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

how we show that $q(A,n)\leq n$