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I'm reading an article and I saw the following affirmation without proof:

Let $u \in H^1(\mathbb{R}^2)$ and $\alpha>0$, then

$$\int_{\mathbb{R}^2}(e^{\alpha u^2}-1)dx<+\infty.$$

Is this claim really true? If yes, if possible, can anyone recommend me an reference for study? thank you so much in advance!

Edit: The article I'm reading is "A nonhomogeneous elliptic problem involving critical growth in dimension two" by João Marcos do Ó, Everaldo Medeiros and Uberlandio Severo. The claim is written in Lemma 2.1 of the article and it is said that the proof can be found in [9], [13] or [21] at the references of the article, However I only found at this references the proof of the part "Moreover..."

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  • $\begingroup$ As @Aleksei Kulikov mentionned below, $\alpha$ must depend on $u$, and this is precised in the reference given in my answer. $\endgroup$ Commented Sep 1 at 9:21

2 Answers 2

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Yes, this is called Moser-Trudinger inequality. See for instance Theorem 1.67 of this book.

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  • $\begingroup$ I believe it is slightly different from the way OP wrote it -- $\alpha$ should depend on $u$ I think... But maybe I am wrong. $\endgroup$ Commented Sep 1 at 6:38
  • $\begingroup$ that is exactly what i'm looking for! thank you! $\endgroup$
    – Victor
    Commented Sep 1 at 12:58
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    $\begingroup$ @Victor you maybe actually need the result stated and proved by Guido below (in that case do not hesitate to attribute him the answer!). $\endgroup$ Commented Sep 1 at 17:32
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This is more a long comment, but it might be useful. A somehow overlooked fact is that

Theorem. For every function $u\in H^1(\mathbb R^2)$ and for every $\alpha\ge 0$, the integral $$ \int_{\mathbb R^2} (e^{\alpha u^2}-1) dx <+\infty $$

That is, the integral is finite no matter what $\alpha$. Note however, that to ensure an uniform upper bound on the integral, $\alpha$ has to depend on (the norm of) $u$.

The proof of this fact easily follows the following three steps:

  1. By classical Moser Trudinger inequality, for every $\alpha >0$ there exists $R>0$ such that $$ \int_{\mathbb R^2} (e^{\alpha u^2}-1) dx \le 1 $$ for all $u \in B_R=\{u \in H^1: \|u\|_{H^1}\le R\}$.

  2. For all $\alpha >0$ and for all $v\in C_c^\infty$ $$ \int_{\mathbb R^2} (e^{\alpha v^2}-1) dx <+\infty. $$

  3. Let $X$ be a normed space and let $F: X \to (-\infty,+\infty]$ be a convex function such that:

    (a). There exists $R>0$ such that $B_R\subset\{F\le 1\}$.

    (b). There exists a (norm) dense set $\mathcal D$ such that $F(v)<+\infty$ for all $v \in \mathcal D$.

    Then $F$ is finite everywhere.

Item 3 above should be well known, but I am not aware of any reference (and I would actually appreciate one if someone knows it). The proof goes as follow:

Let $u \in X$ and let $v \in \mathcal D$ and $w\in B_{R}$ be such that $$ 2u=v+w. $$ Then $$ F(u)=F\Big(\frac{v+w}{2}\Big)\le \frac 1 2 F(v)+\frac 1 2 F(w)<+\infty. $$

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  • $\begingroup$ Ciao Guido :) I have checked the article in reference of the OP and in fact it may be this result (rather than the standard Moser-Trudinger inequality) which is used, as Aleksei somehow suspected above. I didn't know of this fact, thanks ! $\endgroup$ Commented Sep 1 at 17:31

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