I'm reading Existence of solutions to a higher dimensional mean-field equation on manifolds and get stuck on Lemma6. When $\lambda>\Lambda_1$, with $\Lambda_1=(2 m-1) ! \operatorname{vol}\left(S^{2 m}\right)$. They proved that the following functional is not bounded below so they wanted to find the saddle point $$ I_\lambda(u)=\frac{1}{2} \int_M|\Delta_g^{m/2} u|^2 d \mu_g-\frac{\lambda}{2 m} \log \left(\int_M e^{2 m u} d \mu_g\right) $$ on $$ E:=\left\{u \in H^m(M): \int_M u d \mu_g=0\right\}. $$ And equip $E$ with the norm $\|u\|:=\left(\int_M\left|\Delta_g^{\frac{m}{2}} u\right|^2 d \mu_g\right)^{\frac{1}{2}}$. The critical point of this functional is the solution of the mean-field equation $$\left(-\Delta_g\right)^m u+\lambda=\lambda \frac{e^{2 m u}}{\int_M e^{2 m u} d \mu_g}$$ on a unit volume closed Riemannian manifold $(M, g)$ of dimension $2 m$.

In Lemma5, they have proved that there exists a bounded sequence $\left(u_n\right)$ in $E$ such that $I_\mu^{\prime}\left(u_n\right) \rightarrow 0$ and $I_\mu\left(u_n\right) \rightarrow c_\mu$.

Then they assume that $u_n$ converges weakly in $E$ and almost everywhere to a function $u$, and proved that $e^{2 m u_n}$ and $e^{2 m u}$ are uniformly bounded in $L^4$.

My question arises in the next step, they wrote that :

by dominated convergence one has for $N>0$ $$\tag{1} \min \left\{e^{2 m u_n}, N\right\} \rightarrow \min \left\{e^{2 m u}, N\right\} \quad \text { in } L^2\left(M, d \mu_g\right) $$ as $n \rightarrow \infty$ and that $$ \sup _{n \in \mathbb{N}}\left\|\min \left\{e^{2 m u_n}, N\right\}-e^{2 m u_n}\right\|_{L^2}^2 \leq \frac{1}{N^2} \sup _{n \in \mathbb{N}}\left\|e^{2 m u_n}\right\|_{L^4}^4 \rightarrow 0 \quad \text { as } N \rightarrow \infty, $$ we infer that $e^{2 m u_n} \rightarrow e^{2 m u}$ in $L^2$.

I wonder why they need to choose a $N$ and wrote (1) ? Actually I don't even know how the dominated convergence is used here.


2 Answers 2


As your title suggests, the dominated convergence theorem is a powerful tool, which often we can apply to cases to where the assumption of domination is not directly verified, or nor immediate to verify .

Let's call $f_n$ the non-negative function $f_n:=e^{2mu_n}$. The first statement is a direct application of the dominated convergence theorem: since $f_n$ converges a.e. to $f:=e^{2mu}$, for every given $N$ the sequence $f_n\wedge N$ converges a.e. to $f\wedge N$, and it is dominated by the constant function $N$ (which is in $L^2(M)$, just because here $M$ has finite measure). Therefore, $f_n\to f $ in $L^2(M)$. For the other statement, note that, since, quite obviously, $f_n>N$ in the set where $f_n$ is larger than $N$,

$$\|f_n\wedge N- f_n\|_2^2=\int_{\{f_n>N\}}f_n^2dx\le \int_{\{f_n>N\}}\frac{f_n^2}{N^2} f_n^2dx\le \frac1{N^2}\|f_n\|^4_4.$$ If now the conclusion is not clear to you, write
$$\|f-f_n\|_2\le \|f-f\wedge N\|_2+\|f\wedge N-f_n\wedge N\|_2+\|f_n\wedge N-f_n\|_2\le$$$$\le \|f-f\wedge N\|_2+\|f\wedge N-f_n\wedge N\|_2+\frac1N \sup_{n\in\mathbb N}\|f_n\|^2_4$$ which is true for every $n>0$ and every $N>0$. Keep $N$ fixed and let $n\to\infty$, so by the first statement the middle term vanishes in the limit, and this yields to $$\limsup_{n\to\infty}\|f-f_n\|_2\le \|f-f\wedge N\|_2+\frac1N \sup_{n\in\mathbb N}\|f_n\|^2_4.$$ Since this is true for all $N>0$, you can take the infimum of the RHS, which is $0$, proving $f_n\to f$ in $L^2(M)$.


[edit] A more general statement: For a finite measure $\mu$, for $1<p\le \infty$, and for a bounded sequence $(u_n)_{n\ge0}\subset L^p(X,\mu)$ converging a.e. to $u$, one has $u_n\to u$ in $L^q$ for all $q<p$. One can prove it analogously --you can also easily reduce to the case $u_n\ge0$, $u=0$, $q=1$. (Incidentally, also $u_n\to u$ weakly* $L^p$).

  • 1
    $\begingroup$ Here I'm referring to the dominated convergence in $L^p$ spaces, for $p=2$: for a sequence $(f_j)_{j\ge0}\subset L^p$, convergence a.e. to $f$ and $|f_j|\le g\in L^p$ implies convergence to $f$ in $L^p$ norm. This is a plain consequence of the case of $L^1$; it holds true for every $1\le p<\infty$. $\endgroup$ Commented Dec 28, 2023 at 15:54

Given that $u_n$ converges almost everywhere (a.e.) to $u$, we have $$\big(\min (e^{2mu_n},N)-\min(e^{2mu},N)\big)^2\to0$$ a.e. (as $n\to\infty$). Also, $$\big|\big(\min (e^{2mu_n},N)-\min(e^{2mu},N)\big)^2\big|\le N^2$$ and $\int_M N^2\,d\mu_g<\infty$ (provided that the measure $\mu_g$ is finite). So, by dominated convergence $$\int_M d\mu_g\, \big(\min (e^{2mu_n},N)-\min(e^{2mu},N)\big)^2\to0; $$ that is, $\min (e^{2mu_n},N)\to\min(e^{2mu},N)$ in $L^2(M,\mu_g)$.

Here, given that the $e^{2 m u_n}$'s are uniformly bounded in $L^4$, essentially the paper proves a special case of the de la Vallée-Poussin theorem; cf. relation to convergence of random variables.


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