Is the weak derivative of any function of the form $f(\lfloor g(x) \rfloor)$ $0$ everywhere?

Question

I recently found this thing called a weak derivative. I think it is what I have been trying to express for quite some time, but it's definition makes it mildly hard for me to prove the following (which is probably quite trivial... or not):

Let $f$ and $g$ be continuous functions and $x'$ denote the weak derivative, then $\forall_{x \in R} f(\lfloor g(x) \rfloor)'=0$.

I am not real sure how to prove it if it is true. The form I gave is my definition of piecewise constant. Mostly, I am concerned on whether or not it is what has this property as I've been trying for quite some time to formally define such a 'derivative'. If it does, then my next thing to ask is whether or not there is an associated weak integral as that is more what I care about.

Answer 1

In order to define a weak derivative you typically want to use some real or complex valued bilinear form $\langle \phi,g \rangle$ where $\phi \in {\cal D}$ is your 'test' function for a suitable nice class ${\cal D}$ (vector space) and $g\in {\cal D'}$ the objects (called distributions) in some (vector space) class you want to differentiate.

A couple of important points:

(a) The set ${\cal D}$ should be large enough so that you separate points in ${\cal D'}$ .

(b) You should be able to differentiate all $\phi\in {\cal D}$ and best if the derivative $\phi'$ again belongs to ${\cal D}$. [I bypass some topological/semi-norm aspects that introduce other constraints on this construction].

Then you may (hopefully) define the weak derivative of $g$ by declaring it to be the 'object' $Dg$ such that $$\langle \phi,Dg \rangle = -\langle \phi',g \rangle$$ for all $\phi\in {\cal D}$. The standard example is $\phi \in {\cal D}=C^\infty_c({\Bbb R})$ and the objects a class that in particular contains locally integrable functions $g\in L^1_{\rm loc}({\Bbb R})$. The bilinear form used in that case is:

$$ \langle \phi,g \rangle = \int_{\Bbb R} \phi(x) g(x) \; dx $$

It then make perfectly sense to differentiate any 'nasty' (though locally integrable) function $g$ be declaring that $Dg$ is the object that verifies: $$ \langle \phi,Dg \rangle = -\int_{\Bbb R} \phi'(x) g(x) \; dx $$ for all $\phi \in {\cal D}$. The crucial point that gives sense to the definition is that it extends our usual notion of differentiation on functions that are $C^1$. You see this by partial integration, which is allowed in that case. Now in your case, if I understand correctly (and I just consider $f(x)=x$), you want to differentiate e.g. $\lfloor g(x) \rfloor$ which is a slightly 'nasty' but locally integrable function for $g$ continuous. Now, the 'derivative' should verify the essential property for all $\phi\in {\cal D}$: $$ \langle \phi,D(\lfloor g \rfloor) \rangle = -\int_{\Bbb R} \phi'(x) \lfloor g(x)\rfloor \; dx .$$ This, however, is very far from being the 'zero' object (for which the right hand side should be identically zero). For example if $g(x)=x$ then try to integrate against $\phi'(x)$ where $\phi(x)= (1-x)^2 (1+x)^2$ for $x\in [-1,1]$ zero outside and you get $1$. (admitted, this $\phi$ is not $C^\infty$ at $\pm 1$ but it still works). If you carry out the details you find that for any $\phi\in {\cal D}$: $$ \langle \phi,D(\lfloor \cdot \rfloor) \rangle = \sum_{k\in {\Bbb Z}} \phi(k) .$$ This corresponds to $D\lfloor x \rfloor=\sum_{k\in {\Bbb Z}} \delta_k$ being a sum of delta functions at the integers. Note that this actually is no longer in $L^1_{\rm loc}$ so the space of distributions ${\cal D'}$ in the end will be quite a lot bigger and also contain e.g. derivatives of delta-functions and a lot more.

You may show that any distribution $g$ for which $Dg\equiv 0$ is necessarily (equivalent to) a constant.

Answer 2

Your hypothesis as stated cannot be correct. There is no unique weak derivative. As the Wikipedia article you cited states clearly, weak derivatives (if they exist) are only unique up to equality almost everywhere. So it is not meaningful to claim (as you did in a comment) that the weak derivative of $f(\lfloor g(x) \rfloor )$ is the constant zero function. Also, as two other commenters have stated, given the fact that the step function has no weak derivative, and that it is of the form "$f(\lfloor g(x) \rfloor )$" where $f(x) = x$ and $g(x) = \tanh(x)+1$ for every real $x$, it implies that your question is not valid since "$f(\lfloor g(x) \rfloor )$" does not always have a weak derivative.

Answer 3

Differentiation, $D,$ is an elliptic operator on functions and distributions on $R.$ So if $$ Dh=0 $$ then $h$ is smooth, since for elliptic operators a weak solution is a strong solution. So $h$ must be a smooth function with zero derivative in the ordinary sense. So $h$ is a constant.

Clearly if we take $f$ and $g$ to be the identity maps then your function is not constant. In fact, its weak derivative is a sum of delta distributions since it jumps at each integer.