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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.

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    $\begingroup$ Watch out: The way you construct $f$ from $g$ implies that $f$ and $g$ represent the same equivalence class in $H^k(\mathbb{R}^m)$. So the fact that $f$ is discontinuous does not imply anything about $(f_n) \not\to f$ in $H^k(\mathbb{R}^m)$ here. Otherwise your statement $f \in H^k(\mathbb{R}^m)$ would also be bogus. $\endgroup$
    – Hannes
    Oct 21 '18 at 11:58
  • $\begingroup$ @Hannes : looks like you have missed the key word "pointwise" in my question. There are two types of convegwnce i am talking about in this question. One pointwise and other in the norm and i have mentioned them appropriately. $\endgroup$
    – Rajesh D
    Oct 21 '18 at 12:49
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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 $C_0(\Omega)$.

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This is not an answer but to show what I have worked till now, so that it would help someone who wants to answer. (I am not hoping for any upvotes).

I can show this, when the points of discontinuity are not dense. I just need to prove for single discontinuity at it would suffice for any number of discontinuities but only when they are not dense. But if I say that it would suffice for the case of points of discontinuity being dense, then it would be a clear case of hand-waving rather than proof. The question puzzling me is, how can I extend it to dense points, by doing a few extra things.

proof :

Consider a sequence of continuous functions $h_n \in C^0\cap H^k(\mathbb{R}^m)$ and let $h_n\to g $ pointwise, where $g$ is a continuous version of $f$ (I mean from equivalence class of $f$). Now we try to construct $f_n$ from $h_n$ by way of adding shrinking bumps of appropriate amplitude, so that $f_n \to f$ pointwise. (Idea is to add a small perturbation in the form of a shrinking bump, to produce a simple discontinuity in the limit function). Lets add a small bump function $\psi(n\boldsymbol{x})$ to $h_n(\boldsymbol{x})$ to form the desired new sequence $$f_n(\boldsymbol{x}) = \psi(n\boldsymbol{x}) + h_n(\boldsymbol{x}) $$ Now we show that, in doing so, we blow up the norm. For simplicity, assume $\psi_n(\boldsymbol{x}) = \psi(n\boldsymbol{x})$ is radially symmetric. With a change of variable $\boldsymbol{t} = n\boldsymbol{x}$ we can easily see that $$\|f_n(\boldsymbol{x}) + \psi(n\boldsymbol{x})\|_{L^2(\mathbb{R}^m)} \to \|f\|_{L^2(\mathbb{R}^m)}$$ But when we consider the other term of the norm, again with a change of variable $\boldsymbol{t} = n\boldsymbol{x}$, we can see that $$\begin{align} \int_{\mathbb{R}^m} |\frac{\partial^k{f_n}}{\partial{x_i^k}}|^2 \mathop{}\!\mathrm{d}^m\boldsymbol{x} & = \int_{\mathbb{R}^m}|\frac{\partial^k{h_n}}{\partial{x_i^k}}|^2 \mathop{}\!\mathrm{d}^m\boldsymbol{x} + 2\int_{\mathbb{R}^m}\frac{\partial^k{h_n}}{\partial{x_i^k}} \frac{\partial^k{\psi_n}}{\partial{x_i^k}}\mathop{}\!\mathrm{d}^m\boldsymbol{x} + \int_{\mathbb{R}^m}|\frac{\partial^k{\psi_n}}{\partial{x_i^k}}|^2 \mathop{}\!\mathrm{d}^m\boldsymbol{x} \\\\ & = \|\frac{\partial^k{h_n}}{\partial{t_i^k}}\|_{L^2}^2 + O(n^{(k-m)}\|\frac{\partial^k{h_n}}{\partial{t_i^k}}\|_{L^2} \|\frac{\partial^k{\psi}}{\partial{t_i^k}}\|_{L^2}) + n^{(2k-m)}\|\frac{\partial^k{\psi}}{\partial{t_i^k}}\|_{L^2}^2\end{align} $$

The last term becomes unbounded, when $k > \frac{m}{2}$.

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