A transforms converts an unbounded sequence into bounded

Question

Recently, I read a paper(here the link) about variational methods. It said that

Then we study a maximizing sequence $\{(u_k,v_k)\}$ of the functional $I(u,v)$. Such a sequence may be unbounded. Using the family of transforms $T_{\lambda}$, it can be converted into a bounded sequence that converge to a limit $(u_0,v_0)$ in $\tilde{G}\times G$.

Here is some definition: $S$ be the unit circle. For each $\lambda\in(0,\infty),$ define $\psi_{\lambda}:R\rightarrow R$ by $$\psi(\theta)=\sqrt{\lambda^2\cos^2\theta+\frac{1}{\lambda^2}\sin^2\theta}$$ and $m_{\lambda}:R\rightarrow R$ by $$m_{\lambda}(\theta)=\int_0^{\theta}\frac{dt}{\psi_{\lambda}^2(t)}.$$ Here some property of $m_{\lambda}$: $m_{\lambda}(\theta+2\pi)=m_{\lambda}(\theta)+2\pi.$ For $0\leq\theta\leq \frac{\pi}{2}$, $$m_{\lambda}(\theta)=\arctan\left(\frac{1}{\lambda^2}\tan\theta\right).$$ For any measurable function $u$ on $S$, define the transform $T_{\lambda}$ by $$(T_{\lambda}u)(\theta)=u(m_{\lambda}(\theta))\psi_{\lambda}(\theta).$$ Besides, $$G=\{v:v>0,\int_S\frac{1}{v^3}d\theta<\infty,\int_S\frac{1}{v^3}\cos\theta d\theta=0=\int_S\frac{1}{v^3}\sin\theta d\theta\}$$ and $$\tilde{G}=\{u\in G:u\in H^1(S)\}.$$Define on $\tilde{G}\times G,$ $$I(u,v)=\frac{\{\int_S\frac{1}{v^2}d\theta\}^3\{\int_S[u^2-(u')^2]d\theta\}}{\{\int_S\frac{u}{v^3}d\theta\}^2}.$$ It is in Hilbert space $H^1(S)$ with norm $$\|u\|_{H^1(S)}=\left(\int_S[u^2+(u')^2]d\theta\right)^{\frac{1}{2}}.$$ So I wonder how the transform $T_{\lambda}$ convert unbounded sequence into bounded one?
Any ideas would be welcome! Thanks!

*update
I want to show $\|T_{\lambda}u\|_{H^1(S)}<\infty$. In this way, I can show $T_{\lambda}u$ always be bounded. But I was stuck in the step $$\|Tu\|=\int_S[(Tu)^2+[(Tu)']^2]d\theta=\int_S [u^2(2\psi^2-\frac{1}{\psi^2})+2u'u\frac{\psi'}{\psi}+(u')^2]d\theta.$$ Here we denote $T$ instead of $T_{\lambda}$. Is there any problem of my thought?

Answer 1

You seem to be misunderstanding the strategy.

The key technical notions are the observations that

• for any positive $t,s$, you have $I(tu,sv) = I(u,v)$. (Simple scaling)
• for any $(u,v)$, you have $I(T_{\lambda}u, T_{\lambda}v) = I(u,v)$. (Lemma 5.1)

The simple scaling implies that whenever you take a maximizing sequence, you can always assume that the maximizing sequence has norm 1. So you never need to prove by hand uniform boundedness of the maximizing sequence.

The $\lambda$ transformation serves to "localize" the functions $u$ and $v$ (see Remark 5.2.)

More precisely, if you have $u_k, v_k$ any maximizing sequence, you can always replace them by

$$ \tilde{u}_k = \frac{T_{\lambda_k} u_k}{\|T_{\lambda_k} u_k\|}, \quad \tilde{v}_k = \frac{T_{\lambda_k} v_k}{\|T_{\lambda_k} v_k\|} $$

for any sequence of positive $\lambda_k$ and have that $$ I(u_k,v_k) = I(\tilde{u}_k, \tilde{v}_k) $$

You have that $(\tilde{u}_k, \tilde{v}_k)$ is therefore a maximizing sequence with norm 1, that is suitable localized.

How is this used? What the authors wanted to prove is not that $(u_k, v_k)$ is bounded in $H^1$, as you have incorrectly interpreted. What they wanted to prove is that $\sup I(u,v) < \infty$. So they start by taking an arbitrary maximizing sequence, and they argue if this maximizing sequence leads to $I(u_k, v_k) \nearrow +\infty$ you get a contradiction. This is shown with the help of the localization property on the bottom of page 14 and top of page 15.