How to show that this limit converges in the distributional sense to a dirac delta function

Question

Let $$\begin{eqnarray}\nonumber f(y, t) &=& \frac{C}{\sigma ^2 t} \left[\frac{(1-\alpha) (b-y)}{\alpha t^{\alpha}} \, _1F_1\left[\frac{\alpha+1}{2 \alpha};\frac{3}{2};-\frac{ (b-y)^2}{2 \sigma^2 t^{2 \alpha}}\right]- \sqrt{2} \sigma \frac{\Gamma \left(\frac{3}{2}-\frac{1}{2 \alpha}\right)}{\Gamma \left(1-\frac{1}{2 \alpha}\right)} \, _1F_1\left[\frac{1}{2 \alpha};\frac{1}{2};-\frac{ (b-y)^2}{2 \sigma^2 t^{2 \alpha}}\right]\right] \end{eqnarray}$$ be defined in $f(y, t) \in (-\infty, b]$, where $0 < \alpha < 1$, $\sigma > 0$ and $t$ represents time and $C$ is some scalar value. Also where $_1F_1(a,b,z)$ denotes the Kummer confluent hypergeometric function with series expansion $$_1F_1 (a,b,z)=\sum _{k=0}^\infty \frac{(a)_k}{(b)_k}\left(\frac{z^k}{k!}\right),$$ where $(c)_k$ denote Pochhammer’s Symbol \begin{eqnarray}\nonumber (c)_0 &=& 1\\\nonumber (c)_n &=&c(c+1)(c+2)\cdots(c + n -1),\\\nonumber (c)_n &=&\frac{\Gamma(c + n)}{\Gamma(c)}\quad c \neq 0, -1, -2, \cdots \end{eqnarray} and $\Gamma(z)$ denotes the Gamma function which satisfies $$\Gamma (z)=\int _0^{\infty } e^{-t} t^{z-1} dt.$$

I am interested in showing convergence in the distributional sense, that for some $y_0$ the following limit holds $$\lim_{t \rightarrow 0}f\left(y, t\right) = \delta\left(y - y_0\right), \quad y_0 \in \left(- \infty, b\right).$$

The presence of Kummer confluent hypergeometric function have made evaluating Fourier transform of $f(y,t)$ or applying dominated convergence theorem too hard. I am having trouble proving this analytically. Any help?

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Some observations which may be helpful $$\int_{-\infty}^b f(y, t) \, \mathrm{d}y = C t^{\alpha-1}.$$ $$\int_{-\infty}^\infty f(y, t) \, \mathrm{d}y = 0.$$ Contour integral representation of Kummer confluent hypergeometric function $$_1F_1 (a,b,z)= \frac{\Gamma (b)}{2 \pi \iota \Gamma (a)} \int_{\gamma -\iota\infty}^{\gamma +\iota\infty} \frac{(-z)^{-s} \Gamma (s) \Gamma (a-s)}{\Gamma (b-s)} \, \mathrm{d}s$$ where $ 0 <\gamma < \operatorname{Re}(a) \wedge |\arg(-z)| < \frac{\pi}{2}.$

Denoting $$ I(a,b,z) = \int_{\gamma -\iota\infty}^{\gamma +\iota\infty} \frac{(-z)^{-s} \Gamma (s) \Gamma (a-s)}{\Gamma (b-s)} \, \mathrm{d}s$$ Then the function of interest can be rewritten as \begin{eqnarray} f(y, t) &=& \frac{C}{\sigma ^2 t} \left[\frac{\sqrt{\pi }(1-\alpha) (b-y)}{2 \alpha t^{\alpha} \Gamma \left(\frac{\alpha+1}{2 \alpha}\right)} \, I\left[\frac{\alpha+1}{2 \alpha};\frac{3}{2};-\frac{ (b-y)^2}{2 \sigma^2 t^{2 \alpha}}\right] \right. \\[6pt] & & \left. {} - \sqrt{\frac{2}{\pi}} \sigma \sin \left(\frac{\pi }{2 H}\right) \Gamma \left(\frac{3}{2}-\frac{1}{2 \alpha}\right) I\left[\frac{1}{2 \alpha};\frac{1}{2};-\frac{ (b-y)^2}{2 \sigma^2 t^{2 \alpha}}\right] \vphantom{\frac{(1)}{\Gamma\left(\frac\alpha\alpha\right)}} \right] \end{eqnarray}

Solution attempt:

Let $\varphi \in C_{c}^{\infty}(\mathbb{R})$. Using the substitution $x = \frac{b-y}{\sigma t^\alpha}$ we obtain \begin{eqnarray}\nonumber f_t(x) &=& \frac{C}{\sigma t} \left[\frac{(1-\alpha)x}{\alpha} \, _1F_1\left[\frac{\alpha+1}{2 \alpha};\frac{3}{2};-\frac{x^2}{2}\right]- \sqrt{2} \frac{\Gamma \left(\frac{3}{2}-\frac{1}{2 \alpha}\right)}{\Gamma \left(1-\frac{1}{2 \alpha}\right)} \, _1F_1\left[\frac{1}{2 \alpha};\frac{1}{2};-\frac{x^2}{2}\right]\right] \\ \langle f_t,\varphi \rangle &=& \int_{-\infty}^\infty I_{\left(-\infty, b\right]} f(y,t) \varphi(y) \, \mathrm{d}y \\ &=& -\sigma t^\alpha\int_{-\infty}^\infty I_{\left[0, \infty\right)} f_t(x) \varphi(b - x \sigma t^\alpha) \, \mathrm{d}x \end{eqnarray} As $\varphi$ is continuous and compactly supported, this integrand is dominated by $? \|\varphi\|_{\infty}$ which integrates to $?\|\varphi\|_\infty$.

