# The $C^{\infty}$ Urysohn lemma

Jordan Bell
September 11, 2015

Define $\eta:\mathbb{R}\to\mathbb{R}$ by

 $\eta(t)=e^{-1/t}1_{(0,\infty)}(t).$

It is a fact that $\eta$ is $C^{\infty}$. This is proved by showing that for each $k\geq 1$ there is a polynomial $P_{k}$ of degree $2k$ such that $\eta^{(k)}(t)=P_{k}(t^{-1})e^{-1/t}$ for $t>0$, and that $\eta^{(k)}(0)=0$, which together imply that $\eta\in C^{k}$.

Define $\psi:\mathbb{R}^{d}\to\mathbb{R}$ by

 $\psi(x)=\eta(1-|x|^{2})=\begin{cases}e^{\frac{1}{|x|^{2}-1}}&|x|<1\\ 0&|x|\geq 1.\end{cases}$

Because $x\mapsto 1-|x|^{2}$ is $C^{\infty}:\mathbb{R}^{d}\to\mathbb{R}$, the chain rule tells us that $\psi$ is $C^{\infty}$.

For a function $\phi$ on $\mathbb{R}^{d}$ and for $t>0$, we define

 $\phi_{t}(x)=t^{-d}\phi(t^{-1}x).$

We now construct bump functions.11 1 The following construction of a bump function follows Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, second ed., p. 245, Lemma 8.18.

###### Theorem 1 ($C^{\infty}$ Urysohn lemma).

If $K$ is a compact subset of $\mathbb{R}^{d}$ and $U$ is an open set containing $K$, then there exists $\phi\in C^{\infty}(\mathbb{R}^{d})$ with $0\leq\phi\leq 1$, $\phi=1$ on $K$, and $\mathrm{supp}\,\phi\subset U$. Moreover, if $K$ is invariant under $SO(d)$ then the function $\phi$ constructed here is radial.

###### Proof.

Let

 $\delta=d(K,U^{c}),$

which is positive because $K$ is compact and $U^{c}$ is closed. Let

 $V=\left\{x\in\mathbb{R}^{d}:d(x,K)<\frac{\delta}{3}\right\}=K+B_{\delta/3},$

and define $f$ on $\mathbb{R}^{d}$ by

 $f=\left(\int_{\mathbb{R}^{d}}\psi(x)dx\right)^{-1}\psi_{\delta/3},$

whose support is

 $\mathrm{supp}\,f=\mathrm{supp}\,\psi_{\delta/3}=\overline{B_{\delta/3}}.$

Finally define $\phi$ on $\mathbb{R}^{d}$ by

 $\phi=1_{V}*f.$

Because $V$ is bounded and $f$ is $C^{\infty}$, the function $\phi$ is $C^{\infty}$. The support of $\phi$ is

 $\mathrm{supp}\,\phi=\mathrm{supp}\,(1_{V}*f)\subset\overline{\mathrm{supp}\,1_% {V}+\mathrm{supp}\,f}=\overline{V+\overline{B_{\delta/3}}}=K+\overline{B_{2% \delta/3}}\subset U.$

Because $1_{V}$ and $f$ are nonnegative, so is their convolution $\phi$. For any $x$,

 $\phi(x)=\int_{\mathbb{R}^{d}}1_{V}(x-y)f(y)dy\leq\int_{\mathbb{R}^{d}}f(y)dy=1,$

so $0\leq\phi\leq 1$. For $x\in K$, if $y\in V^{c}$ then $|x-y|\geq\delta/3$. But $f(u)=0$ for $|u|\geq\delta/3$, so in this case $f(x-y)=0$. This implies that for $x\in K$ the functions $y\mapsto 1_{V}(y)f(x-y)$ and $y\mapsto f(x-y)$ are equal, hence

 $\phi(x)=\int_{\mathbb{R}^{d}}1_{V}(y)f(x-y)dy=\int_{\mathbb{R}^{d}}f(x-y)dy=% \int_{\mathbb{R}^{d}}f(y)dy=1.$

This shows that $\phi=1$ on $K$, verifying all the assertions made about $\phi$.

The function $\psi$ is radial and so $f$ is too. If $V$ is invariant under $SO(d)$, then the indicator function $1_{V}$ is radial. Thus, if $K$ is invariant under $SO(d)$ then $1_{V}$ is radial, and the convolution of two radial functions is also radial, which means that $\phi$ is radial in this case. ∎

For example, take $d=1$, take $K$ to be the closed ball of radius $1$, and take $U$ to be the open ball of radius $2$. Then $\delta=d(K,U^{c})=1$ and $V=B_{4/3}$. In Figure 1 we plot the bump function $\phi$ constructed in the above theorem.