An inclusion in Sobolev-Bochner spaces for an open set $\Omega \subset \mathbb{R}^2$ smooth enough

Question

Let $\Omega \subset \mathbb{R}^2$ be a bounded open set with smooth boundary.

Assume that

$u \in C(0,T;H^1(\Omega)) \bigcap L^2(0,T;H^2(\Omega))$ and $\partial_tu \in L^2(\Omega \times (0,T)).$

Why does it hold that $$ u \in L^\infty(\Omega \times (0,T)) $$

As I understand, Aubin-Lions compactness Lemma and the assumption on $\Omega$ only guarantee that $$ L^2(0,T;H^2(\Omega)) \hookrightarrow L^2(0,T;L^\infty(\Omega)) $$ Am I missing something?

Answer 1

I beleive so,The Aubin-Lions compactness lemma, in combination with the assumption on $Ω$, guarantees the embedding $L^2(0,T;H^2(Ω)) \hookrightarrow L^2(0,T;L^{\infty}(Ω))$. However, this does not directly imply that $u ∈ L^\infty(Ω×(0,T))$.

The assumption $u \in C(0,T;H^1(Ω)) \cap L^2(0,T;H^2(Ω))$ implies that $u$ is continuous in time and belongs to the space H1(Ω) with its time derivative belonging to $L^2(Ω×(0,T))$. However, this does not provide sufficient regularity to conclude that $u$ is bounded in $Ω × (0, T)$ and thus in $L^{\infty}(Ω × (0, T))$.

To establish the $L^{\infty}(Ω × (0, T))$ regularity of $u$, you would typically need additional assumptions on the problem or the solution itself, such as elliptic regularity estimates, further regularity of the boundary $∂Ω$, or additional regularity assumptions on the coefficients of the underlying partial differential equation.

the statement that $u ∈ L^{\infty}(Ω × (0, T))$ cannot be deduced solely from the given assumptions. Further assumptions or regularity results specific to the problem at hand would be needed to establish the desired regularity.

Answer 2

Here's an explicit counterexample.

Let $a_k$ be a sequence of positive real numbers satisfying:

• $\sum a_k^2 2^{4k}$ converges
• $\sum a_k 2^{2k}$ diverges

(for example, you can choose $a_k 2^{2k} = \frac1k$.)

Let $\phi_0$ be a smooth $\mathbb{R}\to[0,1]$ function satisfying $$ \phi_0(t) = \begin{cases} 1 & t \leq 1 \\ 0 & t \geq 4 \end{cases}$$ and define $\phi_k(t) = \phi_0(2^{2k} t)$; note that $\phi_k(t) = 0$ for all $t \geq 2^{2-2k}$.

Let $\chi_k$ be $\mathbb{R}^2\to\mathbb{R}$ given by $$ \chi_k(\xi) = \begin{cases} 1 & |\xi| \in [2^k, 2^{k+1}) \\ 0 & \text{otherwise} \end{cases} $$ Note that $\chi_k$ has $L^2$ norm $2^k$; and hence also its Fourier transform.

Finally, define $$ u(t,x) = \sum_{k = 1}^\infty a_k \phi_k(t) \check{\chi_k}(x) $$ where $\check{f}$ denotes the inverse Fourier transform of $f$.

Notice that for any $t_0 > 0$, the restriction $u |_{[t_0, 1)\times \{|x| < 1\}}$ is a bounded smooth function (since it is a sum of finitely many smooth functions). Set $\Omega$ to be the unit ball in $\mathbb{R}^2$.

We can compute (taking advantage of the fact that $\chi_k$ are pairwise orthogonal) that the $L^2(0,1;H^2(\Omega))$ and $L^\infty(0,1;H^1(\Omega))$ norms of $u$ are both no more than $\left(\sum a_k^2 2^{4k}\right)^{\frac12}$.

Similarly, noting that $\frac{d}{dt} \phi_k(t)$ has size bounded by $2^{2k}$ and supported on a set of length $2^{-2k}$, we find that the $L^2(0,1;L^2(\Omega))$ norm of $\partial_t u$ is also bounded by $\left(\sum a_k^2 2^{4k}\right)^{\frac12}$.

But for a given $t\in (0,1)$, let $K$ be the smallest integer such that $2^{2-2K} \leq t$, and we find $$ u(t,0) = \sum_{k = 1}^K a_k \int_{\mathbb{R}^2} \chi_k(\xi) ~d \xi = \sum_{k = 1}^K a_k 2^{2k} $$ As we send $t\searrow 0$ we find $K\nearrow +\infty$ and hence $u$ is not a bounded function on $(0,1)\times \Omega$.