[Math] Weak Derivatives and Lp spaces

functional-analysislp-spacesweak-derivatives

I just want to know how to prove two properties of weak derivatives and $L^{p}$ spaces if they are true, the first one just involves weak derivatives:

If we have a locally summable function $u: U \rightarrow \mathbb{R}$ and we have its weak derivative $D^{
\alpha}u$ for $|\alpha| = k$ then how would you show that that $D^{\alpha}u$ exists for $|\alpha| < k $.

Secondly, if $D^{\alpha}u \in L^{p}(U)$ for $|\alpha| = k$ then how would you show that $D^{\alpha}u \in L^{p}(U)$ for $\alpha < k$?

Thanks!

Best Answer

We start with a fact:

Let $u,v$ be locally integrable in $U$. Then $v=D^\alpha u$ if and only if there exists a sequence of $C^\infty(U)$ functions $u_n$ converging to $u$ in $L^1_{loc}(U)$ whose derivatives $D^\alpha u_n$ converge to $v$ in $L^1_{loc}(U)$.

Now, just to fix ideas consider $U\subset\mathbb{R}^2$. We have that $u,D_{11}u,D_{12}u,D_{21}u,D_{22}u\in L^1_{loc}(U)$. Take $u_n$ which satisfies the above fact for $v=D_{11} u$. Note that this $u_n$ is the same for $D_{12}u,D_{21}u,D_{22}u$.

For each $x_0\in U$ take a ball $B(x_0,r)$ such that $\overline{B(x_0,r)}\subset U$. Denote this ball by $K$. For $x\in K$, let $(D_1u_n)_K=D_1u_n$ and for $x$ outside $k$ let it be zero. Note that $$\int_U |(D_1u_n(x))_K|dx=\int_K |D_1 u_n(x)|\leq\int_K|\nabla D_1 u_n(\theta)||x-y|+|D_1u_n(y)|dx \tag{1}$$

where $\theta\in [x,y]\equiv\{z\in K:\ z=tx+(1-t)y,\ ~~ \mbox{for some}\ ~~ t\in [0,1]\}$ and $y\in K$ is choosen in such way that $|D_1u_n(y)|$ is bounded independently of $n$.

We conclude from $(1)$ that there is a constant $M>0$ depending only on $K$ such that $$\int_U |(D_1u_n(x))_K|dx\leq M\tag{2}$$

On the other hand $$\int_U |(D_1u_n(x+y))_K-(D_1u_n(x))_K|dx=\int _{U\setminus K}|(D_1 u_n(x+y))_K|dx+\\ + \int_ K|(D_1u_N(x+y))_K-D_1u_n(x)|dx\tag{3}$$

Given $\epsilon>0$, if we choose a small $y$ and use a argument similar to $(1)$ we conclude from $(3)$ that

$$\int_U |(D_1u_n(x+y))_K-(D_1u_n(x))_K|dx\leq \epsilon \tag{4}$$

independently of $n$. Now we use $(2)$, $(3)$ to apply Kolmogorov-Riesz Theorem (see this paper of Hanche-Olsen and Holden) to conclude that there is a subsequence of $D_1u_n$ which converges for for some functions $w$ in $L^1_{loc}(U)$. Without loss of generality assume that this subsequence is the whole sequence $u_n$. Note that for all $\varphi\in C_0^\infty U)$ we have that $$\int_U D_1 u_n\varphi=-\int_U u_nD_1\varphi,\ \forall n\tag{5}$$

We pass the limit in $(5)$ to conclude that $$\int_U w\varphi =-\int uD_1\varphi,\ \forall\ \varphi\in C_0^\infty (U)$$

This conclude the proof.

Now I would like to adress some remarks: Although I have solved the problem to a particular case, it can be easily seen that the calculations are the same for every case. You start to prove the result with the higher derivatives, i.e. with derivatives of order $k-1$. You need to do this because of the Mean Value Theorem, which I have used in the proof.

Note that the fact stated in the beggining works also in $L^p(U)$, hence your second question can be solved by a similar arugment.

To conclude, I would like to say that maybe there is a more straightforward argument, something like integrate distributions.