A Question on certain Hilbert space of continuous functions, and a characteristic of convergence in it

Define $T^k(\Omega)$, $\Omega$ an open subset of $\mathbb{R}^m$ (with a smooth boundary), as a space of function equivalance classes, with the norm defined as $$\|f\|_{T^k(\Omega)}^2 = \|f\|_{L^2(\Omega)}^2 + \|(\sum\limits_{i=1}^m(\frac{\partial^{k}f}{\partial x_i^{k}})^2)^{\frac{1}{2}}\|_{L^2(\Omega)}^2$$

It can be easily noted that $T^k(\Omega)$ is a Hilbert space. Also note that this norm is not a Sobolev norm, as we don't consider cross derivatives.

Consider the set $$M = C^0(\bar{\Omega}) \cap T^k(\Omega)$$

Prove that :

If $k > \frac{m}{2}$, every sequence $\{f_n\},f_n \in M$, that converges in the norm $\|.\|_{T^k(\Omega)}$, also converges in the norm $\|.\|_{C^0(\bar{\Omega})}$ (and to a limit $f \in M$)

Proof :

Consider a sequence $f_n \in M$ and let $f_n\to f \in M$ in the norm $\|.\|_{T^k(\Omega)}$ Idea is to add a small perturbation in the form of a shrinking bump, to produce a simple discontinuity in the limit function $f$. Lets add a small bump function $\psi(n\boldsymbol{x})$ to $f_n(\boldsymbol{x})$ to form a new sequence $$\phi_n(\boldsymbol{x}) = \psi(n\boldsymbol{x}) + f_n(\boldsymbol{x})$$ Now we show that, in doing so, we blow up the norm and spoil the convergence of the sequence. 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 $$\|f_n(\boldsymbol{x}) + \psi(n\boldsymbol{x})\|_{L^2(\Omega)} \to \|f\|_{L^2(\Omega)}$$ 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_{\Omega} |\frac{\partial^k{\phi_n}}{\partial{x_i^k}}|^2 \mathop{}\!\mathrm{d}^m\boldsymbol{x} & = \int_{\Omega}|\frac{\partial^k{f_n}}{\partial{x_i^k}}|^2 \mathop{}\!\mathrm{d}^m\boldsymbol{x} + 2\int_{\Omega}\frac{\partial^k{f_n}}{\partial{x_i^k}} \frac{\partial^k{\psi_n}}{\partial{x_i^k}}\mathop{}\!\mathrm{d}^m\boldsymbol{x} + \int_{\Omega}|\frac{\partial^k{\psi_n}}{\partial{x_i^k}}|^2 \mathop{}\!\mathrm{d}^m\boldsymbol{x} \\\\ & = \|\frac{\partial^k{f_n}}{\partial{t_i^k}}\|_{L^2}^2 + O(n^{(k-m)}\|\frac{\partial^k{f_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 blows up, when $k > \frac{m}{2}$. So one cannot produce a discontinuity by way of adding a shrinking bump. Hence all sequences in $M$ that converge in the norm $\|.\|_{T^k(\Omega)}$ also converge in the norm $\|.\|_{C^0(\bar{\Omega})}$

Other Cases

1. For a jump discontinuity, we let the bump have a flatter region and we shrink only the transition region. Same logic applies here.
2. For a blow up situation, consider $\phi_n(\boldsymbol{x}) = g(n)\psi(n\boldsymbol{x})$, where $g(n) = \omega(1)$ (Bachmann–Landau notations), which means $g(n)$ grows faster than 1, or $\lim\limits_{n\to\infty}g(n) = \infty$. In this case, one can see that the last term in the RHS of the last equation in my proof, blows up when $k\ge\frac{m}{2}$. Hence blow up discontinuity is also ruled out in case of $k>\frac{m}{2}$.
3. Case of Oscillatory: For oscillatory case consider $\phi_n(\boldsymbol{x}) = \sin(n\boldsymbol{x})\psi_(n\boldsymbol{x})$

Question : Is the above proof complete?(if at all it makes sense) Is there any simpler proof with a direct application of a well known theorem? Similar references appreciated.For $\Omega = [0,1)^m$, with periodic boundary conditions, this norm is equivalent to Sobolev norm, hence by general Sobolev inequality, the result follows. But I am seeking a direct proof as in this approach, without invoking Sobolev inequality.

• You've only ruled out one possible mechanism by which $C^0$ convergence could fail. – user101142 Aug 15 '18 at 15:25
• @user37208 : You mean a jump discontinuty? For a jump, we let the bump have a flatter region and we shrink only the transition region. Same logic applies there if I am not wrong. – Rajesh Dachiraju Aug 15 '18 at 15:41
• @user37208 I need some thinking about the possibility of blow up. – Rajesh Dachiraju Aug 15 '18 at 16:02
• As it turns out, it is not necessary to use "mixed" partial derivatives to define (global) Sobolev spaces. Powers of the Laplacian suffice. But/and I think you may have some typos, since you'd want the $L^2$ Sobolev norms to behave Hilbert-space-ly, so you'd need something like $|f|^2_1=\langle (1-\Delta)f,f\rangle$, that is, with slightly different exponents than you'd written. Beyond that, isn't the question you're asking exactly about a Sobolev inequality? Is it something different that I'm not catching? – paul garrett Aug 16 '18 at 0:44
• @paulgarrett : Its not yet proven that it is equivlent to Sobolev norm for a bounded domain, even if it has a smooth boundary. I am interested in $(0,1)^m$, and have to wait for response from Willie to see if the equivalence to Sobolev norm holds in this case. In that case. – Rajesh Dachiraju Aug 16 '18 at 0:53