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I have a question that consists of the characterization of all functions $f(x)$ and all constants $k\in\mathbb{R}$ satisfying:

$$f:\mathbb{R}^+\rightarrow (0,1)$$

$$k-\int_4^x\frac{f(t)}{t}dt\leq\log(2)-\frac{1}{2}\log(x),\ \ \forall x\geq 4$$

Does someone have an idea about the second inequatily? Thanks a lot for your help!

I have found a sufficient condition. If we express $\log(x)$ by integral, then we have

$$k-\int_4^x\frac{f(t)}{t}dt\leq-\int_4^x\frac{1}{2t},\ \ \forall x\geq 4$$

which leads to a sufficient condition

$$k\leq 0$$


Does someone have another idea to make this characterizaton more accurate? Thanks a lot

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The equality can be made a bit more handy: $$ \int_4^x\frac{f(t)}{t}dt\geq c-\frac{1}{2}\log(x) $$ – Andrew Jul 17 '12 at 8:04
Just curious, where does the question come from? – Ashok Jul 17 '12 at 8:31
This question comes from a characterization of implicit volatility surface. In order to avoid arbitrage, we have found the conditions for some $\varphi(x)$ that satisfy the previous conditions. – Higgs88 Jul 17 '12 at 9:12
up vote 0 down vote accepted

This inequality can only hold if $k\leqslant0$, as the limit when $x\to4^+$ shows, and it holds for $k=0$ as soon as, for every $x\geqslant4$, $$ \int_4^x2f(t)\,\mathrm dt\geqslant x-4. $$ This condition is strictly weaker than the condition that $2f(x)\geqslant1$ for every $x\geqslant4$. For any given $k\lt0$, a weaker condition is to ask that, for every $x\geqslant4$, $$ \int_4^x2f(t)\,\mathrm dt\geqslant x-4+8k. $$ Finally, for any given $k\lt0$, a more general condition is to ask that, for every $x\geqslant4$, $$ \int_4^x2f(t)\,\mathrm dt\geqslant x-4+8k+g(x), $$ where the function $g$ is such that $$ \inf\limits_{x\geqslant4}\left(g(x)+x\int_4^x\frac{g(t)}{t^2}\,\mathrm dt\right)\geqslant0. $$

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