Orlicz–Sobolev spaces

Question

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 $q(A,n)\leq n$?

Answer 1

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 $$q(A,n)\leq n$$