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let $A$ an N-function such that

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

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

we define $$A^{-1}_{\ast}(t)=\int^t_0\frac{A^{-1}(\tau)}{\tau^{\frac{n+1}{n}}} d\tau$$ how we show that $\sigma(t)=(A_{\ast}(t))^{\frac{n-1}{n}}$ is Lipschitz on $\mathbb{R}$?

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