Let $\alpha \in (0,1)$ and $M_{\alpha}$ be a class of Borel nonnegative measures $\mu$ on $\mathbb{R}^n_{+}$ such that there exists $r(\mu) > 0$: $$ \int\limits_{\mathbb{R}^{n}_{+}} \exp\left(r(\mu)(x_1^2+\ldots+x_n^2)^{\alpha}\right) \, \mu(dx) < \infty. $$ Let $M_{\alpha}'$ be a class of Borel nonnegative measures $\mu'$ on $\mathbb{R}^n_{+}$ such that there exists $r'(\mu') > 0$: $$ \int\limits_{\mathbb{R}^n_{+}} \exp \left(r'(\mu')(x_1^{2\alpha}+\ldots+x_n^{2\alpha}) \right) \, \mu'(dx) < \infty. $$ Since $\left(x_1^{2\alpha}+\ldots+x_{n}^{2\alpha}\right)^{\frac{1}{\alpha}} \geqslant x_1^2 +\ldots+x_n^2$ for $\alpha \in (0,1)$ we have $M_{\alpha}' \subseteq M_{\alpha}$. But I can't find an example of measure $\nu$ such that $\nu \in M_{\alpha} \setminus M_{\alpha}'$. Help me please to find such measure of to prove that $M_{\alpha} = M_{\alpha}'$.
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Using $|s+t|^p\leqslant |s|^p+|t|^p$ for $0<p<1$, we have $$\left(\sum_{j=1}^nx_j^2\right)^\alpha\leqslant \sum_{j=1}^nx_j^{2\alpha}.$$ As the map $t\mapsto |t|^\alpha$ is concave, we have $$\left(\sum_{j=1}^nx_j^2\right)^\alpha=n^\alpha\left(\sum_{j=1}^nn^{-1}x_j^2\right)^\alpha\geqslant n^{\alpha}\sum_{j=1}^nn^{-1}x_j^{2\alpha},$$ which proves that $M_\alpha=M'_\alpha$.
Davide Giraudo
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