Does the convergence to 0 in $L^2(0,T;L^2(B))$ where $|\mathbb{R}^{d}-B|<\infty$ imply the convergence in $L^2(0,T;L^2(\mathbb{R}^{d}))$?

Question

Definition: One denotes $L^p(a,b;X)$, $1\leq p<+\infty$, the set of (classes of) measurable functions $u:(a,b)\rightarrow X$ such that $$\|u \|_{L^p(a,b;X)}:= \left(\int_{a}^{b}\|u(t)\|^p_X dt\right)^{\frac{1}{p}}<\infty .$$

Let $(f_n)$ be a sequence in $L^2(0,T;L^2(\mathbb{R}^{d}))$ such that:

• $\|f_n\|_{L^2(0,T;L^2(\mathbb{R}^{d}))} \leq C_T$ for all $n \in \mathbb{N}$; (or $\|f_n(t)\|_{L^2(\mathbb{R}^{d})}\leq C_T$ for all $t \in (0,T))$
• $f_n \rightarrow 0$ a.e. in $(0,T)\times \mathbb{R}^{d}$;
• $f_n \rightarrow 0$ in $L^2(0,T;L^2(K))$ for all compact $K \subset \mathbb{R}^{d}$.
• $f_n \rightarrow 0$ in $L^2(0,T;L^2(B))$ where $B \subset \mathbb{R}^{d}$ and $|\mathbb{R}^{d}-B|<\infty;$

My question: Is it true that $f_n \rightarrow 0$ in $L^2(0,T;L^2(\mathbb{R}^{d}))$ or that there is a subsequence of $f_n$ that converges to $0$ in $L^2(0,T;L^2(\mathbb{R}^{d}))$? Or what additional hypothesis do I need for this convergence to be true?

Note that $$\int_{0}^{T}\int_{\mathbb{R}^{d}} |f_n(x,t)|^2\,dxdt=\int_{0}^{T}\int_{B} |f_n(x,t)|^2\,dxdt+\int_{0}^{T}\int_{\mathbb{R}^{d}-B} |f_n(x,t)|^2\,dxdt=I_1+I_2.$$ Observe that $I_1 \rightarrow 0$ by virtue of the fourth hypothesis above.

It remains to prove that $I_2 \rightarrow 0$. From Egorov's Theorem, given $\varepsilon>0$ there exists a closed subset $F \subset (0,T)\times (\mathbb{R}^{d}-B)$ such that $|((0,T)\times (\mathbb{R}^{d}-B))-F|<\varepsilon|$ and $f_n\rightarrow 0$ uniformly in $F$, which implies that $f_n \rightarrow 0$ in $L^2(F)$. Thus, we obtain \begin{eqnarray} I_2&=&\int_{0}^{T}\int_{\mathbb{R}^{d}-B} |f_n(x,t)|^2\,dxdt\\ &=& \int_{F} |f_n(x,t)|^2\,dxdt+\int_{((0,T)\times(\mathbb{R}^{d}-B))-F} |f_n(x,t)|^2\,dxdt. \end{eqnarray}

Is there any way to prove that the $\int_{((0,T)\times(\mathbb{R}^{d}-B))-F} |f_n(x,t)|^2\,dxdt$ integral converges to 0?

Answer 1

Here's a counterexample demonstrating that the given assumptions aren't enough to ensure that $f_n \to 0$ in $L^2(0, T; L^2(\mathbb{R}^d))$, or even that a subsequence converges to $0$.

First, let $(y_n)$ be a sequence of points in $\mathbb{R}^d$ with, say, $| y_n | = 2^n$, and let $B_n = B(y_n, 2^{-n})$. Then define $f_n = 2^{nd/2} \chi_{(0, T) \times B_n}$ and $B = \left( \bigcup_n B_n \right)^c$.

These functions have constant $L^2(0, T; L^2(\mathbb{R}^d))$ norm, and it's not too hard to check the other conditions are also satisfied. In particular, the $f_n$ are zero off of $B$, so we trivially have $f_n \to 0$ in $L^2(0, T; L^2(B))$. But because of the constant norms, neither the full sequence nor any subsequence goes to $0$.

Heuristically, this is a case of mass escaping to infinity, and so some additional conditions would be needed to preclude that possibility.

For example, if $B^c$ were a bounded set (i.e. with compact closure), that would make the result true using the assumption about convergence involving compact sets. Another possibility would be if you could dominate the $f_n$ by a single $L^2(0, T; L^2(\mathbb{R}^d))$ function.

Those both strike me as relatively crude conditions for ensuring the desired convergence, though, in that they would allow at least one of the bulleted assumptions to be dropped.