## Thursday, February 11, 2010

### A college math problem

Is there a positive and twice differentiable function f defned on [0,∞) such
that f(x)f’’(x) ≤ -1 on [0,∞) ? Why or why not???

Suppose the statement is true. Then $f(x)>0, f''(x)<0$ for all $$x>0$$. Moreover, $\lim_{x\rightarrow +\infty} f''(x)<0$ if it exists. Let $$f(0)>0, f'(x)$$ is monotonically decreasing, then $$\lim_{x\rightarrow +\infty} f'(x)$$ is either $$-\infty$$ or a constant.
If $$\lim_{x\rightarrow +\infty} f'(x)=-\infty$$, then $$f$$ is monotonically decreasing on $$[T,\,+\infty)$$ for some $$T$$ large enough and $$\lim_{x\rightarrow+\infty} f(x)$$ must be negative. Otherwise, $$\lim_{x\rightarrow+\infty} f(x)=c_0\geq 0$$ and there exists a unbounded sequence $$\{x_n\}$$ so that $$\lim_{n\rightarrow +\infty} f'(x_n)=0$$. Contradiction;
if $$\lim_{x\rightarrow +\infty} f'(x)=c$$, where $$c$$ is a constant, then there exists a unbounded sequence $$\{x_n\}$$ so that $$\lim_{n\rightarrow +\infty} f''(x_n)=0$$, but we already have $$\lim_{n\rightarrow +\infty} f''(x_n)<0$$. Contradiction. P. S. Note that the assumption is $$f(x)\cdot f''(x)\leq -1$$ on $$[0,\,+\infty)$$. If we change it into $$f(x)\cdot f''(x)<0$$, then we cannot have $$\lim_{x\rightarrow +\infty} f''(x)<0$$ and in this case the statement is true: e.g.: Let $$f(x)=\ln (x+2)$$. Then we have [LEFT] \begin{align*} f'(x)&=1/(x+2),\\ f''(x)&=-1/(x+2)^2. \end{align*} [/LEFT] Therefore, for all $$x\geq 0$$, we have [LEFT]\begin{align*}f(x)\cdot f''(x)&=-\frac{\ln (x+2)}{(2+x)^2}<0 \\ \intertext{and} f(x)&>0. \end{align*}[/LEFT][/QUOTE]