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

Question

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.

Answer 1

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 $\varepsilon\to 0$.

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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.