Answer 1

Here is an attempt at a solution.

Continuing with your idea: Substitute $\xi:=\frac{b-y}{\sqrt{2}\sigma t^\alpha}$ to make everything nicer: $$\begin{align} f_{t}(\xi) &= \frac{\sqrt{2}C}{\sigma t} \left[\frac{(1-\alpha)}{\alpha} \xi\, _1F_1\left(\frac{\alpha+1}{2 \alpha};\frac{3}{2};-\xi^2\right)-\frac{\Gamma \left(\frac{3}{2}-\frac{1}{2 \alpha}\right)}{\Gamma \left(1-\frac{1}{2 \alpha}\right)} {_1F_1}\left(\frac{1}{2 \alpha};\frac{1}{2};-\xi^2\right)\right] \tag{1}\\ &= \frac{\sqrt{2}C}{\sigma t} \left[(-\tfrac{1}{2}+\tfrac{1}{2\alpha}) 2\xi {_1F_1}\left(\tfrac{1}{2}+\tfrac{1}{2 \alpha};\tfrac{3}{2};-\xi^2\right)-\frac{(\tfrac{1}{2}-\tfrac{1}{2\alpha})\Gamma(\tfrac{1}{2}-\tfrac{1}{2 \alpha})}{\Gamma(1-\frac{1}{2 \alpha})} {_1F_1}\left(\tfrac{1}{2 \alpha};\tfrac{1}{2};-\xi^2\right)\right] \tag{2}\\ &= \frac{\sqrt{2}C}{\sigma t} (\tfrac{1}{2}-\tfrac{1}{2\alpha})\left[-2\xi {_1F_1}\left(\tfrac{1}{2}+\tfrac{1}{2 \alpha};\tfrac{3}{2};-\xi^2\right)-\frac{\Gamma(\tfrac{1}{2}-\tfrac{1}{2 \alpha})}{\Gamma(1-\frac{1}{2 \alpha})} {_1F_1}\left(\tfrac{1}{2 \alpha};\tfrac{1}{2};-\xi^2\right)\right] \tag{3} \end{align}$$ Now we use the magic formula $$ {_1 F_1}(a;b;-z) = e^{-z} \cdot {_1 F_1}(b-a;b;z) \tag{4}$$ and get $$f_t(\xi) = \frac{\sqrt{2}C}{\sigma t} (\tfrac{1}{2}-\tfrac{1}{2\alpha})e^{-\xi^2}\left[-2\xi {_1F_1}\left(1-\tfrac{1}{2 \alpha};\tfrac{3}{2};\xi^2\right)-\frac{\Gamma(\tfrac{1}{2}-\tfrac{1}{2\alpha})}{\Gamma(1-\frac{1}{2 \alpha})} {_1F_1}\left(\tfrac{1}{2}-\tfrac{1}{2\alpha};\tfrac{1}{2};\xi^2\right)\right] \tag{5}$$

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So far not much has happened. But now comes the part where I like the other sign more. If the original function indeed contains a sign error and we are really talking about $$f_t(\xi) = \frac{\sqrt{2}C}{\sigma t} (\tfrac{1}{2}-\tfrac{1}{2\alpha})e^{-\xi^2}\left[-2\xi {_1F_1}\left(1-\tfrac{1}{2 \alpha};\tfrac{3}{2};\xi^2\right) \color{red}{+}\frac{\Gamma(\tfrac{1}{2}-\tfrac{1}{2\alpha})}{\Gamma(1-\frac{1}{2 \alpha})} {_1F_1}\left(\tfrac{1}{2}-\tfrac{1}{2\alpha};\tfrac{1}{2};\xi^2\right)\right] \tag{6}$$ then we can use generalised Hermite polynomials $$H_\nu(z) = 2^\nu \sqrt{\pi} \left(\frac{{_1 F_1}(-\frac{\nu}{2};\frac{1}{2};z^2)}{\Gamma(\frac{1-\nu}{2})} - \frac{2z \cdot {_1 F_1}(\frac{1-\nu}{2};\frac{3}{2};z^2)}{\Gamma(-\frac{\nu}{2})} \right) \tag{7}$$ with $\nu:=1-\tfrac{1}{\alpha}$ ($\iff \tfrac{1-\nu}{2}=1-\tfrac{1}{2\alpha} \iff -\tfrac{\nu}{2}=\tfrac{1}{2}-\tfrac{1}{2\alpha}$). This gives us: $$f_t(\xi) = \frac{\sqrt{2}C}{\sigma t} (-\tfrac{\nu}{2})e^{-\xi^2} \Gamma(-\tfrac{\nu}{2}) 2^{-\nu} \pi^{-1/2} H_\nu(\xi) \tag{8}$$

In other words: The whole function is a constant multiple of $\tfrac{1}{t}e^{-\xi^2} H_\nu(\xi)$. Note that this is a true polynomial iff $\alpha\in\{1,\tfrac{1}{2},\tfrac{1}{3},\tfrac{1}{4},\ldots\}$ which may or may not be the case anyway in your situation.

Anyhow: Using the $\xi$-substitution, we find $$\langle f_t,\phi\rangle = const \int_0^\infty \frac{1}{t} e^{-\xi^2} H_\nu(\xi) \phi(b-\sqrt{2}\sigma t^\alpha \xi) d\xi$$ For $\alpha=1$ this goes to $const\cdot\phi(b)$ so that $f_t \to const\cdot \delta_b$ as desired (and the constant will probably work out alright). However, if $\alpha<1$, the integral goes to $sgn(\phi(b))\cdot\infty$.