13
$\begingroup$

For each $\epsilon > 0$, define $f_\epsilon:\mathbb R\to \mathbb R$ as follows: \begin{align} f_\epsilon(k) = \frac{1}{\pi}\frac{\epsilon}{\epsilon^2+k^2}. \end{align} How does one rigorously show (in the sense of distributions) that $f_\epsilon(k) \to \delta(k)$ as $\epsilon\to 0$?

I think I have the essential structure of an argument using contour integration, but it's missing some details that I don't have the expertise to fill in.

For each $a>0$, let $C_a$ be the CCW contour consisting of a straight segment between $-a$ and $a$ on the real axis, and a semicircular segment in the upper half plane of radius $a$. Then morally speaking, I'd hope the following steps are correct: \begin{align} \lim_{\epsilon\to 0}\int_{-\infty}^\infty f_\epsilon(k)\varphi(k)\, dk &= \lim_{\epsilon\to 0} \lim_{a\to\infty}\int_{C_a} \frac{1}{\pi}\frac{\epsilon}{(z-i\epsilon)(z+i\epsilon)}\varphi(z) \, dz \\ &= \lim_{\epsilon\to 0} (2\pi i) \frac{1}{\pi}\frac{\epsilon}{(i\epsilon + i\epsilon)}\varphi(i\epsilon) \\ &= \varphi(0) \end{align} However, I'm most concerned about the details of extending the test function $\varphi$ to a sufficiently nice function on $\mathbb C$ in order to perform the contour integration. In the book I'm studying, a test function is defined as a function in $C^\infty(\mathbb R)$ such that it and all its derivatives are $O(|x|^{-N})$ for all $N$ as $|x|\to\infty$.

Does any test function have a nice continuation to $\mathbb C$ that makes the above steps valid? Perhaps there is a way of doing this without contour integration so that one doesn't have to worry about continuation?

Edit. I was made aware by Mister Benjamin Dover below that there is in fact a quite general way to argue convergence to $\delta$ without complex analysis. I'm most interested at this point in determining if there is some way to make my manipulation above rigorous -- those sorts of arguments abound in the physics literature.

$\endgroup$
2
$\begingroup$

I think it is easier to do it without complex analysis, just using elementary results on integration theory. The following is a more general, often useful result.

Let $K\in L^1(\mathbf{R})$, for $\varepsilon>0$ let $K_\varepsilon$ be given by $K_\varepsilon(x)=\frac{1}{\varepsilon}K(x/\varepsilon)$; assume also that the integral of $K$ over all of $\mathbf{R}$ is equal to $1$. Then if $\phi\in L^\infty(\mathbf{R})$ we have $\phi_\varepsilon=\phi\ast K_\varepsilon\rightarrow\phi$ for $\varepsilon\downarrow 0$ at every point of continuity of $\phi$. (This is theorem 9.8 in Wheeden's measure and integral.)

Now for your case take the poisson kernel $$K(x)=\frac{1}{\pi}\frac{1}{1+x^2},$$ and $\phi$ be of class $C^\infty_0(\mathbf{R})\subset L^\infty(\mathbf{R})$. Then just notice that $$\phi_\varepsilon(0)=(\phi\ast K_\varepsilon)(0)=\frac{1}{\pi}\int_{\mathbf{R}}\phi(x)\frac{\varepsilon}{\varepsilon^2+x^2}dx.$$

For the proof of the result cited above (which is not that hard and a standard argument anyway): enter image description here

(As you can see, there is some notational conflict between the notation used in the book and the notation which you use in the question; I suggest we ignore this.)

$\endgroup$
  • 1
    $\begingroup$ +1: This is great, thank you. I'm still hoping, however, that someone will be able to tell me how to make the approach that uses complex analysis work as a matter of interest. $\endgroup$ – joshphysics Feb 20 '15 at 1:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.