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Question

Suppose $f: \mathbb{R}_+ \rightarrow \mathbb{R}_+$ with $f(0) = 0$, $f'(x) > 0$ for all x, and $f''(x) > 0$ for all $x$.

Is $f'(x)^2 > f(x) f''(x)$ for all $x$?

My Attempts

Attempt 0: This is true for the special case in which $f(x) = x^\alpha$ for $\alpha \geq 2$. $\alpha^2 x^{2 \alpha - 2} > \alpha (\alpha - 1) x^{2 \alpha - 2}$, which reduces to $1 > 0$.

Attempt 1: $f'(0)^2 > f(0) f''(0)$ because $f'(0) > 0$ and $f(0) = 0$. If the derivative of the left hand side is weakly greater than the derivative of the right hand side, then the proposition is true. The LHS derivative is weakly greater if $f'(x) f''(x) > f(x) f'''(x)$, but we don't know anything about $f'''(x)$.

Attempt 2: Since we know this is true for $x = 0$, we can focus on the case where $x > 0$. For $x > 0$, this is proposition is equivalent to $\frac{f'(x)}{f(x)} > \frac{f''(x)}{f'(x)}$. This is equivalent to $\frac{d}{dx} \log f(x) > \frac{d}{dx} \log f'(x)$. Using the second fundamental theorem of calculus and $f(0) = 0$, $f(x) = f(x) - f(0) = \int_{0}^{x} f'(y) dy$. Substituting, we need $\frac{d}{dx} \log \int_{0}^{x} f'(y) dy > \frac{d}{dx} \log f'(x)$. This seems like it should be useful, but I wasn't able to figure out how to proceed from here.

Physor
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Christian Fong
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1 Answers1

2

Even the non-strict inequality does not hold.

Take $f(x) = e^{x^2}-1$. Then $f$ is increasing and strictly convex and $f(0) = 0$. We have $f'(x) = 2x e^{x^2}$ and $f''(x) = (2 + 4x^2)e^{x^2}$. Thus $f(x) f''(x) = (2+4x^2) e^{2x^2} - (2+4x^2)e^{x^2}$ and $f'(x)^2 = 4x^2 e^{2x^2}$. So, we have $$\forall x \geqslant 0,\ f'^2(x)-f(x)f''(x) = (2+4x^2)e^{x^2}-2e^{2x^2}.$$ The right-hand side is clearly negative for large $x$.

halbaroth
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