Application of the Arzela-Ascoli Theorem

Set-Up

$\Omega$ is an open subset of $\mathbb{R}^{n}$. The family $\{\phi_{j}\}_{j \in \mathbb{N}}$ is a Complete Orthonormal Set in $L^{2}_{\sigma}(\Omega) = \{f : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n} \ | \ f \in L^{2}(\Omega), \ \text{div}(f) = 0 \}$. In particular the closure $\overline{\text{span} \{\phi_{j}\}_{j \in \mathbb{N}}} = L^{2}_{\sigma}(\Omega) $

$\{u_{m}\}_{m \in \mathbb{N}}$ is a family of functions $u_{m} : \mathbb{R}^{n} \times [0,T] \rightarrow \mathbb{R}^{n}$, $u_{m} \in L^{2}([0,T] ; H^{1}(\Omega) \cap L^{2}_{\sigma}(\Omega))$ such that:

The family $\{ (u_{m}(\cdot,t) , \phi_{j} ) \}_{m \in \mathbb{N}}$ is uniformly bounded and equitcontinuous with respect to $t \in [0,T]$, for each $j$ fixed. Here $(\cdot , \cdot)$ denotes the inner product of $L_2(\Omega)$, ie $(u(\cdot, t),f) = \int_{\Omega} u(x,t) \cdot f(x) \text{d}x$.

Then, by the Arzela-Ascoli Theorem, it holds that there exists a subsequence $\{ (u_{k}(\cdot,t) , \phi_{j} ) \}_{k \in \mathbb{N}}$ which converges uniformly to some function on $[0,T]$.

We also have the following inequality on $u_{m}$:

$||u_{m}(t)||^2 + \int^{t}_{0} ||\nabla u_{m}(\tau)||^2 \text{d}\tau \leq M_{T}$, for all $0 \leq t < T$, where $M_{T}$ is a constant independent of $m, t$.

Problem

In the paper I am reading, it is stated that from these two facts, we can use "the usual diagonal argument" to show that:

There exists a subsequence $\{u_{k}\}$ such that $u_{k}(t)$ converges to some $u(t)$ uniformly on $[0,T]$, in the weak topology of $L^{2}_{\sigma}(\Omega)$.

I understand, and have been able to confirm my self, that the sequence $\{ (u_{k}(\cdot,t) , \phi_{j} ) \}_{k \in \mathbb{N}}$ does converge uniformly to a function on $[0,T]$, but how do I show from this that the functions $u_{k}$ converge weakly in $L^{2}_{\sigma}$?

Is it possible to show that there exists some $u \in L^{2}_{\sigma}$ such that $\{ (u_{k}(\cdot,t) , \phi_{j} ) \}_{k \in \mathbb{N}} \rightarrow \{ (u(\cdot,t) , \phi_{j} ) \}$, for all $\phi_{j}$? If we could show this, then we definitely have weak convergence of $u_{k}$.

Attempted Answer

Due to the definition of $\{\phi_{j}\}_{j \in \mathbb{N}}$, we can show that $\{ (u_{m}(\cdot,t) , f ) \}_{m \in \mathbb{N}}$ converges uniformly with respect to $t \in [0,T]$ to some function in $L^{2}$, for arbitrary $f \in L^{2}(\Omega)$.

We claim that this limit is of the following form: $\{ (u_{m}(\cdot,t) , f ) \}_{m \in \mathbb{N}} \rightarrow \{ (u(\cdot,t) , f ) \}$ for some $u \in L^{2}_{\sigma}$. In which case, it is clear that $u_{m}$ converges to this $u$ weakly in $L^{2}_{\sigma}(\Omega)$.

We consider $u_{m}(\cdot, t)$ as a function $(u_{m}(\cdot, t), \cdot)$ in the dual space $(L^{2})^{\ast}$. Since $\{ (u_{m}(\cdot,t) , f ) \}_{m \in \mathbb{N}}$ converges uniformly with respect to $t \in [0,T]$, for all $f \in L^{2}(\Omega)$, we conclude that $(u_{m}(\cdot, t), f)$ converges pointwise with respect to $f$ to some function $u_{t}(f)$.

Is it possible from this to conclude that the function $u_{t}(f)$ is in fact in $(L^{2})^{\ast}$???

Solution 1:

Let us first summarise what you have: You have an equicontinous and uniformly bounded sequence $u_m: [0,T]\to L_\sigma^2(\Bbb R^n)$ and an ONB $\phi_i$ of $L^2_\sigma(\Bbb R^n)$. Then define:

$$\theta_{mi}:[0,T]\to \Bbb R, \quad t\mapsto (u_m(t),\phi_i).$$

Now for any fixed $i$ this gives a bounded family of equicontinuous functions. By the diagonal argument a sub-sequence $u_{m_k}$ so that $\theta_{m_ki}(t)$ converges to some $\theta_i(t)$ for any fixed $i$ as $k\to\infty$ (that is pointwise convergence in $t$). By Azerla Ascoli this convergence is uniform in $t$.

We define $u:[0,T]\to L^2_\sigma(\Bbb R^n)$ by $u(t)=\sum_i \theta_i(t)\,\phi_i$. (I will not check that this does define an element of $L^2$ for any $t$, that should be achievable by a standard argument.)

Now the question remains:

Let $f\in L^2_\sigma(\Bbb R^n)$, why does $(f,u_{m_k}(t)\,)$ converge to $(f,u(t)\,)$ uniformly in $t$?

To that end denote with $P_n$ the projection $$P_n(f)= \sum_{i≥n}(f,\phi_i)\,\phi_i,$$ note that $P_n(f)\to0$ for any $f\in L^2$ (ie this sequence of projection converges to zero in the strong operator topology). Further denote with $f_i=(f,\phi_i)$.

Now lets get dirty with the $\epsilon$'s. Denote with $M$ the uniform bound of $\sup_{t\in [0,T]}\|u_m\|_{L^2}$. Let $N$ be such that $\|P_n(f)\|<\frac\epsilon M$ for any $n>N$. Then:

$$|(f,u_{m_k}(t)-u(t)\,)|=\left|\sum_{i=0}^N f_i\,(\theta_{m_ki}(t)-\theta_i)(t) + (P_{N+1}(f),u_{m_k}(t)-u(t)\,)\right|$$ apply the triangle inequality and Cauchy-Schwarz to get: \begin{align} |(f,u_{m_k}(t)-u(t)\,)|&≤\sum_{i=0}^N |f_i|\,|\theta_{m_ki}(t)-\theta_i(t)|+\|P_{N+1}(f)\|\cdot\|u_{m_k}(t)-u(t)\|\\ &≤\sum_{i=0}^N |f_i|\,\sup_{\tau\in[0,T]}\|\theta_{m_ki}(\tau)-\theta_i(\tau)\|+\frac\epsilon M (2M). \end{align}

Now the sum on the left is a finite sum of things converging to $0$ in a way independent of $t$, and may thus independently of $t$ be bounded by $\epsilon$. The term on the right is $2\epsilon$. Thus you have a bound of $|(f, u_{m_k}(t)-u(t)\,)|$ by $3\epsilon$ which works independently of $t$, meaning that it converges to $0$ uniformly. This works for any $f$, so according to your definition this means that $u_{m_k}$ converges uniformly in the weak topology to $u$.