What is the major difference between real analytic and test function (smooth compact supported functions). Can we find a real analytic function $f$ on $\mathbb R^n$ which is also smooth compact supported? If not then need a proof.

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    $\begingroup$ Dear Sara, I know this is not directly what you asked about, but there is a more subtle relationship: the Fourier transform of a test function will be an analytic function (in fact an entire function of a complex variable). Regards, $\endgroup$
    – Matt E
    May 17, 2013 at 5:02
  • $\begingroup$ Oh that is really a elusive relationship. But I didn't take any course in Fourier transform so don't know much about it. Can you see my proof that i just attempted. Thank you. $\endgroup$ May 17, 2013 at 5:24

4 Answers 4


Assume $f$ is a real-analytic test function which is not identically zero; this will lead to a contradiction. Let $K$ denote the support of $f$. The idea (also present in OP's attempt) is to expand $f$ into a power series centered at some boundary point $x_0$ of the support; find that the coefficients are all zero, obtain a contradiction. Key steps:

  1. A nonempty bounded set must have nonempty boundary (because the only sets with empty boundary are $\varnothing$ and $\mathbb R^n$).
  2. Let $x_0$ be a boundary point of $K$. Note that $x_0\in K$ and $x_0$ is also a limit point of the complement of $K$.
  3. All derivatives of $f$ are identically zero on the complement of $K$
  4. They are also continuous on all of $\mathbb R^n$. By continuity they are equal to zero at $x_0$.
  5. The power series of $f$ centered at $x_0$ is $0+0(x-x_0)+\dots$
  6. Therefore, $f$ is equal to zero in some neighborhood of $x_0$. But this contradicts the definition of support of $K$.
  • $\begingroup$ I really appreciate it. Thank you so much my friend. $\endgroup$ May 17, 2013 at 7:19

If by "test function" you mean a smooth function with compact support (as opposed to a Schwarz function), then the answer is no. Consider an analytic function $f \in C_c^\infty(\mathbb R^n)$ and a point $x$ outside of its support. Because $f$ is analytic, it must equal its own Taylor series developed at $x$, i.e. $0$, and since the radius of convergence is infinite, $f$ must be the zero function.

The upshot of this is that analytic functions are very rigid objects. Loosely speaking, perturbing an analytic function in a single point causes global changes to "ripple" out from the perturbation in order to preserve analyticity.

  • $\begingroup$ So @kahen you mean if a function is real analytic and smooth with compact support then it must be equivalent to zero function. Let me check it. thank you $\endgroup$ May 15, 2013 at 23:29
  • $\begingroup$ So, do you mean that test function is flexible but analytic is not. If we disturb a test function on some point then doesn't it effect the rest of the function? How can we see it through some example. $\endgroup$ May 15, 2013 at 23:46
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    $\begingroup$ @SaraTancredi, any smooth and compact supported function is a test function. Draw any weirdly shaped bump without corners that goes to zero outside some interval, and it's basically a test function. An analytic function, however, has very strict requirements; it has to be a power series, so that controls how the function grows and shrinks along the real line. $\endgroup$ May 16, 2013 at 0:16
  • $\begingroup$ @kahen , Please check my proof. Thank you. $\endgroup$ May 17, 2013 at 4:46
  • $\begingroup$ @kahen: Can't you do away with that problem by saying $f$ is analytic only within the compact interval K? $\endgroup$
    – MSIS
    Jan 16, 2020 at 20:41

A test function is a smooth and compactly supported function, while a real analytic function is smooth and given by a power series uniformly converging at every point.

Perhaps I will only give a hint as an answer to your second question: A test function must be equal to zero everywhere outside some compact set. If it were also analytic, then what would the coefficients of its power series have to be?

  • $\begingroup$ Thank you for providing me this hint. Let me workout. $\endgroup$ May 15, 2013 at 23:36
  • $\begingroup$ A.Wong , Please check my proof. Thank you in advance. $\endgroup$ May 17, 2013 at 4:47

I am trying to prove that if some test function $f$ in $\Bbb R^n$ is real analytic then it must be equivalent to zero function. Please correct me if there is any problem.


Let us define a test function,

$ f_\epsilon (x)= \begin{cases} \frac{C^{-1}}{\epsilon^n} e^{{\frac{-\epsilon^2}{{\epsilon^2}-|x|^2}}}, &\text{for |x|< $\epsilon$ } \\ 0, & \text{everywhere else} \\ \end{cases} $

$\color{red}{\text{Edited in: This was a false start}}$. You should work with a general test function, not a particular one. That is, simply say: let $f:\mathbb R^n\to \mathbb R$ be a test function. By definition of a test function, $f$ has compact support. The goal is to show that if $f$ is real-analytic, then the support is empty.

We know, this function has derivatives of all orders and also has a compact support.

Now, Let us suppose that it is also a real Analytic function.

If this function is a real analytic then it must has a Taylor series expansion about any point $x_0$ in its domain which will converge to $f_\epsilon (x)$ for a nbdd. of $x_0$.

Now in particular take Taylor expansion about $\pm\epsilon$

$f_\epsilon (x) \approx 0 + 0x + 0x^2 + . . .$

The Taylor series expansion does in fact converge to the function $f_\epsilon^˜ (x) = 0$.

We can see that $f_\epsilon^˜ (x)$ and $f_\epsilon (x)$ are two different functions for |x|< $\epsilon$. But they must be equal since $f_\epsilon (x)$ is real analytic, which is only possible when $f_\epsilon (x) \equiv 0$.

  1. Is it enough to prove it. Personally I don't feel that its sufficient...
  2. If I need to elaborate it by writing its Taylor series then calculations of derivatives is again a problem for me. I don't know how to write derivatives of $f_\epsilon (x)$. And in above proof, the way I approximate its Taylor series at $\epsilon$ looks quite tricky... isn't it? Please give me your suggestions and feel free to edit this proof.

Thanks you so much.

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    $\begingroup$ The problem here is not with series or derivatives, but with abstract logic. It's like proving the statement "every Republican supports gun control" by saying "my uncle is a Republican and he supports gun control, that's the proof." You get off track the moment you replace a class of functions (or people) with the particular function $f_\epsilon$ (or your uncle). $\endgroup$
    – 75064
    May 17, 2013 at 5:02
  • $\begingroup$ @75064 Isn't it okay, if we take any random republican and prove (by some special reasoning) that it must be in the favor of gun control. Of course reasoning will be the attitude of the entire class of republicans towards guns;... their principles, characteristics, special behavior or stand on some particular issue. I think it makes sense that way. I mean my above approach. Please elaborate your point if you think i am not on the right track. $\endgroup$ May 17, 2013 at 5:53
  • $\begingroup$ @SaraTancredi I edited in a little remark, and posted my answer separately. $\endgroup$
    – 75064
    May 17, 2013 at 6:14

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