Here $U$ is an open subset of $\mathbb{R}^n$.
Above is a theorem regarding approximation of Sobolev functions in Evans's Partial Differential Equations. When I tried to the recover the proof on my own, I found that the proof might be much shorter than the one in the book. But it looks too simple to be true and I'm wondering if there is a big gap there.
Here is my argument. Suppose $u\in W^{k,p}(U)$. Then $u\in L^p(U)$ by definition and thus $u\in L^1(U)$ since $U$ is bounded. Now according to the answer and comments to the following questions:
- convolutions and mollification of functions in $L^1_{\text{loc}}(\Omega)$
- Properties of mollification for integrable functions
one can define the mollification $u^\epsilon=\eta_\epsilon*u$ on $U$ such that $u_\epsilon\in C^\infty(U)$. Moreover, since $$ D^\alpha u^\epsilon=\eta_\epsilon*D^\alpha u \quad\textrm{in } U, $$ one has $$ \|u^\epsilon-u\|_{W^{k,p}(U)}^p=\sum \|D^\alpha u^\epsilon-D^\alpha u\|_{L^p(U)}^p\to 0. $$
Could anyone identify if there is any serious mistake in the above argument?
[Added:]A possible naive analogy I make in the above argument is as the following. First of all, I have the following facts
- $D^\alpha u^\epsilon=\eta_\epsilon*D^\alpha$ in $\color{blue}{U_\epsilon}$, where the definition can be seen in the linked question.
- Also, $D^\alpha u^\epsilon\to D^\alpha u$ in $L^p(V)$ for any $V\Subset U$.
Now that I can do mollification on the entire domain $U$ instead of just $U_\epsilon$, I just guess one might have
- $D^\alpha u^\epsilon=\eta_\epsilon*D^\alpha u$ in $\color{blue}{U}$.
- and $D^\alpha u^\epsilon\to D^\alpha u$ in $L^p(U)$.
For the original "long" proof of THEOREM 2 by Evans, see this question.
NOTE. The numbering of equations is not progressive because of successive edits. I have left the original numbering so that comments stay understandable.
If you perform the mollification in the brutal way $$ u^\epsilon(x)= \int_{\mathbb R^n} \eta_\epsilon(x-y)u(y)\mathbf 1_{y\in U}\, dy,\quad x\in U $$ that is, setting $u$ to be equal to $0$ outside $U$, then the last two properties mentioned in your question need not hold in the whole domain $U$.
For a concrete example consider the space $W^{1,2}(-1, 1)$ and the constant function $u(x)=\mathbf 1_{|x|\le 1}.$ You have that $$\frac{du}{dx} \equiv 0\quad \text{in } U.$$ When you perform the brutal mollification you get $$\frac{d}{dx} \left( u\ast \eta_\epsilon\right)(x)= \eta_\epsilon(x+1) - \eta_\epsilon(x-1)\tag{2}$$ This already shows that the desired property $$\tag{!!!}\frac{d}{dx}\left( u\ast \eta_\epsilon\right) = \frac{du}{dx}\ast \eta_\epsilon\quad \text{in }U$$ needs not hold in $U$. Moreover, you can easily check that the $L^2(-1, 1)$ norm of (2) blows up to infinity as $\epsilon\downarrow 0$. This shows that the desired property $$\tag{!!!}\left \| \frac{d}{dx}u\ast\eta_\epsilon - \frac{du}{dx}\right\|_{L^2(U)}\to 0$$ needs not hold either.
Note that if you exclude a small neighborhood of $\{-1, +1\}$, that is, if you consider $U_\delta= (-1+\delta, 1-\delta)$ instead of $U$, then both properties $(!!!)$ do hold (the first holds provided $\epsilon < \delta$). That's why Evans worries about taking mollifications an epsilon away from the boundary of $U$.
Final remark. The distributional point of view provides some more insight. Note that, if you extend $u$ to be zero outside of $U$and consider the result as a distribution on $\mathbb R$, then its derivative is equal to $$\frac{ du}{dx} = \delta(x+1)-\delta(x-1).\tag{1}$$ The brutal extension produced two Dirac deltas in the first derivative. Those are responsible for the failure of properties $(!!!)$.