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

$$2f(x)+f”(x)=-xf'(x)$$

**show that:**

- Let $a_{i} \in\mathbb{R}$ ($i=1,2,\dots,n$), and $f(x)=\sum_{i=0}^{n}a_{i}x^i$ such that if $|x|\leqslant 1$, then $|f(x)|\leqslant 1$. Prove that:
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$f(x)$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!

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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.

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