# Why is this test function admissible? [Paper explanation]

Reading Non-linear Elliptic and Parabolic Equations Involving Measure Data by Boccardo$$\&$$Gallouet , I had trouble understanding the following:

Why is $$\psi(u_n)\chi_{(0,t)}$$ admissible as a test function in $$(62)$$?

The reason I'm asking is because characteristic\indicator functions have no smooth derivatives and plus I don't understand in which function space of test functions the authors define the weak formulation of $$(59)$$.

Below you can have a quick look at the set up of the problem and the point in which I'm stuck. Throughout this paper $$\Omega$$ is a bounded open set of $$\mathbb R^N$$ with $$N \ge 2$$ and $$p$$ is such that $$p \in (2-1/(N+1),\infty)$$.

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Any help is much appreciated. Thanks in advance!

The reason I'm asking is because characteristic\indicator functions have no smooth derivatives and plus I don't understand in which function space of test functions the authors define the weak formulation of (59).

I think we can agree that whatever the authors mean by "$$\psi(u_n(t)\chi_{(0,t)}$$", they must realise that it isn't a 'test function' in the traditional (smooth, compactly supported) sense, given its insufficient smoothness in both the $$x$$ and $$t$$ variables. However, it is possible that 'test function' means something more general here. The sequences $$(u_n)$$ and $$(f_n)$$ being considered here are test functions in the traditional sense, so we can still plug things into the natural bilinear pairing between $$\mathscr{D}(Q) = C_0^\infty(Q)$$ and its dual $$\mathscr{D}'(Q)$$ (the space of distributions on $$Q$$) and draw conclusions from this.

The fact that $$P(u_n) = f_n$$ in $$\mathscr{D}(Q)$$ implies (tautologically) that

$$\langle \Psi, P(u_n)\rangle = \langle \Psi,f_n\rangle$$

for all distributions $$\Psi\in \mathscr{D}'(Q)$$. We can therefore learn things about $$u_n$$ and $$f_n$$ by making clever choices of 'test distribution' $$\Psi$$. When $$\Psi$$ is integration against a function, said function may arguably be termed a 'test function', which I think may be what is being proposed here.

This sort of terminology occurs in other settings where one has a non-degenerate bilinear pairing; 'test function' often just means 'thing we're free to choose which will fit into the other slot of the pairing from what we already have'.

• First of all, thank you very much for your time and your reply. Now, if I understood correctly, in the case of this paper $\psi(u_n(t))\chi_{(0,t)}$ is a distribution and since $(u_n)$ and $(f_n)$ are considered as test functions in the traditional sense, the duality pairing is satisfied. Thus $\psi(u_n(t))\chi_{(0,t)}$ is basically the function we use to integrate against $P(u_n),f_n$ and for that reason the wide term "test function" is adopted. Am I right? May 9 '19 at 8:20
• "Thus $\psi(u_n(t))\chi_{(0,t)}$is basically the function we use to integrate against $P(u_n), f_n$ and for that reason the wide term "test function" is adopted." - that is my understanding yes.
– DCM
May 9 '19 at 17:44