# A problem about how dominated convergence is used in the analysis of variation

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.

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 $$L^2(M)$$.

$$*$$

 A more general statement: For a finite measure $$\mu$$, for $$1, and for a bounded sequence $$(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. 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* $$L^p$$).

• 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 $1\le p<\infty$. 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.