Unboundedness of the Sobolev norm of a sequence of functions.

Question

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.

Answer 1

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 $k > \frac{m}{2}$.