Show that the limit of distributions is Dirac delta

Question

I would like to show that the following statement is true

$ \lim_{a\searrow 0} \theta(x)\frac{x^{1-a}}{\Gamma(a)} = \delta(x). $

$\Gamma$ is the gamma function. The above limit is in the sense of distribution, i.e.

$ \lim_{a\searrow 0} \int d x\, f(x) \theta(x)\frac{x^{1-a}}{\Gamma(a)} = f(0) $

for any test function $f$.

Answer 1

Let a test function $\phi$ with support in $[-1,1]$. Then the integral writes $$\int_{0}^1\frac{x^{1-a}}{\Gamma(a)}\phi(x)dx $$

Obviously, for $a \le 1 $ $$0\le \int_{0}^1\frac{x^{1-a}}{\Gamma(a)}\phi(x)dx \le\frac{1}{\Gamma(a)} \int_{0}^1 \phi(x)dx\to 0\text{ as }a\to 0^+,$$

hence your hypothesis is false.

Answer 2

As it was already explained above, the sequence of function-type distributions, i.e. distributions in $\mathscr D'(\mathbb R)$ which are in fact functions in $L^1_{\text{loc}}(\mathbb R)$, $$ \tau_a=\theta(x)\frac{x^{1-a}}{\Gamma(a)}, \text{ for }a>0, $$ does not converge to $\delta(x)$ as $a\to 0.$ A very quick, informal way to see this is to note that, letting $$ \theta(x)= \begin{cases} 1\text{ if }x\ge0\\ 0\text{ if }x<0, \end{cases} $$ then, for an arbitrary but fixed $x\in\mathbb R$, $$ \lim_{a\to0 }\theta(x)\frac{x^{1-a}}{\Gamma(a)}=0, $$
from here we see that the sequence has no hope of converging to $\delta$.

This argument is made rigorous as follows: notice that the definition of $\theta$ function is not restrictive, since altering is value only at the origin cannot affect the value of any integral it appears in. Then, letting $\varphi\in\mathscr D(\mathbb R)$, with support enclosed in the ball $B_R$ of radius $R$ centred at the origin, $$ \langle \tau_a,\varphi\rangle=\int_{-\infty}^{+\infty}\tau_a(x)\varphi(x)dx $$ and we can pull the limit $a\to0$ under the integral sign thanks to Lebesgue's theorem, getting $0$, since the integrand is dominated for small $a$ by $$ R\chi_{B_R}(x) $$ which is summable ($\chi_A$ is the characteristic function of the set $A$).

As an exercise, one can follow your example and look for a sequence of distributions involving the $\Gamma$ function and which do converge to $\delta$. Consider the functions $$ \rho_\varepsilon(t,x) \equiv \frac{e^{-|t|/\varepsilon}|t|^{x-1}}{2\Gamma(x)}\varepsilon^{-x},\text{ for }x>0. $$ Then $$ \int_{\mathbb R}\frac{e^{-|t|/\varepsilon}|t|^{x-1}}{2\Gamma(x)}\varepsilon^{-x}\varphi(t) dt= \int_{\mathbb R}\frac{e^{-|s|}|s|^{x-1}}{2\Gamma(x)}\varphi(\varepsilon s)ds; $$ by Lebesgue's theorem, since the integrand is dominated by $$ \sup_{\text{supp}\varphi}|\varphi|\frac{e^{-|s|}|s|^{x-1}}{2\Gamma(x)} $$ and $$ \int_{\mathbb R}\frac{e^{-|s|}|s|^{x-1}}{2\Gamma(x)} ds = \frac{1}{\Gamma(x)}\int_{0}^{+\infty}e^{-\rho}\rho^{x-1}d\rho=1, $$ finally $$ \lim_{\varepsilon\to0}\rho_\varepsilon(t,x)=\delta(t)\text{ for }x>0. $$