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Assume that $f$ is twice diffentiable on $R$,and such

$$2f(x)+f''(x)=-xf'(x)$$

show that:

$f(x)$and $f'(x)$ are bounded on $R$

My try:since $$2f(x)+f''(x)+xf'(x)=0$$ and following I can't any work,Thank you very much!

1 Answers1

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Define $$g=f^2+\frac{1}{2}(f')^2.$$ By definition, $g$ is non-negative and differentiable; moreover, $$g'(x)=f'(x)\cdot\left(2f(x)+f''(x)\right)=-x\cdot (f'(x))^2,\quad\forall x\in \Bbb R.$$ Therefore, $g$ is increasing on $(-\infty,0]$ and decreasing on $[0,+\infty)$, so $g(\Bbb R)\subset [0, g(0)]$. The conclusion follows.

Hu Zhengtang
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  • Thank you,Then How prove $f(x)$ is bounded on $R$,and $f'(x)$ –  Oct 31 '13 at 12:08
  • @nanchangjian: We have shown that $f^2+\frac{1}{2}(f')^2\le g(0)$ on $\Bbb R$. Starting from this, can you see both $f$ and $f'$ are bounded by yourself? – Hu Zhengtang Oct 31 '13 at 12:24