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?


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.

  • 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

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