Context: I have proved Weierstrass' theorem (polynomials are dense in $C[a,b]$) in two ways: one using Bernstein polynomials, and one using convolutions. You can also use Stone-Weierstrass theorem, however I do not quote that result since I have not proved it.

My question, however, is of a different nature: I am not allowed to use Stone-Weierstrass, but am asked to prove the following:

Given a continuous function $f \in C(\mathbb R)$ such that $\lim_{|x| \to \infty} f(x) = 0$, and given $\epsilon>0$, there exists a polynomial $p$ such that $|f(x) - p(x)e^{-|x|}| < \epsilon$ for all $x \in \mathbb R$. Which is to say, the set $\{ p(x)e^{-|x|} : p \in \mathscr P\}$ is dense in the set of continuous functions on $\mathbb R$ decaying to zero (in the supremum norm).

You cannot try to approximate $f(x)e^{|x|}$ by polynomials, because that will give you a right side bound dependent on $x$, which is not the case in uniform convergence. Furthermore, working over $\mathbb R$, breaking it into compact intervals and applying Weierstrass is getting me nowhere.

Hence, I need help on this question. I request an incomplete "fill-in-the-blank" answer, if that's possible!

  • $\begingroup$ You mean $p(x)e^{-|x|}$ correct? $\endgroup$ – s.harp Feb 21 '17 at 12:05
  • $\begingroup$ @s.harp Oh, but of course! Thank you. $\endgroup$ – Teresa Lisbon Feb 21 '17 at 12:06
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    $\begingroup$ Has anyone suggested rephrasing the question to the case where we replace R with the unit interval. It is identifying R with open unit interval, and the condition at infinities corresponds to having 0 at both ends, so any function defined as above on R corresponds to a function defined on closed interval with value equal to 0 on both end-points. $\endgroup$ – Behnam Esmayli Mar 12 '17 at 5:50
  • $\begingroup$ @Behnam I thought of that idea, and while it did not give me an immediate response, pushing the measure forward along this correspondence (or homeomorphism) that you have suggested gives a new measure, which is on a bounded set, making things easier. You can see my answer below for more details. Thank you for your input! $\endgroup$ – Teresa Lisbon Mar 12 '17 at 5:53

Consider using the power series for $e^{|x|} = \sum_{n\ge0} \frac{|x|^n}{n!}$ and use the fact that partial sums are polynomials and they converge uniformly in compact sets. Such a partial sum combined with $e^{-|x|}$ will be almost $1$ for a sufficiently large set.

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  • $\begingroup$ I apologize, but can you elaborate further? That is to say, do I have to split $\mathbb R$ (using the decay condition) into a compact set and an unbounded set, and apply Weierstrass on the bounded set? If so, how do I tackle the unbounded set? $\endgroup$ – Teresa Lisbon Feb 21 '17 at 12:27
  • $\begingroup$ Exactly. Outside the closed interval the decay of $e^{-|x|}$ will dominate and since it goes to $0$ at infinity, as well as $f$, you're done. $\endgroup$ – lzralbu Feb 21 '17 at 12:30
  • $\begingroup$ Ok, so I will be a little more specific in this post. Let $r $ be such that $|f(x)| < \frac{\epsilon}{2}$ if $|x| > r$. So, inside $[-r,r]$, there is a polynomial $p$ such that $p(x) - f(x)e^{|x|} < \epsilon$. This can be used to construct a polynomial approximating $f$ on $[-r,r]$. Now, $e^{-|x|}$ dominates beyond a certain interval, let's say $p(x)e^{-|x|} < \epsilon$ for $|x| > s$, so that the difference between $f$ and $p(x)e^{-|x|}$ is less than $\epsilon$. Can I take $\max \{r,s\}$ and finish the proof? Do you see any errors in this logic? $\endgroup$ – Teresa Lisbon Feb 21 '17 at 12:34
  • $\begingroup$ My idea was that inside $[-r, r]$, you just take a polynomial $p(x)$ such that $|p(x) - f(x)| < \eps$. After that, let $s_n$ be the nth-partial sum of $e^{|x|}$. It approximates $e^{|x|}$ uniformly on $[-r, r]$. Then for large $n$, $s_n(x) e^{-|x|}$ is almost $1$ in $[-r, r]$. Since $p(x) s_n(x)$ is still a polynomial, you get the result. $\endgroup$ – lzralbu Feb 21 '17 at 12:43
  • $\begingroup$ Ah, that is smart! All right, thank you so much! This question was troubling me for some time, so I'm very happy that with your grateful help I've tided over it. Brilliant. Once again, thanks. $\endgroup$ – Teresa Lisbon Feb 21 '17 at 12:47

.All right, here is the answer.

I'll restrict myself to $[0,\infty)$, and so the set to be shown dense in the question will become $p(x)e^{-x}$. A similar proof will follow for the other space.

Note that $\{p(x)e^{-x}\}$ doesn't span $C_0[0,\infty)$ if and only if there is a non-zero linear functional $L$ over $C_0[0,\infty)$ , on whose application, all functions of the form $p(x)e^{-x}$ vanish. (This is by the Hahn-Banach theorem).

What is a linear functional over $C_0 [0,\infty)$? Well, it corresponds to some measure (with some other properties like regularity which I will ignore) $\mu$, that is to say, there exists a measure $\mu$ so that $$L(f) = \displaystyle\int_{0}^\infty f d\mu$$

for all $f \in C_0[0,\infty)$. This follows from the Riesz-Markov-Kakutani theorem, although not used in it's full strength.

Now, all we need to do, is to push this measure to $(0,1]$. That is to say, we have a homeomorphism from $[0,\infty)$ to $(0,1]$, given by $x \to e^{-x}$, which pushes the measure $\mu$ into a measure $\nu$ on $(0,1]$ (in a manner which I leave the reader to figure out, it's called the push-forward of a measure).

Now, if one does this, it will be found, after explicit computation of the push-forward, that $\displaystyle\int_0^1 p(x)d\nu = 0$ for all polynomials $p$. Then, by the uniqueness of moment theorem for bounded sets (because $[0,1)$ is bounded), it will follow that $\nu \equiv 0$, so $\mu \equiv 0$, so $L \equiv 0$, showing the density of the given set.

Alternatively, I have used a technique and computation similar to the one showed in an answer above, to see that it is enough that $e^{-\lambda x}, \lambda > 0$ be dense in $[0,\infty)$. This then will follow from a similar logic to above, but the computations are much easier.

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    $\begingroup$ It is clever using the substitution $x \mapsto e^{-x}$, but I think that the pushforward computation results in $\int_0^1 x p(- \log x) d \nu(x) = 0$ for all polynomials $p$. This relation is satisfied when $\nu$ is the Dirac delta at $0$, and so there is some work remaining to complete the proof. $\endgroup$ – A Blumenthal Mar 15 '17 at 4:46
  • $\begingroup$ Yes, there is some work left to do. But I am half-happy with this answer, because somebody else did the computations for me. $\endgroup$ – Teresa Lisbon Mar 15 '17 at 4:49

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