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Let $f$ be a continuous function vanishing at $+\infty$ and $-\infty$. Construct a sequence of functions in the space of infinitely times differentiable functions on $\Bbb R$ with compact support so that this sequence converges uniformly to $f$.

Thanks for any help.

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There is a standard trick that you may need to use for this problem. Do you know how to construct a compactly supported smooth function $\varphi(x)$ with $0 \le \varphi(x) \le 1$, such that $\varphi(x) = 1$ when $x \in [-1,1]$ and $\varphi(x) = 0$ when $|x| > 1 + \epsilon$ for arbitrarily small $\epsilon$? – Christopher A. Wong Nov 26 '12 at 19:46
@Wong No,can you please elaborate and explain? – Ester Nov 26 '12 at 19:52
up vote 3 down vote accepted

There are many ways to do this, but there is one really concrete way of constructing smooth approximations to a function. This way is called mollification. You can see the Wikipedia article on the subject for more details. I will summarize the most important points here:

  1. We need a smooth function called a mollifier. First, it is compactly supported, smooth, positive, and has integral over $\mathbb{R}$ equal to $1$; let us label this function $\varphi(x)$. We also want it to be the case that the rescaling $\varphi_{\epsilon}(x): \epsilon^{-1} \varphi(x/\epsilon)$ approaches the Dirac delta function $\delta(x)$ as $\epsilon \rightarrow 0$. This last condition gives us the intuition for what a mollifier is: it is a smooth approximation to the Dirac delta function. A standard, explicit mollifier (also the one used on the Wikipedia page) is given by first taking $$ g(x) = \begin{cases} e^{1/(1 - |x|^2)} & |x| < 1 \\ 0 & |x | \ge 1 \end{cases}$$ and then setting $\varphi(x) = g(x)/(\int g \, dx)$ to normalize.

  2. Let us recall what the Dirac delta function does: it is really a distribution, not a function, and it obeys the property that $\delta \ast f = f$. This motivates the following strategy: we will instead use the rescaled mollifier $\varphi_{\epsilon}$ as a surrogate for $\delta$, and we will look at the convolution $\varphi_{\epsilon} \ast f$. Of course, then $\varphi_{\epsilon} \ast f \neq f$, but we will get something very close. Furthermore, the function $\varphi_{\epsilon} \ast f$ is smooth because $\varphi_{\epsilon}$ is smooth. This gives us a way to smoothly approximate any function $f$, by considering its mollification $\varphi_{\epsilon} \ast f$, and taking $\epsilon > 0$ small.

  3. We are looking for a sequence of approximations to $f$ that are compactly supported, so taking mollifications of $f$ is not sufficient. Instead, let us consider the functions $f_n(x) = f \cdot \chi_{[-n,n]}$; these are just the function $f$, chopped off at the endpoints $|x| = n$. Now we can mollify these functions, and we can check that mollifying $f_n$ yields a smooth, compactly supported function, since $f_n$ and $\varphi_{\epsilon}$ are both compactly supported. Now, to get one whole sequence that simultaneously converges to $f$ and is compactly supported and smooth, we can also pick $\epsilon = 1/n$, so then we get the sequence of approximations $$ \{ \varphi_{1/n} \ast f_n \}$$

Some remarks: If you haven't seen this approach, it would be, in my opinion, quite challenging to think of it. But I cannot think of any less general way of doing it, for your problem. Also, notice that while I've given you the full method, I've not proven anything, so if what you're doing is a homework exercise, you'll need to prove many things, namely that (1) a mollification is in fact smooth, and (2) these approximations will converge uniformly.

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@Wong Can you suggest any simpler method? I mean is there any other way to prove that the uniform closure of inf. diff. comp. supp. func. is the space of func. vanishing at +-infinity? – Ester Nov 26 '12 at 21:48
Yes, I think another possibility is that you can use a sequence of polynomials to approximate $f$ uniformly on $[-n,n]$, and then smoothly decay those polynomials to zero quickly for $|x| > n$. – Christopher A. Wong Nov 26 '12 at 22:01
@Wong Yes,I was trying exactly that,but cannot rigorously prove the result.Can you please help me? – Ester Nov 26 '12 at 22:06
You need to use the fact that polynomials uniformly converge to any continuous function on a fixed finite interval (this is the Weierstrass approximation theorem). Then, show that the "decaying" part outside the polynomials must have a small error, since $f(x)$ is small for large $x$. – Christopher A. Wong Nov 26 '12 at 22:12

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