# Simplest smooth ($C^{\infty}$) approximation to Dirac's $\delta$ with bounded support.

I'm looking for a function $$f(x)$$ that approximates Dirac's $$\delta$$ distribution that has the following properties:

1. $$f\in C^\infty(\mathbf R)$$, it has finite derivatives of all orders.
2. $$f$$ has a simple definition (in increasing complexity: polynomial $$\rightarrow$$ rational $$\rightarrow$$ exponentials $$\rightarrow$$ trigonometric/hyperbollic ). Ideally it has a simple expression, but if necessary it can also be defined piecewise.
3. $$f$$ has bounded support, its only nonzero values occur within a closed interval centered at the origin.
4. $$f$$ approximates the $$\delta$$ distribution in the sense that it has total integral of $$1$$, it is a symmetric function, and it has a parameter $$\lambda$$ that can be adjusted such that in the limit of either $$\lambda\to0$$ or $$\lambda\to\infty$$ you have $$f(0)\to\infty$$ and $$f(x\neq0)\to0$$.

Does such a function exist? If so, what's the simplest one you know of?

I need this function for a program I'm writing, (3) and (4) are very necessary for me. If the restrictions are too harsh it's OK as long as $$f$$ is at least continuously differentiable or even if $$f$$ doesn't satisfy (2) in the sense that it's defined in terms of more complicated transcendental functions I could work around it.

Example of functions that approximate the $$\delta$$ distribution but don't satisfy all requirements are:

• $$\frac{1}{2\lambda}[1-\tanh^2(x/\lambda)]$$ is $$C^\infty$$, simple, but doesn't have bounded support.
• The function $$\frac{1}{\lambda}\zeta(x/\lambda)$$ in this other MSE post has compact-, and therefore bounded-, support but it doesn't have a simple expression as it is defined in terms of a transcendental function.
• The piecewise defined function $$f(x)=\{0, ~\text{if}~ |x|>\frac{\pi\lambda}{2}; \quad\frac{2}{\pi\lambda}\cos^2(x/\lambda), ~\text{if}~ |x|\leq\frac{\pi\lambda}{2}$$. It does have bounded support but its expression is not so simple because it's defined piecewise and it also violates (1), since it's second derivative is discontinuous.

Thanks.

Such a function exists. Inspired by Are there other kinds of bump functions than $$e^\frac1{x^2-1}$$?, consider the bump function $$\Psi(x)=\begin{cases} \frac{4\left( {x}^{2}+1 \right)\ {{\rm e}^{{\frac {4x}{{x}^{2}-1}}}}}{\left( \left( {x}^{2}-1 \right) \left( 1+{{\rm e}^{{\frac {4x}{{x}^{ 2}-1}}}} \right)\right)^2},&x\in (-1,1) \\ 0,&\mathrm{otherwise}. \end{cases}$$ One has that $$\Psi\in L^1_{\mathrm{loc}}(\mathbb{R})$$ and this function has an elementary antiderivative with $$\int_{\mathbb{R}} \Psi(x)=1$$. Then by Dirac's $$\delta$$ distribution smooth approximation (your linked MSE post), the family of functions $$f_{\varepsilon}(x)=\frac{1}{\varepsilon}\Psi\left(\frac{x}{\varepsilon}\right)$$ satisfy all the necessary properties with $$f_{\varepsilon}(x)\to \delta(x)$$ as $$\varepsilon\to 0$$.
Note: You could have replaced $$\Psi$$ by any other normalized bump function. For instance as suggested by @TSF, you could have started by considering the function $$\Phi(x)=\begin{cases} \exp\left(-\frac{1}{1-x^2}\right),&x\in (-1,1) \\ 0,&\mathrm{otherwise}. \end{cases}$$ However, here the function does not have a normalization in terms of elementary functions $$C:=\int_{\mathbb{R}} \Phi(x)~dx=e^{-1/2}\left(K_1(1/2)-K_0(1/2)\right)\approx 0.4439938161680794,$$ where $$K_{\alpha}$$ is a modified Bessel function, which you have mentioned in your post that you would like to avoid.
• If it's ok that the integral does not have to be exactly $1$ (which might be fine since you said you are using it for some program), then you could consider the function $$\Psi(x):=\frac{1}{C}\Phi(x),$$ where you use the approximate value of $C$ and define $f_{\varepsilon}$ similarly. This gives you a simpler formula. Oct 19, 2021 at 1:30