Let $A$ be an N-function and suppose that $$\int^{+\infty}_1\frac{A^{-1}(\tau)}{\tau^{1+\frac{1}{n}}}d\tau=+\infty. $$ We denote by $\widehat{A}$ an N-function equal to $A$ near infinity and $\widehat{A}$ satisfies

$$\int^{1}_0\frac{\widehat{A}^{-1}(\tau)}{\tau^{1+\frac{1}{n}}}d\tau<+\infty. $$

Now define an N-function $\widehat{A_1}$ by: $$\widehat{A_1}^{-1}(t)=\int^t_0\frac{\widehat{A}^{-1}(\tau)}{\tau^{1+\frac{1}{n}}}d\tau $$

and finally denote by $A_1$ an N-function equal to $A$ near 0 and to $\widehat{A_1}$ near infinity.

If $$\int^{+\infty}_{1}\frac{A^{-1}_1(\tau)}{\tau^{1+\frac{1}{n}}}=+\infty $$

we repeat the same construction, then we pose: $A_2=(A_1)_1$, etc.

Let $q(A,n)$ be the smallest integrer $q\geq0$ such that $$\int^{+\infty}_1\frac{A_q^{-1}(\tau)}{\tau^{1+\frac{1}{n}}}d\tau<+\infty. $$

How do we show that $q(A,n)\leq n$?