Is it possible to uniformly approximate every continuous function $f: \mathbf{R} \rightarrow \mathbf{R}$ by smooth functions?

In other words, is it true that for each continuous function $f: \mathbf{R} \rightarrow \mathbf{R}$ there exists a sequence $(f_n)$ of smooth functions $f_n: \mathbf{R} \rightarrow \mathbf{R}$ such that $f_n \rightarrow f$, as $n\rightarrow \infty$, uniformly on $\mathbf{R}$?


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    $\begingroup$ If $f$ is bounded you can convolve with an approximate identity and that will converge uniformly to your original function. $\endgroup$ – Jonas Teuwen Sep 5 '11 at 18:14

Yes. Approximate by polynomials $p_k(x)$ on each interval $[k,k+2]$, and put them together using a smooth partition of unity: if $\phi(x)$ is a smooth function with $\phi(x) = 0$ for $x \le 0$, $\phi(x) = 1$ for $x \ge 1$, and $0 \le \phi(x) \le 1$ for $0 \le x \le 1$, take $g(x) = \phi(x-k) p_k(x) + (1 - \phi(x-k)) p_{k-1}(x)$ for $k \le x \le k+1$.

  • $\begingroup$ It is a very nice answer. Thanks. $\endgroup$ – Richard Sep 5 '11 at 19:18

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