Does $\lim_{x\rightarrow 0^+}Cxy^{\frac{-4}{3}}e^{\frac{-x^3}{9y}} (x>0, y>0)$ converge to delta function?

Question

Does $P(x,y) = Cxy^{\frac{-4}{3}}e^{\frac{-x^3}{9y}} (x>0, y>0)$ converge to $\delta(y)$ as $x \rightarrow 0^+$ for some constant C?

It's easy to check that $\lim_{x\rightarrow 0^+}Cxy^{\frac{-4}{3}}e^{\frac{-x^3}{9y}} = 0$.

When it comes to prove $\lim_{x\rightarrow 0^+}\int_{-\infty}^{+\infty} Cxy^{\frac{-4}{3}}e^{\frac{-x^3}{9y}}dy = 1$ or not, I don't know how to deal with it because it's elementary integral.

Any advice would be appreciated!

Answer 1

As it stands there is a problem. We have functions defined only on $(0,\infty)$ but want to show that they converge as distributions to $\delta$ which has its support outside of this interval. This doesn't fit well with the definition of distributions and limits of such. I therefore take the approach of extending the functions to take the value $0$ on $(-\infty, 0]$. In the following I have changed the variables; $x$ is used for the variable of integration ($y$ in the question), and $\epsilon$ for the parameter ($x$ in the question).

First, for $\epsilon>0$, I define $u_\epsilon : \mathbb{R} \to \mathbb{R}$ by $$ u_\epsilon(x) = \begin{cases} \epsilon \, x^{-4/3} \, e^{-\epsilon^3/(9x)} & (x>0) \\ 0 & (x \leq 0). \end{cases} $$

Then I take $\phi \in C_c^\infty(\mathbb{R})$ and start calculating, using a substitution: $$\begin{align} \langle u_\epsilon, \phi \rangle &= \int_{0}^{\infty} \epsilon \, x^{-4/3} \, e^{-\epsilon^3/(9x)} \, \phi(x) \, dx = \left\{ \xi := \frac{\epsilon^3}{9x}, \ x = \frac{\epsilon^3}{9\xi} \right\} \\ &= \int_{\infty}^{0} \epsilon \, (\frac{\epsilon^3}{9\xi})^{-4/3} \, e^{-\xi} \, \phi(\frac{\epsilon^3}{9\xi}) \, (-\frac{\epsilon^3}{9\xi^2}) \, d\xi \\ &= 9^{1/3} \int_{0}^{\infty} \xi^{-2/3} \, e^{-\xi} \, \phi(\frac{\epsilon^3}{9\xi}) \, d\xi \\ &\to 9^{1/3} \int_{0}^{\infty} \xi^{-2/3} \, e^{-\xi} \, \phi(0) \, d\xi \\ &\to 9^{1/3} \left( \int_{0}^{\infty} \xi^{-2/3} \, e^{-\xi} \, d\xi \right) \phi(0) \\ &= \langle 9^{1/3} \, \Gamma(1/3) \, \delta, \phi \rangle. \end{align}$$

Thanks to $\phi(x)$ vanishing for $|x|$ big enough, $\phi(\frac{\epsilon^3}{9\xi})$ vanishes for $\xi$ close to $0$. On the other hand, for big $|\xi|$ it might happen that $\phi(\frac{\epsilon^3}{9\xi})$ does not vanish; it actually tends to $\phi(0)$ as $|\xi| \to \infty.$ Luckily, it is bounded, which allows us to take the limit as $\epsilon \to 0$ inside the integral.

Thus, taking $C = \frac{1}{9^{1/3} \, \Gamma(1/3)},$ we have that $C u_\epsilon \to \delta$ when $\epsilon \to 0+$.

Answer 2

Solution

The answer to the question of the OP is "no".

But it is very easy to extend the definition of your interesting function which then goes to a delta function, and in this manner reconciliate the different positions in the discussion. Just replace $y$ by $|y|$ and you get the full symmetric delta function in the limit:

$$u_{s}(y,\epsilon) = \frac{\epsilon \left| y\right| ^{-4/3} \exp \left(-\frac{\epsilon ^3}{9 \left| y\right| }\right)}{2\ 3^{2/3} \Gamma \left(\frac{1}{3}\right)}$$

It is normalized

$$i_n = \int_{-\infty}^\infty u_{s}(y,\epsilon)\,dx = 1$$

and the integral over the test function $f(x)$ defined in the original post gives

$$i_f = \int_{-\infty}^\infty u_{s}(y,\epsilon)f(x)\,dx = \frac{3}{2}$$

This is the same result as that for the integral with the delta function insted of $u_{s}(y,\epsilon)$

Remark:

With the same idea but even simpler (and aesthetically pleasing) is this function

$$v(x,\epsilon)= \sqrt{\frac{\epsilon }{\pi }}\frac{1}{x^2} \exp \left(-\frac{\epsilon }{x^2}\right)$$

Original post

My answer to the question "Does $P(x,y) = Cxy^{\frac{-4}{3}}e^{\frac{-x^3}{9y}} (x>0, y>0)$ converge to $\delta(y)$ as $x \rightarrow 0^+$ for some constant C?" was given in a comment: "The one sided $y$-integral is in fact constant: for $x>0$ we have $\int_0^\infty Cxy^{\frac{-4}{3}}e^{\frac{-x^3}{9y}} \,dy=3^\frac{2}{3} \Gamma(\frac{1}{3})$. But a true Delta function would be symmetric in $y$, and this is not the case here."

So the brief answer is "no".

A lengthy discussion followed up. Here I wish to clarify my point with a simple example, hoping that we can come to a common understanding. Altervatively, I can learn what is wrong with my argument.

I start by defining two functions containing a parameter $w>0$ so that the limit $w \to 0$ each function goes to a delta functions with symmetry properties corresponding to theses functions.

One function is symmetric in $x$

$$s(x,w) = \frac{1}{2w} \left\{ \begin{array}{ll} 1 & |x| \lt w\\ 0 & \text{else} \\ \end{array} \right. $$

the other one is unsymmetric

$$u(x,w) = \frac{1}{w} \left\{ \begin{array}{ll} 1 & 0\lt x \lt w\\ 0 & \text{else} \\ \end{array} \right. $$

Both functions are normalized

$$n_s = \int_{-\infty}^\infty s(x,w)\,dx =1$$ $$n_u = \int_{-\infty}^\infty u(x,w)\,dx =1$$

Next we define a test function $f(x)$ which intentionally is not continuous at $x=0$ but has a jump

$$f(x) = \left\{ \begin{array}{ll} 2 & x \gt 0\\ \frac{3}{2} & x =0\\ 1 & x \lt 0 \\ \end{array} \right. $$

Now we calculate two integrals

$$i_s = \int_{-\infty}^\infty s(x,w) f(x)\,dx = \frac{3}{2}$$

$$i_u = \int_{-\infty}^\infty u(x,w) f(x)\,dx = 2$$

Finally we compare the results with that over the standard Dirac delta function in a CAS (here Mathematica) which is

$$i_{\delta} = \int_{-\infty}^\infty \delta(x) f(x)\,dx = \frac{3}{2}$$

We conclude that the symmetric approximation $s(x,w)$ gives the correct result while the unsymmetric does not.

As the function $u_{\epsilon}(y)$ of the OP (considered normalized) is not symmetric the integral over $u_{\epsilon}(y) f(y)$ (is equal to 2 and not the Delta function result $\frac{3}{2}$. Hence I have justified my initial answer.