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Let $00, f'(x)>0, \left(\frac{f(x)}{x}\right)'\leq0, f(x)\sim x (x\rightarrow 0^+)$$ and satisfies the differential inequality $$xf'(x)\leq a f(x), \forall x\in (0,1/2)$$

Note: I attempted to consider examples of the form $f(x)=1-\exp(-2bx)$ where $b>0$ is fixed, but one can show that $f$ is strictly monotone over $(0,1/2]$ with $f(0^+)=1$, and thus such form of $f$ does not fufill the condition $01$ fixed, and it turns out one also needs $a\geq 1$ to make it work, which is not allowed.

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    If $f(x)/x\to 1$ as $x\to 0^+,$ then choose $a$x.$ This contradicts $f(x)/x\to 1.$2017-01-28

1 Answers 1

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There does not exist any function $f$ with the given properties. Proof: the differential inequality is equivalent to $$(x^{-a}f(x))'\leq 0,~\forall 0

Since $f(x)\sim x ~(x\rightarrow 0^+)$ and $00 ~(x\rightarrow 0^+),$$ which contradicts to the differential inequality near a neibourhood of $0^+$.