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More specifically, suppose $f$ is continuous on $[a,b]$ with $f(a)=f(b)=0$ and $x^2f^{\prime \prime}(x)+4xf^{\prime}(x)+2f(x)\geq 0$ for $x\in (a,b)$. Prove that $f(x)\leq 0$ for $x\in [a,b]$.

I'm not sure how to approach this question. Any help would be greatly appreciated.

Mael
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  • Solution will be of the form $x^n$ so you could try to substitute it in and get the result for $f(x)$ but I guess there's another way... – user88595 Oct 24 '13 at 20:34
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    @user88595 There are many other solutions besides $x^n$..... – N. S. Oct 24 '13 at 20:56

1 Answers1

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Hint: $\left(x^2 f(x)\right)'' \ge 0$ means $x^2 f(x)$ is convex.

njguliyev
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