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Suppose $g:\mathbb{R} \rightarrow [0,\infty)$ is a strictly increasing function such that $\lim g(x) = \infty$ as $x \rightarrow \infty$. Suppose $h:\mathbb{R} \rightarrow [0,\infty)$ is a strictly decreasing function such that $\lim h(x) = 0$ as $x \rightarrow \infty$. Consider the product function $f(x)=g(x)h(x)$. Is it possible to construct such $g,h$ such that $f$ does not have a limit as $x \rightarrow \infty$?

Even better, can such an $f$ be constructed if we insist that $g(x)=x$?

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Generally the way to do these kinds of questions is to construct f first. What have you tried? –  Qiaochu Yuan Dec 13 '10 at 20:38
    
If you do not want to have your product diverging either, the only way is $g$ having infinitely many saddle-points, says my intuition. Is this possible and sufficient? –  Raphael Dec 13 '10 at 21:03
    
Do you consider a limit that is equal to $\infty$ to exist or not to exist? (In my opinion, the latter). –  Arturo Magidin Dec 13 '10 at 21:12

3 Answers 3

Take

$$ g(x) = x 2^{\lfloor x/2\rfloor}, h(x) = \frac{1}{x} 2^{-\lfloor (x+1)/2\rfloor} $$

The product oscillates between 1 and 1/2: let $f = g\cdot h$, $f(2n) = 1$, $f(2n+1) = 1/2$ for $n$ an integer.

Edit: If you want $g(x) = x$. Take $h(x)$ to be the following (continuous) function: let $h(e^{2n}) = e^{-2n}$, and let $h(e^{2n+1}) = e^{-2n - \frac12}$, and linearly interpolate. Then $f(e^{2n}) = 1$ and $f(e^{2n+1}) =\sqrt{e} \neq 1$ for all natural numbers $n$. So $f$ does not have a limit.

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These functions don't seem to have the right behavior for x negative. –  Qiaochu Yuan Dec 13 '10 at 23:00
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But they can be patched up below x=1 and show the same behavior as x gets large. –  Ross Millikan Dec 13 '10 at 23:17
    
Like Ross says, you can modify the function arbitrarily for $x < x_0$ for any fixed $x_0$ to satisfy any monotonicity, positivity condition you want, while keeping the limiting behaviour the same. –  Willie Wong Dec 13 '10 at 23:40

Define $h(x) = 3e^{-x}$ for $x < 1$, and $h(x) = 1/x$ for $x \geq 1$. Hence, $h$ satisfies the conditions in the question. Define $g(x) = (1/3) e^x$ for $x<1$, and for $x \geq 1$ define it piecewise as follows: $g(x) = x + a_n$ for $n \leq x < n+1$, $n=1,2,3,\ldots$, where $a_n$ is a strictly increasing sequence with $a_1 = 0$. Hence, $g$ satisfies the conditions in the question. Now define $f(x)=g(x)h(x)$. Thus, for any $n \in \mathbb{N}$, $f(n)=(n+a_n)/n = 1 + a_n / n$. We can choose $a_n$ so that $a_n / n$ does not have a limit. For example, consider arbitrarily large $n$, and set $a_n = 2n$. Then, let $a_{n+k} = (2n+k)$, for $k = 1,\ldots,n^2$. Then, $a_n / n = 2$, but $a_{n+n^2}/(n+n^2) \approx 1$. Continue the same way to construct a sequence with $\lim \sup _{n \to \infty}(a_n/n) = 2$ but $\lim \inf _{n \to \infty}(a_n/n) = 1$. We conclude that $f(x)$ does not have a limit as $x \to \infty$.

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You are greatly helped by the fact that $f$ and $g$ are not required to be analytic so you can easily look at $f$ and $g$ in some neighbourhood of infinity $(-\infty, -\varepsilon) \cup (\varepsilon, \infty)$ and not really care about their behaviour outside that neighbourhood :)

Fix $g$. In general, you need a function $f$ such that $\lim_{x\to\infty} f(x)g(x) < \infty$, which is to say that $f(x) = \frac{\alpha(x)}{g(x)}$, where $\alpha$ is bounded (in your terms, $\alpha = h/g^2$). It is not very hard to prove that such an $f$ exists for any continuous $g$. Just construct $f$ as $\frac{\alpha}{g}$ in some neighbourhood of infinity where $g(x) \neq 0$ and then continue $f$ to the rest of the domain (it is possible to do this in $C^\infty$ way).

This covers all monotonic functions in $C^0, C^1, \ldots, C^\infty$. The most interesting stuff is about the rest of them :)

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