Continuity Question?

Question

Hi, I really need help on this question.
----------
Let f:R->R be a continuous function such that f(x)>0 for all x in R, and lim[x->∞] f(x)=0=lim[x->-∞] f(x).

Prove that there exists a real number c>0 such that
f(R)=(0,c]
------
Thank you so much

Answer 1

This problem becomes really easy if you extend the function to the extended real line. Let cl(R) = [-∞, ∞] denote the extended real line. Define a function g:cl(R) → R by g(x) = {f(x) if x∈R, 0 if x=∞ or x=-∞}. Since f is continuous, g is continuous at every point of R, and since [x→∞]lim g(x) = [x→∞]f(x) = 0 = g(∞) and [x→-∞]lim g(x) = [x→-∞]f(x) = 0 = g(-∞), g is continuous at ∞ and -∞ as well. So g is a continuous function.

Since cl(R) is a compact connected set, it follows that g(cl(R)) must also be a compact connected set. Since the only compact connected sets in R are closed bounded intervals, it follows that g(cl(R)) = [a, b] for some real numbers a and b. Since 0∈g(cl(R)), it follows that a≤0, and since ∀x∈cl(R), g(x)≥0, it follows that a≥0. So in fact a = 0 and g(cl(R)) = [0, b] for some real number b.

Now, consider g(R). This contains every point of g(cl(R)) except possibly g(∞) = g(-∞) = 0. And indeed, since for every x∈R, g(x) = f(x) > 0, g(R) does not contain 0, so g(R) = g(cl(R))\{0} = (0, b]. But since on R, g=f, we have f(R) = g(R) = (0, b]. So we are done.