Let's consider a domain $\Omega\in \mathbb{R}^n$, $n\in\{1,2,3\}$.

$u:\Omega \rightarrow \mathbb{R}$ is a smooth function.

How to demonstrate the following equality:

$$ u(x)=\int_\Omega u(x')\delta(x-x')d\Omega_{x'}$$ with $$ \delta(x-x')=\left\{ \begin{array}{lc} \infty, & x=x' \\ 0, & x\neq x'\end{array}\right.$$ I think that this is very simple to demonstrate but i can't see it throw.

Thanks in advance for your help.

  • 2
    $\begingroup$ The equality is more or less the definition of the Dirac delta-"function". $\endgroup$
    – mrf
    Commented Oct 14, 2017 at 14:39
  • $\begingroup$ Laplace transform ensures ${\cal L}(u)={\cal L}(u){\cal L}(\delta)$ with convolution method! $\endgroup$
    – Nosrati
    Commented Oct 14, 2017 at 15:37
  • 1
    $\begingroup$ Note that the definition of $\delta$ you quote is just a handwavy way of depicting how $\delta$ works (for one thing $\infty$ is not a number so $\delta(0) = \infty$ is not a well defined thing to say). See e.g. en.wikipedia.org/wiki/Dirac_delta_function#Definitions for more rigorous definitions. $\endgroup$
    – Winther
    Commented Oct 14, 2017 at 16:09

3 Answers 3


You can do this using mollifiers. A mollifier $\varphi(x)$ is a smooth compactly supported function on $\Omega$ such that $$ \int_\Omega \varphi(x) dx = 1, $$ and $$ \lim_{\epsilon \to 0} \epsilon^{-n}\varphi_\epsilon(x/\epsilon) = \lim_{\epsilon\to 0}\varphi_\epsilon(x) = \delta(x). $$ The limit to $\epsilon$ shrinks the size of the support, so that the value of the integral stays 1 but the function itself goes to the dirac delta mass. You can find an example of these functions on wikipedia https://en.wikipedia.org/wiki/Mollifier.

Try considering the integral $$ \int u(x') \varphi_\epsilon(x'-x)dx $$ and using integration by parts and dominated convergence theorem.

  • $\begingroup$ Thanks for the answe. I thinks that using a mollifier is the rigourous answer i was looking for. Although i asked for a detailed solution, this was sufficient enough for me. But i don't see how to use the integration by parts as you suggested $\endgroup$
    – Varazda
    Commented Oct 28, 2017 at 12:24

The Dirac Delta Distribution is defined by the property $\int_{\mathbb R^n} f(x) \delta(x) dx = f(0)$. The so called convolution of two functions is defined as:

$$ [f*g](x) = \int_{\mathbb R^n} f(y)g(x-y)dy $$

What you want to demonstrate is that $[f*\delta] = f$, i.e. the Dirac Delta is an identity element of this operation. This follows immediately from its definition and a substitution:

$$\int f(y)\delta(x-y)dy \underset{z = x-y}{=} \int f(x-z)\delta(z)dz = f(x)$$

Another nice way to arrive at this conclusion, at least in $\mathbb R^1$, is by using the fact that the Dirac Delta can be seen as the derivative of the Heaviside step function $H(x) = \begin{cases}1:&x\ge 0 \\ 0:& x<0\end{cases}$. Then, for compactly supported $f$ holds:

$$ [f*\delta](x) = [f*H'](x) = [f'*H](x) = \int_{-\infty}^{+\infty} f'(y) H(x-y)dy = \int_{-\infty}^{x}f'(y)dy = f(x)$$


I am afraid this is false, the problem is that a point $x$ is a set of zero measure, so the integral is zero! Imagine the case $\Omega = \mathbb{R}$

$$ \int_{-\infty}^{+\infty}{\rm d}x'~u(x')\delta(x-x') = \int_{-\infty}^x{\rm d}x'~u(x')\underbrace{\delta(x - x')}_{\color{red}{0}} + \int_{x}^{+\infty}{\rm d}x'~u(x')\underbrace{\delta(x - x')}_{\color{red}{0}} = 0 $$

The point here is that $\delta(x)$ is not the Dirac delta the way you defined it above

  • $\begingroup$ Yes, i edited the question so the function is the delta function $\endgroup$
    – Varazda
    Commented Oct 14, 2017 at 13:27
  • $\begingroup$ @MyGlasses I appreciate your feedback, but would you please be more specific? Do you think the answer is wrong? Thanks $\endgroup$
    – caverac
    Commented Oct 14, 2017 at 14:29

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