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 $n \in \mathbb{N}$;
- $f_n \rightarrow 0$ a.e. in $(0,T)\times \mathbb{R}^{d}$;
- $f_n \rightarrow 0$ in $L^2(0,T;L^2(K))$ for all compact $K \subset \mathbb{R}^{d}$.
My question: Is it true that $f_n \rightarrow 0$ in $L^2(0,T;L^2(\mathbb{R}^{d}))$?
I was trying to work with the sequence of compacts $B[0,k+1]\setminus B(0,k)$ and write $(0,T)\times \mathbb{R}^{d}=\bigcup_{k=1}^{\infty} (0,T) \times (B[0,k+1]\setminus B(0,k))$ to get an estimate like $$\int_{(0,T)\times (B[0,k+1]\setminus B(0,k))} |f_n(x,t)|^2 \, dx dt <\frac{\varepsilon}{2^k} \tag{*}$$ for all $\varepsilon>0$ in order to use the following result:
Theorem: If $\int_E f$ exists and $E=\bigcup_k E_k$ is the countable union of disjoint measurable sets $E_k$, then $\int_E f=\sum_k \int_{E_k} f$.
But I don't know if I can use this result for vector integrations. Moreover, the estimate $(*)$ only holds for a $k\geq k_0(n)$, that is, the constant $k_0$ depends on $n$.
This isn't necessarily true. We can just let $\phi$ be some smooth compactly supported function on $(0, T) \times \mathbb{R}^d$ and then let $f_n(t,x) = \phi(t, x - nv)$ where $v$ is some nonzero vector in $\mathbb{R}^d$ be the translations of $\phi$ in space. Then: \begin{align*} &(1) \ \ \ \ \|{f_n}\|_{L^2([0, T]; L^2(\mathbb{R}^d))}^{2} = \int_0^T \|f_n(t)\|_{L^2(\mathbb{R}^d)}^{2}\, dt = \int_0^T \|\phi(t)\|_{L^2(\mathbb{R}^d)}^{2}\, dt =: C_T \\ &(2) \ \ \ \ f_n \to 0 \ \text{a.e., since for any} \ (t, x) \in [0, T] \times \mathbb{R}^d, f_n(t, x) \to 0 \\ &(3) \ \ \ \ f_n \to 0 \ \text{in} \ L^2([0, T]; L^2(K)) \ \forall K \subset \mathbb{R}^d \ \text{compact}. \end{align*} The last point follows since $f_n(t)|_K = 0$ for all $t$ for $n$ large enough (depending on $K$), as $\phi$ has compact support.
The computation in item (1) shows that the $L^2([0, T]; L^2(\mathbb{R}^d))$ norm of each $f_n$ is always $C_T$ hence the sequence cannot approach zero in this space.