Show that $\lim_n \|\partial^s (f_n - g_n)\|_p = 0$ (no homework...) the setting is as follows: 
Let $\Omega \subset \mathbb{R}^m$ be open and consider some $L^p(\Omega)$ which I will shortly write as just $L^p$ from now on.
Furthermore let (for some $k \in \mathbb{N}$) $\widetilde{X}$ denote the smallest completion of $$X:=\left \{ F \in C^\infty(\Omega,\mathbb{R}) ;\ \|f\|<\infty,\ \|f\|:=\sum_{|s|\leq k} \|\partial^sf\|_p \right\}$$
$\widetilde{X}$ is a subset of $W^{k,p}$, that is the set of functions in $L^p$ that satisfy:
$$\forall \varphi \in C^\infty_0(\Omega,\mathbb{R}): \forall s \in \mathbb{N}^m \  \text{with}\ |s|\leq k: \exists f^{(s)} \in L^p:
\int_\Omega f\partial^s\varphi\,d\lambda=\int_\Omega \varphi f^{(s)}\,d\lambda$$
In H.W. Alt's Textbook "Lineare Funktionalanalysis" and in my lecture the statement is, that 
$J: \widetilde{X} \rightarrow W^{k,p}, [(f_n)] \mapsto \lim_n f_n$, where the limit in $L^p$ is meant and $[(f_n)]$ denotes the equivalence class belonging to the Cauchy sequence obtained by the completion process of $X$, is injective.
So one has to show that 
$$J([(f_n)])=J([(g_n)]) \implies [(f_n)]=[(g_n)] \overset{\text{by}\, \text{def}.}{\iff} 0=\lim_n \|f_n-g_n\|\\=\lim_n \sum_{|s|\leq k} \|\partial^s(f_n-g_n)\|_p$$
But how does $\lim_n f_n = \lim_n g_n$ in $L^p$ imply this?
Obviously, by definition of the norm on $X$, the title statement has to be shown (with the setting provided in this description...).
Can anybody help me with the proof?
I've tried to prove this myself, but didn't get any results... please note, that we haven't had convolution yet - but if there is no other way to prove this than using convolution I will still be glad to hear an answer, of course, just so I know I can't yet understand the proof. 
It is kind of hard to look for this kind of question, I've tried though and did not find anything. I guess this could be interesting for anyone learning functional analysis...
 A: For (any Cauchy-sequence) $(f_n) \in {X}^\mathbb{N}$ and arbitrary multiindex $|s|\leq k$, the definition of the norm on $X$ quite directly implies the existence of $f^{(s)}$, s.t. $\|\partial^{(s)} f_n - f^{(s)}\|_p \overset{n \rightarrow \infty}{\longrightarrow} 0$, especially $\| f_n - f^{(0)}\|_p \overset{n \rightarrow \infty}{\longrightarrow} 0$. Define $f:=f^{(0)}$.
It can be shown, that $$\forall \varphi \in C^\infty_0\colon \int_\Omega f \partial^{(s)}\, \varphi\, d\lambda = (-1)^{|s|}\int_\Omega \varphi\,f^{(s)}\,d\lambda$$
These $f^{(s)}$ are uniquely determined by $f=\lim_n f_n=\lim_n g_n$ (this is another theorem, which will not be proven here, see H.W. Alt's book which is also available in English). Using the above for $(g_n)$ thus, due to this uniqueness property, yields 
$$\|\partial^{(s)}f_n - \partial^{(s)}g_n\|_p\leq \|\partial^{(s)}f_n - f^{(s)}\|_p + \|f^{(s)} - \partial^{(s)}g_n\|_p\overset{n \rightarrow \infty}{\longrightarrow} 0\,,$$ since $\partial^{(s)}f_n\,,\,\partial^{(s)}g_n\overset{n \rightarrow \infty}{\longrightarrow}f^{(s)}$.
It may be noteworthy to add (for people like me, who are new to this...): 


*

*In $W^{k,p}$, equality means equality $\lambda$-almost everywhere.

*For continuous functions, equality $\lambda$-almost everywhere is equivalent to equality everywhere. So for each partial derivative of any $f \in C^\infty$, the equivalence class of $f$ - with respect to the equivalence of being identical $\lambda$-a.e. - can be identified with $f$ itself, so that $\partial[f]:=[\partial f]$ is well-defined.

