# what do test function mean?

I am trying to learn weak derivatives. In that we call $\mathbb{C}^{\infty}_{c}$ function as test function and we use this function in weak derivatives. I want to understand why these are called test functions and why the functions with these properties are needed. i have some idea about these but couldn't understand them properly.

Also i'll be happy if any one can suggest some good reference on this topic and sobolev spaces.

-
It's just a jargon, used to make it clear that the distribution is defined by the way it acts when "tested" against those functions. – Giuseppe Negro Feb 14 '13 at 19:12
The naming question is duplicate, see math.stackexchange.com/questions/47433/… If I may add my opinion, I think test functions $\varphi$ are called this way because they are regular enough to be "tested" against $f$ (i.e. one can give a mathematical sense to $\int \varphi f$) for very irregular objects $f$ such as distributions. The regularity of $\varphi$ "absorbs" the irregularity of $f$. – Tom-Tom Jan 9 '14 at 16:31

Suppose you want to find the solution of a differential equation, $f'' = gf$ for example.
Take any solution $f$ of this equation, then if you take any function $\psi \in \rm C^{\infty}_c$ it is true to write $$f'' = gf \Longrightarrow \psi f'' = \psi gf \Longrightarrow \int \psi f'' = \int \psi g f \Longrightarrow \int \psi'' f = \int \psi g f$$
Conclusion : any solution of the differential equation $f$ satisfies $\int \psi'' f = \int \psi g f$ but it is possible that functions that are only continuous satisfies the same equation ! These solutions will be called weak solutions because they are solutions of a weaker problem.
Now, why have we chosen the functions $\psi$ to be in $\rm C^{\infty}_c$ ? It is because we transfered the derivation operation from $f$ to $\psi$ by integration by part ; this integration by part goes well only if you suppose that $\psi$ has compact support. I encourage you to do the details of the last implication and it will become clearer.