I am trying to learn weak derivatives. In that, we call $\mathbb{C}^{\infty}_{c}$ functions as test functions and we use these functions 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.

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    $\begingroup$ 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. $\endgroup$ Feb 14, 2013 at 19:12
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    $\begingroup$ 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 $f$. $\endgroup$
    – Tom-Tom
    Jan 9, 2014 at 16:31

3 Answers 3


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.

  • $\begingroup$ My teacher made us do all this calculation and why the compact support is needed. But i wasn't convinced why they are called test functions(i.e, if they play a larger role than they are playing in this context). Also how they work with distributions. $\endgroup$
    – Phani Raj
    Feb 17, 2013 at 6:50

To understand why we are calling them test functions, we have to understand what distributions are and where they come from.

Usually, to evaluate a function, we compute its value at the point where we want to know it. But remember that there are spaces of functions (or equivalence classes of functions) such as $L^p$ spaces where the value on a point is not a good representation of the underlying function (it may even have no sense at all). Then, why not trying to evaluate the function as some sort of a weighted mean ? And an integral of $f$ times a function wisely chosen can be seen as a weighted mean.

We define :

$$T_f(\phi)=\int_{\mathbb{R}}f(x) \phi(x) dx.$$

In fact, test functions are just a new way to "know" a function. Because we have the choice of the space from which we take our tests functions, why not taking one with very good properties ? So we do the choice of $\mathbb{C}^{\infty}_{c}$. By doing so, we can proceed to integrate a large variety of functions and we also have linearity and continuity (compared to the topology of $\mathbb{C}^{\infty}_{c}$) of the mapping $T_f$ when it has a meaning. We can also define a lot of useful operations by reporting every problem one would meet with $f$ on the tests functions.

We just have contructed a continuous linear application from $\mathbb{C}^{\infty}_{c}$ to $\mathbb{R}$. Those kind of applications are known as linear forms. So we define that a distribution is a continuous linear form over $\mathbb{C}^{\infty}_{c}$ to $\mathbb{R}$.

This way, we can take $\phi$ being narrower around the point we are considering with constant aera to know the "value" of $f$ near it.

Note that this is just the tip of the icerberg, and that the OP may have understood why we are calling them this way by now. But I wanted to share this POV on this site.

  • $\begingroup$ What are the "OP" and "POV"? $\endgroup$
    – null
    Jan 29, 2020 at 19:16
  • $\begingroup$ OP=Original Poster=Guy who started the thread POV=Point of view $\endgroup$
    – handy
    Jul 29, 2020 at 14:46

Specifically why it is called test function. (Essentially not different from another answer.)

Strichartz liked to introduce the concept with physical intuition.


Suppose you want to measure the temperature at a point in the room. You hold out your thermometer, but the tip of it is actually a blob of glass with some liquid inside, so are you really measuring the temperature at any actual point? At best it is a weighted average in a neighborhood of the point.


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