# Butterfly Arbitrage condition

I hope anybody can help me. According to Gatheral and Jacquier (https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2033323) no Butterfly Arbitrage can be expressed like this:

Define the function $$\displaystyle g(k):= \left( 1- \frac{k \omega'(k)}{2 \omega(k)}\right)^2- \frac{\omega'(k)}{4} \left(\frac{1}{\omega(k)}+\frac{1}{4} \right) + \frac{\omega''(k)}{2}$$

A slice is free of Butterfly Arbitrage $$\Leftrightarrow$$ \ the function $$g(k) \geq 0 \ \forall k \in \mathbb{R}$$ and $$\lim \limits_{k \rightarrow \infty} d_+(k)=-\infty$$. The second condition here is equivalent to call prices converging to 0 as $$k \to \infty$$ (if I saw this right this is needed to have a density). Now I'm asking is there an arbitrage strategy to "see" the arbitrage if this point isn't fullfilled?

• The condition $g(k)\ge 0$ basically says that the option price is a convex function of the strike. If this condition is violated, you can constructed an arbitrage strategy as in this answer. – Gordon Jan 17 '19 at 17:19
• But why do I need $\lim \limits_{k \to \infty}d_+(k)=-\infty$? – J.Doe Jan 21 '19 at 11:15
• @J.Doe I recommand reading : The SVI implied volatility modeland its calibration, by Alexander Aurell; pdfs.semanticscholar.org/4584/…. – Marine Galantin Jun 9 at 20:00