Let $\{f\_n\}$ be a sequence of functions that are continuous and lying in $H^k(\mathbb{R}^m)$. Assuming $k>\frac{m}{2}$, and if $f\_n \to f$ pointwise, where $f\in H^k(\mathbb{R}^m)$, such that $f$ has points of isolated disconitnuty on a dense set of measure zero. Sobolev embedding says that $f\_n$ cannot converge to $f$ under the norm $\|.\|\_{H^k(\mathbb{R}^m)}$. I also believe and want to show that $\|f\_n\|\_{H^k(\mathbb{R}^m)}$ grows unbounded. How can I prove that and also $$\|f\_n\|\_{H^k(\mathbb{R}^m)} \to \infty$$ Appreciate some suggestions.


PS : By isolated discontinuties, I mean $f$ is such that it can be obtained from a continuous function $g \in H^k(\mathbb{R}^m)$, by changing its values on a dense set of measure zero.



Unboundedness of the Sobolev norm of a sequence : Does it follow from Sobolev embedding?
In fact, for $k > \frac{m}{2}$, any sequence of continuous functions $f\_n$ which is bounded in $H^k$ norm must have a subsequence converging uniformly on compact sets. This proves your claim, since if $f\_n$ converges pointwise to a discontinuous function $f$, then every subsequence converges pointwise to the same $f$ and the convergence cannot be uniform. (It doesn't matter whether the discontinuities of $f$ are isolated or dense or what have you.)


To see why this fact is true, first, by multiplying by a smooth cutoff function, we may assume that all the $f\_n$ are supported inside some open ball $\Omega$. Now note that Sobolev embedding says that $H^k\_0(\Omega)$ is continuously embedded in some Hölder space $C^{l, \alpha}(\Omega)$ for appropriate values of $l,\alpha$, so $f\_n$ is bounded in $C^{l,\alpha}$ norm. Using Arzelà–Ascoli, you can show that such a sequence has a subsequence converging uniformly on $\overline{\Omega}$.


In other words, $H^k\_0(\Omega)$ is *compactly* embedded in $C\_0(\Omega)$.


