Orlicz-Sobolev Spaces

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 $$\widehat{A_1}$$ 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 $$q(A,n)$$ 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 $$q(A,n)\leq n$$

if $$n=1$$, $$\begin{eqnarray*} \int^{+\infty}_1 \frac{\widehat{A_1}^{-1}(\tau)}{\tau^{1+\frac{1}{n}}}d\tau&=&\bigg[-n\tau^{-\frac{1}{n}}\widehat{A_1}^{-1}(\tau)\bigg]^{+\infty}_1\\ &+&n\int^{+\infty}_1\frac{\widehat{A}^{-1}(\tau)}{\tau^{1+\frac{2}{n}}}d\tau\\ &\leq& \widehat{A_1}(1)+\int^{+\infty}_1\frac{A^{-1}}{\tau^3}d\tau\\ &\leq& \widehat{A_1}^{-1}(1)+\int^{+\infty}_1\frac{1}{\tau^2}d\tau\\ \end{eqnarray*}$$ $$\Rightarrow q(A,1)=1$$
if $$n>1$$ we suppose $$q>n-1$$ we notice that $$\begin{eqnarray*} \widehat{A_2}^{-1}(t)&=&\int^t_0 \frac{\widehat{A_1}^{-1}(\tau)}{\tau^{1+\frac{1}{n}}}d\tau\\ &=& \int^1_0 \frac{\widehat{A_1}^{-1}(\tau)}{\tau^{1+\frac{1}{n}}}d\tau+\bigg[-n\tau^{-\frac{1}{n}} \widehat{A_1}^{-1}(\tau) \bigg]^t_1\\ &+&n\int^t_1\frac{\widehat{A}^{-1}(\tau)}{\tau^{1+\frac{2}{n}}}d\tau\\ \end{eqnarray*}$$
$$\widehat{A_2}(t)\leq k_2\int^t_0\frac{\widehat{A}^{-1}}{\tau^{1+\frac{2}{n}}}d\tau$$
similarly ,we find $$\widehat{A_3}^{-1}(t)\leq K_3 \int^t_0\frac{\widehat{A}^{-1}(\tau)}{\tau^{1+\frac{3}{n}}}d\tau$$ until n we find $$\widehat{A_n}^{-1}(t)\leq k_n \int^t_0\frac{\widehat{A}^{-1}}{\tau^2}d\tau$$ then $$\begin{eqnarray*} \int^{+\infty}_1 \frac{\widehat{A_n}^{-1}(t)}{t^{1+\frac{1}{n}}}dt &\leq& k_n \bigg[-nt^{-\frac{1}{n}}\int^t_0\frac{\widehat{A}^{-1}(\tau)}{\tau^2}d\tau \bigg]^{+\infty}_1\\ &+&n\int^{+\infty}_1\frac{\widehat{A}^{-1}}{\tau^{2+\frac{1}{n}}}d\tau\\ &\leq& k_{n+1}+n\int^{+\infty}_1\frac{1}{t^{1+\frac{1}{n}}}d\tau\\ &\leq& k_{n+1}+n^2\\ \end{eqnarray*}$$
so $$q(A,n)\leq n$$