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 : 


*

*For a jump discontinuity, we let the bump have a flatter region and we shrink only the transition region. Same logic applies here.

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

*Case of Oscillatory discontinuity remains: 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.
 A: I don't think the proof is complete. You have a Banach space $X$ with a norm
and you have a sequence $\{x_{n}\}_{n}$ which converges to some $x$. In
particular, the sequence is bounded so $\Vert x_{n}\Vert\leq M$ for all $n$.
You are considering a sequence $\{y_{n}\}_{n}$ such that $\Vert y_{n}
\Vert\rightarrow\infty$. Then by the triangle inequality
$$
\Vert x_{n}+y_{n}\Vert\geq\Vert y_{n}\Vert-\Vert x_{n}\Vert\geq\Vert
y_{n}\Vert-M
$$
and so you obtain that
$$
\lim_{n\rightarrow\infty}\Vert x_{n}+y_{n}\Vert\geq\lim_{n\rightarrow\infty
}\Vert y_{n}\Vert-M=\infty.
$$
So you have shown that $\Vert x_{n}+y_{n}\Vert\rightarrow\infty$ but that does
not say anything about the original sequence.
The fact that in your case your norm is the sum of two $L^{2}$ norms is not
important. As you computed, if you denote $\psi_{n}(x)=\psi(nx)$, you have
$$
\Vert\psi_{n}\Vert_{T^{k}(\Omega)}\rightarrow\infty,
$$
and so by what I wrote above, $\Vert f_{n}+\psi_{n}\Vert_{T^{k}(\Omega
)}\rightarrow\infty$, but you cannot conclude anything about the original sequence.
Concerning your approach, while for functions of one variable there are simple ways to classify discontinuities (left limit and right limit exist but they are different, left limit and right limit exist and equal but different from the value of the function at the point, one of the left or right limit does not exist), for functions of several variables the situation is much more difficult since you can approach a point in a lot of different ways. In two dimensions you can approach a point along any curve passing through it. So when you say that in the oscillatory case, which I guess means 
$$\liminf_{\bf{x}\to\bf{x}_0}f(\bf{x})<\limsup_{\bf{x}\to\bf{x}_0}f(\bf{x}),$$
all you know is that there are two sequences approaching $\bf{x}_0$ along which the limits are different. So your perturbation would need to be tailored on those two sequences but you have little or no information of where they are. So your "oscillation" $\sin (n \bf{x})$ (which I don't know what it means in the vectorial case) would have to follow the two sequences and so it cannot be that simple.
Also, since $f$ is an equivalence class of functions and not a function, the discontinuities change with the representative you choose. Hence, given an equivalence class of functions $[f]$, you have no way to classify the discontinuities, since given a representative you can modify it on a countable dense set and change all its discontinuity points.  
As for the original problem, it seems that you would need to use the Morrey embedding theorem for the super critical case, but that would imply that you would need to prove that your space coincides with $H^k(\Omega)$. To be honest I would not know how to start to prove it and even if it is true or not (take a look at this article mixed derivatives). The closest thing I can think of for $k=2$ is an identity that relates the Laplacian to the full Hessian. See the reply 
Laplacian-Hessian
but it only applies in $\mathbb{R}^m$ and you need density of smooth functions. There are version of this identity for bounded convex domains (it's in Grisvard book), but you need the function to have trace zero on the boundary and in your space there is no easy way to define a trace. So I don't really know. Sorry I cannot be of more help.
EDIT: If you are in the entire space, say $\mathbb{R}^{2}$, then using Fourier
transforms you have that
\begin{align*}
\int_{\mathbb{R}^{2}}\left\vert \frac{\partial^{2}f}{\partial x^{2}%
}\right\vert ^{2}dxdy  & =\int_{\mathbb{R}^{2}}\xi_{1}^{2}|\hat{f}(\xi_{1}%
,\xi_{2})|^{2}d\xi_{1}d\xi_{2},\\
\int_{\mathbb{R}^{2}}\left\vert \frac{\partial^{2}f}{\partial y^{2}%
}\right\vert ^{2}dxdy  & =\int_{\mathbb{R}^{2}}\xi_{2}^{2}|\hat{f}(\xi_{1}%
,\xi_{2})|^{2}d\xi_{1}d\xi_{2},
\end{align*}
and so
\begin{align*}
\int_{\mathbb{R}^{2}}\left\vert \frac{\partial^{2}f}{\partial x\partial
y}\right\vert ^{2}dxdy  & =\int_{\mathbb{R}^{2}}|\xi_{1}||\xi_{2}||\hat{f}%
(\xi_{1},\xi_{2})|^{2}d\xi_{1}d\xi_{2}\\
& \leq\frac{1}{2}\int_{\mathbb{R}^{2}}(|\xi_{1}|^{2}+|\xi_{2}|^{2})|\hat
{f}(\xi_{1},\xi_{2})|^{2}d\xi_{1}d\xi_{2}\\
& =\frac{1}{2}\int_{\mathbb{R}^{2}}\left\vert \frac{\partial^{2}f}{\partial
x^{2}}\right\vert ^{2}dxdy+\frac{1}{2}\int_{\mathbb{R}^{2}}\left\vert
\frac{\partial^{2}f}{\partial y^{2}}\right\vert ^{2}dxdy.
\end{align*}
But in a bounded domain this is no longer true. Take $f(x,y)=xy$. Then
$\frac{\partial^{2}f}{\partial x^{2}}=\frac{\partial^{2}f}{\partial x^{2}}=0$. 
