Limit of a sequence of test functions

Question

$ B(x,\frac{1}{n}) $ be open balls in $ \mathbb{R}^{N} $ with $ N\geq2 $, and $ (\phi_{n})$ be a sequence of test functions with support in $ B(x,\frac{1}{n}) $ such that $ \phi_{n}=1 $ on $ B(x,\frac{1}{2n}) $, for all $ n $. Suppose there exists a sequence $(z_{n})\in B(x,\frac{1}{n}) $ such that $$ \phi_{n}(z_{n})\not=0, $$ for all $ n $. Can we conclude that $$ \lim_{n\rightarrow\infty} \phi_{n}(z_{n})=1?$$

Answer 1

Heuristically speaking, for each $n$, let us consider the sets \begin{align} W_n:= B\left(x,\frac{1}{2n}\right) \subset V_n:=B\left(x,\frac{3}{4n}\right) \subset U_n:=B\left(x,\frac{1}{n}\right) \end{align} and construct a radial nonnegative test function $\phi_n$ such that $\phi_n \equiv 1$ on $B(x, \frac{1}{2n})$ and $\phi_n \equiv 0$ on $B(x, 1/n)^c$ while $\phi_n(y)\leq \frac{1}{2}$ for all $y \in B(x, \frac{3}{4n})^c$.

If we choose $z_n \in B(x, 1/n)\backslash B(x, \frac{3}{4n})$ such that $\phi_n(z_n) \neq 0$, then $\phi_n(z_n)\leq \frac{1}{2}$ which means \begin{align} \limsup_{n\rightarrow \infty} \phi_n(z_n) \leq \frac{1}{2}. \end{align} Hence we are done if we could show such a family of test functions exists.

Technical Details

Suppose $\varphi(|x|)$ is your standard radial bump function constructed from \begin{align} f(x) = \begin{cases} e^{-1/x^2} & \ \ \text{ if } x>0\\ 0& \ \ \text{ if } x\leq 0 \end{cases} \end{align} with the property $\varphi(|x|)\geq 0, \operatorname{supp}\varphi \subset B(0, 1)$ and $\int \varphi=1$.

For each $n$, we consider $W_n \subset V_n \subset U_n$ as above. Moreover, notice we have \begin{align} d(\bar W_n, V_n^c)= \frac{1}{4n} \ \ \text{ and } \ \ d(\bar V_n, U_n^c) = \frac{1}{4n}. \end{align} For convenience, we shall set $r_n=\frac{1}{4n}$. Next, consider the function \begin{align} \varphi_{r_n}(y) = \frac{1}{r_n^d}\varphi\left(\frac{y}{r_n}\right) \end{align} which has the properties $\operatorname{supp}\varphi_{r_n} \subset B(0, r_n)$ and \begin{align} \int_{\mathbb{R}^d}\varphi_{r_n}(x)\ dx = \int_{\mathbb{R}^d}\varphi\left(\frac{x}{r_n} \right)\frac{dx}{r_n^d} = \int_{\mathbb{R}^d} \varphi(u)\ du = 1. \end{align} Lastly, consider the function \begin{align} \phi_n(x) = \varphi_{r_n}\ast \chi_{V_n}(x)=\int_{\mathbb{R}^d}\varphi_{r_n}(x-y)\chi_{V_n}(y)\ dy. \end{align}

Let us check $\phi_n$ has the desired properties.

Observe if $x \in W_n$, then $\operatorname{supp}\varphi_{r_n} (x-\cdot)\subset B(x, r_n) \subset V_n$ since $d(\bar W_n, V_n^c) = \frac{1}{4n}$. Thus, it follows \begin{align} \phi_n(x) = \int_{\mathbb{R}^d}\varphi_{r_n}(x-y)\chi_{V_n}(y)\ dy = \int_{B(x, r_n)}\varphi_{r_n}(x-y)\ dy = 1. \end{align} Likewise, if $x \in U_n^c$ we see that $\operatorname{supp}\varphi_{r_n}(x-\cdot) \subset V_n^c$ which means \begin{align} \phi_n(x) = \int_{\mathbb{R}^d}\varphi_{r_n}(x-y)\chi_{V_n}(y)\ dy = \int_{B(x, r_n)}\varphi_{r_n}(x-y)\cdot 0\ dy = 0. \end{align} Hence $\phi_n$ is supported in $U_n$. It's also clear that $\phi_n$ is $C^\infty$ since it's obtain from a convolution of a function against a $C^\infty$ function.

The last property to check is $\phi_n(y) \leq \frac{1}{2}$ for all $y \in U_n\backslash V_n$.

Let $x \in U_n\backslash V_n$, then we see $B(x, r_n) \cap V_n$ miss at least half of $B(x, r_n)$ which means \begin{align} \phi_n(x)= \int_{\mathbb{R}^d} \varphi_{r_n}(x-y)\chi_{V_n}(y)\ dy = \int_{B(x, r_n)\cap V}\varphi_{r_n}(x-y)\chi_{V_n}(y)\ dy \leq \frac{1}{2}. \end{align}