# Existence of higher order weak derivative implies existence of lower order if integrable (1 dim). Proof?

I am learning for a pde exam and found the following statement without proof in some old lecture notes:

If for some $$u \in L^1_{loc}(a,b)$$ and some $$n \in \mathbb{N}$$ there exists a weak derivative of order n $$u^n \in L^1(a,b)$$, i.e., for all test functions $$\Phi \in \mathcal{C}^\infty_c(a,b)$$ it holds that $$\int_a^b u(x) \Phi^n(x) \mathrm{d}x=(-1)^n \int_a^b u^n(x) \Phi(x) \mathrm{d}x$$ then all lower order weak derivatives exist and are absolutely continuous.

I know that this is not true if we do not require the integrability of the n-th weak derivative.

I tried to proof the statement, my idea was to construct the lower order derivatives inductively as integrals (like $$u^{k-1}(x)=\int_a^x u^k(y) \mathrm{d}y$$) but I didn't know how to proceed from there and am not sure if this is the right approach at all. Any ideas?

Also, can we generalize this for the multidimensional setting (maybe assume that all weak derivatives with multiindex $$\alpha$$ s.t. $$|\alpha|=n$$ exist and are integrable?)?

As I said in the comment: I figured it out and here is the proof in case somebody else has the same problem in the future.

Proof: For $$1 \leq k \leq n$$ we define $$v_k$$ inductively:

Set $$v_n:=u^n$$. A theorem which was proven earlier in my lecture states that there exists a function $$\Phi_1 \in \mathcal{C}^\infty_c(a,b)$$ with $$\int_a^b \Phi_1(x) \mathrm{d}x=1$$.

For all $$1 \leq k and $$x \in (a,b)$$ we define $$v_k(x):=\int_a^x v_{k+1}(y) \mathrm{d}y+\underbrace{(-1)^k \int_a^b u(z)\Phi_1^k(z) \mathrm{d}z-\int_a^b (\int_a^z v_{k+1}(y) \mathrm{d}y)\Phi_1(z)\mathrm{d}z.}_{=:c}$$ We claim that for all $$1 \leq k : $$v_k \in L^1(a,b)$$ is well defined, a weak derivative of order k of u, and differentiable almost everywhere, i.e., absolutely continuous, with $$v_k^\prime=v_{k+1}$$.

We prove this by induction over $$n-k$$:

Assume that $$v_{k+1} \in L^1(a,b)$$ is well defined and a weak derivative of order k+1 of u.

Firstly, we will show that $$x \mapsto \int_a^x v_{k+1}(y) \mathrm{d}y \in L^1(a,b)$$: $$\int_a^b |\int_a^x v_{k+1}(y) \mathrm{d}y|\mathrm{d}x \leq \int_a^b \int_a^x |v_{k+1}(y)| \mathrm{d}y \mathrm{d}x\leq \int_a^b \underbrace{\int_a^b |v_{k+1}(y)| \mathrm{d}y}_{<\infty,\ v_{k+1} \in L^1(a,b)} \mathrm{d}x<\infty.$$

This implies that $$v_k$$ is well defined and that $$v_k \in L^1(a,b)$$. By the fundamental theorem of calculus we know that $$v_k$$ is differentiable almost everywhere with weak derivative $$v_k^\prime=v_{k+1}$$.

Let $$\Phi \in \mathcal{C}^\infty_c(a,b)$$. There is a theorem that states that there exist $$d \in \mathbb{R}$$ and $$\Phi_0 \in \mathcal{C}^\infty_c(a,b)$$ s.t. $$\Phi=\Phi_0^\prime + d\Phi_1$$. This implies $$(-1)^k \int_a^b v_k(x)\Phi(x)\mathrm{d}x=(-1)^k \int_a^b v_k(x) (\Phi_0^\prime(x)+d\Phi_1(x)) \mathrm{d}x=(-1)^k \int_a^b v_k(x)\Phi_0^\prime(x) \mathrm{d}x+(-1)^k d \int_a^b v_k(x) \Phi_1(x)\mathrm{d}x \underset{v_k^\prime=v_{k+1}}{=} (-1)^{k+1} \int_a^b v_{k+1}(x)\Phi_0(x) \mathrm{d}x+(-1)^k d \int_a^b (\int_a^x v_{k+1}(y) \mathrm{d}y +c)\Phi_1(x) \mathrm{d}x \underset{\text{def. of c and }\int \Phi_1=1}{=} (-1)^{k+1} \int_a^b v_{k+1}(x)\Phi_0(x) \mathrm{d}x + (-1)^k d (-1)^k \int_a^b u(x)\Phi_1^k(x)\mathrm{d}x \underset{v_{k+1} \text{ weak deriv. of order k+1}}{=}\int_a^b u(x)\Phi_0^{k+1}(x) \mathrm{d}x+\int_a^b u(x)d\Phi_1^k(x)\mathrm{d}x=\int_a^b u(x) \underbrace{(\Phi_0^\prime+d\Phi_1)^k}_{=\Phi^k}(x) \mathrm{d}x.$$ This means that $$v_k$$ is a weak derivative of order k of u and thereby finishes the proof.