If A is continuously embedded in B and $||u_{n}||_{A} \rightarrow +\infty$, can I say that $||u_{n}||_{B}\rightarrow +\infty?$

Question

Consider the Sobolev space $H^{1}_{0}(\Omega)$, it is well known that there exists a constant $C>0$ such that

$$|u|_{L^{2}(\Omega)}\leq C||u||_{H^{1}_{0}(\Omega)}.$$

Consequently $H^{1}_{0}(\Omega) \hookrightarrow L^{2}(\Omega)$ continuously, i.e, if $u_{n} \rightarrow u$ in $H^{1}_{0}(\Omega)$ then $u_{n} \rightarrow u$ in $L^{2}(\Omega)$. But, what does happens to $(u_{n})$ in $L^{2}(\Omega)$ if $||u_{n}||_{H^{1}_{0}(\Omega)} \rightarrow +\infty$ ? Does exists $|u_{n}|_{L^{2}(\Omega)} \rightarrow +\infty$? If not, what the counterexample?

I have no clue about the question above, however, the statement seems to be true in fractional Sobolev spaces. As example, consider the fractional Sobolev space $X^{1/2}(\Omega)$ defined as follow.

\begin{equation*} X^{1/2}(\Omega) := \left\lbrace u \in L^{2}(\mathbb{R}); \displaystyle\iint_{\Omega \times \Omega}\dfrac{|u(x)-u(y)|^{2}}{|x-y|^{2}}dxdy < \infty\right\rbrace, \end{equation*}

The space $X^{1/2}(\Omega)$ is a Hilbert space and $X^{1/2}(\Omega)\hookrightarrow L^{2}(\Omega)$ continuously, the norm of $X^{1/2}(\Omega)$ is given by \begin{equation*} ||u||_{X^{1/2}(\Omega)} = \left( \displaystyle\iint_{\Omega \times \Omega}\dfrac{|u(x)-u(y)|^{2}}{|x-y|^{2}}dxdy\right)^{\frac{1}{2}}. \end{equation*}

In this case, if $||u_{n}||_{X^{1/2}(\Omega)} \rightarrow +\infty$, there exists $B \subset \Omega$ such that $|u_{n}(x)| \rightarrow +\infty$ a.e in $B$ (I'm not completely sure about this statement). Then, by Fatou's Lemma, we obtain \begin{equation} \int_{\Omega}|u_{n}|^{2} \geq \int_{B}|u_{n}|^{2} \rightarrow +\infty. \end{equation}

Moreover, can the answer to these questions be generalized for any normed spaces?

Thanks in advance.

Answer 1

In $H^{1}_{0}(\Omega)$ this is not true.

Let $\Omega=(0,1) \subset \mathbb{R}$ and consider the sequence $f_{n}$ of piecewise linear bumps of height $\frac{1}{n}$ on intervals of length $\frac{1}{n^2}$.

We obviously have $||f_{n}||_{L^{2}(\Omega)} \rightarrow 0$ as $n \rightarrow \infty$

On the other hand, you can check that $f_{n}$ is weakly differentiable in $\Omega$ with $|f_{n}'(x)|=2n$ for all $x$ and thus $||f_{n}'||_{L^{2}(\Omega)} \rightarrow \infty$ as $n \rightarrow \infty$

Alltogether, $||f_{n}||_{H^{1}_{0}(\Omega)} \rightarrow \infty$ but $||f_{n}||_{L^{2}(\Omega)} \rightarrow 0$