# Unboundedness of the Sobolev norm of a sequence : Does it follow from Sobolev embedding?

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.

• 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. Commented Oct 21, 2018 at 11:58
• @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. Commented Oct 21, 2018 at 12:49

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 $$C_0(\Omega)$$.

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