4 at 0:13
  • $\begingroup$ @LucasCorrêa you can put the derivatives on the $\phi_{e_{m,k}}$ $\endgroup$ – mathworker21 Mar 4 at 16:12

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