Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do I show that the Dirac delta distribution cannot be represented by a continuous function?

My try is to show that there exists no continuous function $f(x)$ such that $\int f(x) \phi(x) dx = \phi(0)$ for all test functions $\phi(x)$ but I can not figure out how.

Any hints?

share|cite|improve this question
Suppose $f(x_0)$ were nonzero for some nonzero $x_0$. Then $f(x)$ is close to $f(x_0)$ for $x$ close to $x_0$. Let $\phi$ be a bump function with a short bump centered at $x_0$. What can you conclude? – Samuel Apr 2 '13 at 16:07
@Samuel I would rather consider $\phi$ zero at $x_0$ with two small bumps, one on each side of $x_0$. – Did Apr 2 '13 at 16:12
@Samuel Then the integral is not equal to zero, and if the bump function is zero at $0$, the equality does not hold. Hence $f(x) \neq 0$ only at $x=0$ and it is not continuous. But then it remains to ensure that there exists bump functions s.t. $\phi(0) = 0$ and $\phi(x) \neq 0$ for some $x > 0$. Did I got it right? – Nils Apr 5 '13 at 11:10
@Nils: That's right. Do you know of any bump function $\phi(x)$? If you do, you should be able to make the bump as short as you want (take $\phi(N x)$ for some big $N$), and move the bump wherever you want (take $\phi(x-x_0)$). – Samuel Apr 5 '13 at 21:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.