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I want to verify the following claim that I found in some paper. Suppose f is a smooth real-valued function on the real line satisfying $f'(x)x-f(x)\ge x^2$ for all x. Then there is a constant C, s.t. $f(x)\ge \frac{x^2}{2}-C$ for all $x.$

The connection with the title is that the authors claim that any function satisfying the first inequality is of quadratic growth, i.e. satisfies the second inequality.

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Note that $$\left( \frac{f(x)}{x} \right)' = \frac{xf'(x) - f(x)}{x^2} \geq 1,$$ hence the result.

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  • $\begingroup$ Very nice. I always tried to integrate from 0, thus neglecting functions of the form $x^\alpha f(x)$ for negative $\alpha$. Thanks! $\endgroup$
    – Orbicular
    Commented Jun 7, 2011 at 15:21
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Consider what happens when you have equality, and use a comparison argument.

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