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Suppose there is a function $f:\mathbb{R}_+^n\mapsto \mathbb{R}$. Are there any systematic ways to determine whether the range of $f$ is bounded or not?

For example, there is a function $f(x,y)=-x^2+2y\log(1+x)-y\log(y),x>0,y>0$. How can we prove that it is bounded above (as indicated in the simulation results)?

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How are you given $f$? –  Qiaochu Yuan Oct 27 '11 at 17:16
    
If $f$ is differentiable, as in your example, why can't you just look at critical points and then compare that with behavior for large |x|, |y| ? –  Christopher A. Wong Oct 27 '11 at 18:45
    
Critical points are not usually needed. If $f$ is continuous on $[0, \infty)^n$ it's bounded on compact subsets, so you just need to look at the behaviour "at $\infty$", i.e. when $\sum_j x_j$ is large. –  Robert Israel Oct 27 '11 at 21:21
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In your example, write $f(x,y) = -x^2 + y \log((1+x)^2/y) = -x^2 + (1+x)^2 \log(t)/t$ where $t = (1+x)^2/y$. Now $\log(t)/t \le 1/e$ for $t >0$, the maximum occurring at $t=e$, so $f(x,y) \le -x^2 + (1+x)^2/e$ which is easily seen to be bounded above. –  Robert Israel Oct 27 '11 at 21:33
    
True, I suppose I would find it valuable to also have an estimate of the bound. –  Christopher A. Wong Oct 27 '11 at 23:21

1 Answer 1

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There isn't any general method. What you usually need to do in practice is like Robert illustrated. The function is bounded on compact sets on which the function is continuous, so you need to focus on the neighbourhoods of points of non-continuity and on trajectories going to infinity. Note that it is easy to make a continuous function $f(x,y)$ such that $f(x,y)\to 0$ as $x,y\to\infty$ along any straight line, and yet $f(x,y)$ is not itself bounded. So it isn't enough to consider straight trajectories. Other methods that might work on odd occasions include writing the function as an integral (which might be obviously bounded when it isn't obvious in the original formulation), and interpreting the function as something already known to be bounded (such as a probability).

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Thanks a lot. But could you elaborate more on why " it isn't enough to consider straight trajectories"? That is because, previously, I also come out with a solution in which I consider a straight line y=bx and prove that f(x,bx) is bounded above over any b from -infty to +infty, since the set of lines y=bx can cover all points in the plane. –  Allen Nov 2 '11 at 9:11
    
If $f(x,bx)\le c$ for some fixed $c$, and also $f(0,y)\le c$ for all $y$, then certainly $f(x,y)\le c$ everywhere. However, if the maximum of $f(x,bx)$ depends on $b$ and can be arbitrarily large, the function is not bounded. It all depends on the uniformity. –  Brendan McKay Nov 2 '11 at 13:38

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