Convergence weakly to measure?

Question

Let $ w_e (x)=\frac {\partial^2}{\partial x_1^2}\sqrt {x_1^2+e}.$ Show that $ w_e $ converges weakly as $e\to 0$ in the dual of $ C (\bar {B_1}) $ to measure $\mu $

I am that dual $ C (B) $ is borel measure but I did not understand very well the question

Answer 1

By definition of weak convergence, you must show that for every $\phi\in C(\overline B_1)$ the following holds: $$\int_{B_1} \phi w_\epsilon \to \int_{B_1} \phi\,d\mu,\quad \epsilon\to 0 \tag1$$

It helps if you can guess $\mu$ beforehand. Here $\sqrt{x_1^2+\epsilon}$ converges to $|x_1|$ as $\epsilon\to 0$. The first derivative of $|x_1|$ is the sign function $\operatorname{sign}x_1$, and the second derivative is $2\delta_0$, twice the Dirac mass at $x_1=0$ (multiplied by the Lebesgue measure in $x_2,\dots,x_n$). This is what $\mu$ is going to be.

As an aside, if you are familiar with distributional derivatives and the convergence of distributions, then the above paragraph is already a proof: $w_\epsilon\to |x_1|$ as distributions, hence $w_\epsilon''$ converges to the second derivative of $|x_1|$ in the sense of distributions.

But an explicit computation may be more instructive: $$w_\epsilon = \frac{\partial }{\partial x_1} \frac{x_1}{(x_1^2+\epsilon)^{1/2}} = \frac{\epsilon}{(x_1^2+\epsilon)^{3/2}} \tag1$$ There are three things to be taken from (1):

• $w_\epsilon$ is a positive function
• $\int_{-\infty}^\infty w_\epsilon = 2$
• For any $\delta>0$, $$\int_{-\delta}^\delta w_\epsilon = \frac{2\delta}{(\delta^2+\epsilon)^{1/2}}\tag2$$

For any fixed $\delta$, taking $\epsilon\to 0$ in (2) yields $2$ as the limit. Therefore, the contribution from outside $[-\delta,\delta]$ to the integral $\int_{-\infty}^\infty w_\epsilon$ tends to $0$ as $\epsilon\to 0$. Taken together, the above means that $w_\epsilon$ is (twice) an approximation of unity.

Now split the integral $$\int_{B_1} \phi w_\epsilon $$ into the integrals over $B_1\cap \{|x_1|> \delta\}$ and $B_1\cap \{|x_1|\le \delta\}$. The first integral tends to zero. In the second, $\phi(x)$ is close to $\phi(0,x_2,\dots,x_n)$ by the continuity of $\phi$. Arbitrarily close, since you can choose $\delta$ as small as you wish. Hence, $$\int_{B_1} \phi w_\epsilon \to 2 \int_{B_1'} \phi(0,x_2,\dots,x_n) \,dx_2\dots dx_n$$ where $B_1' $ is the unit ball in the hyperplane $x_1=0$.