Weak derivative of a double series

Let $f\left(x\right),\,x>0$ be a continuous and piecewise smooth function and assume that the double series $$f\left(x\right)=\sum_{m\geq1}\sum_{n\geq1}f_{n,m}\left(x\right)$$ converges absolutely and locally uniformly, where $f_{n,m}\left(x\right)$ and $\sum_{n\geq1}f_{n,m}\left(x\right)$ are continuous and piecewise smooth functions. Also assume that $D\left(\cdot\right)$ is the weak derivative operator.

Question 1: Is it true that $$D\left(f\left(x\right)\right)=\sum_{m\geq1}\sum_{n\geq1}D\left(f_{n,m}\left(x\right)\right)?\tag{1}$$

Question 2: Since $f\left(x\right),\,\sum_{n\geq1}f_{n,m}\left(x\right)$ and $f_{n,m}\left(x\right)$ are piecewise smooth functions then they are differentiable a.e., so, assuming that $(1)$, holds, is it possible to conclude $$\frac{d}{dx}f\left(x\right)=\sum_{m\geq1}\sum_{n\geq1}\frac{d}{dx}f_{n,m}\left(x\right)\,\mathrm{a.e.}\tag{2}$$ where $d/dx$ is the canonical derivative? And if is not, when this is true?

My approach is the following: since $f\left(x\right),\,\sum_{n\geq1}f_{n,m}\left(x\right)$ and $f_{n,m}\left(x\right)$ are piecewise smooth functions they can be identified as distributions and so we can differentiate term by term (see this theorem). For proving $(2)$ we can see use the fact that, where exists, the canonical derivative and the weak derivative are the same. I'm quite sure that my argument is wrong but I didn't find anything better.

As we know the weak derivative is action on test function, which has very good regularity, both like the slow-decay space and supp compact space (i do not remember the standard name of these space anyway).

So what we need to do is just to proof for all $g\in C_c(R^n)$, we have: $$\int_{R^n}D(g)f=\int_{R^n}D(g)\sum_{m\geq 1}\sum_{n\geq 1}f_{n,m}...(*)$$

But this is definitely WRONG, counterexample is very easy to find. you can just think you divide a cube into a lots of small cubes in a sequence of smaller and smaller scale. On every scale some cube is bad, that is to say $\sum\sum f_{n,m}$ coverage very slow on it, this could lead the RHS of $(*)$ never meaningful in fact!

If we know $\sum\sum f_{m,n}=f$ coverage uniformly then this is easy to get by a dominate coverage theorem. And may involve of Calderon-Zegmund decomposition could make the assume weaker on $f_{m,n}$.

• But if we identify $f$, $\sum_{n\geq1}f_{n,m}\left(x\right)$ and $f_{n,m}\left(x\right)$ as distributions (and this is true since we assumed that are continuous and piecewise smooth functions) then the termwise differentiation is legitimate since the operator $D$ is continuous. Am I wrong? – user422009 Nov 14 '17 at 12:51
• I don't agree with it. In fact it is easy to construct a counterexample to make $\sum_{n,m\geq 1}f_{n,m}(x)$ is NOT a distribution. Because it is too luxury to let it be continues compatible with the weak coverage. – Hu xiyu Nov 14 '17 at 12:54
• On the other hand , I do not believe and never know distribution have countable-additive property just like measure without any other assumption without it is meaningful. I think I can construct counterexample. – Hu xiyu Nov 14 '17 at 12:56
• Sorry but I don't understand. $\sum_{m\geq1}\sum_{n\geq1}f_{n,m}\left(x\right)=f(x)$ and I assumed that $f$ is a continuous and piecewise smooth function(I don't say it is always with this regularity, it is only the case I studied). Why can not be identified as a distribution? Maybe we have to assume the convergence of the series in the sense of distributions. – user422009 Nov 14 '17 at 12:58
• At once you want to look it as a distribution, you need to proof it is weak continue. Always remember there is a natural topology on the distribution space. and this weak continue may be false in general in your case. – Hu xiyu Nov 14 '17 at 13:01