2
$\begingroup$

The Weierstrass function is a function that is continuous everywhere but nowhere differentiable.
I'm wondering if it has 1-th weak derivative.

According to a book I'm reading, $u(x_1,x_2) = f(x_1) + f(x_2)$ defined in $\Omega = (0,1) \times (0,1)$, where $f$ is the Weierstrass function, actually has 2-th weak derivative. We have $$0 = \int_\Omega(f(x_1)+f(x_2))\frac{\partial^2 \phi}{\partial x_1 \partial x_2}dx _1dx_2$$ for all $\phi \in C_c^\infty (\Omega)$.
Hence weak derivative $D_{x_1}D_{x_2}u = 0.$

So does weak derivate $D_{x_1}u$ exist?

$\endgroup$
1
$\begingroup$

First, a point of terminology: as discussed in Are weak derivatives and distributional derivatives different? a "weak derivative" of $f$ is understood to be a locally integrable function $g$ such that $$ \int_\Omega f\phi' = -\int_\Omega g\phi\quad \forall \phi\in C_c^\infty(\Omega). $$ This property implies that:

  1. $f$ is an absolutely continuous representative: Absolutely continuous representative of Sobolev function
  2. $f$ is differentiable almost everywhere (as a consequence of 1, see Proof that absolute continuity implies differentiability a.e.

The Weierstrass function is nowhere differentiable, and therefore does not have a weak derivative.

As an exercise, you can try to streamline the above argument 1-2 by proving directly that a continuous function $f$ with weak derivative $g$ is differentiable at every Lebesgue point of $g$.

If you asked about distributional derivative, the answer would be yes. Every continuous function is locally integrable, therefore is a distribution, and therefore has a distributional derivative.

$\endgroup$
  • 1
    $\begingroup$ So a continuous function with weak derivatives is supposed to be absolutely continuous? I don't know if I get that right. $\endgroup$ – xixumei Mar 14 '18 at 11:56
  • 1
    $\begingroup$ Yes that is correct. The converse is also true, absolutely continuous functions are weakly differentiable. $\endgroup$ – user357151 Mar 14 '18 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.