The C Urysohn lemma

Jordan Bell
September 11, 2015

Define η: by


It is a fact that η is C. This is proved by showing that for each k1 there is a polynomial Pk of degree 2k such that η(k)(t)=Pk(t-1)e-1/t for t>0, and that η(k)(0)=0, which together imply that ηCk.

Define ψ:d by


Because x1-|x|2 is C:d, the chain rule tells us that ψ is C.

For a function ϕ on d and for t>0, we define


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 Urysohn lemma).

If K is a compact subset of Rd and U is an open set containing K, then there exists ϕC(Rd) with 0ϕ1, ϕ=1 on K, and suppϕU. Moreover, if K is invariant under SO(d) then the function ϕ constructed here is radial.




which is positive because K is compact and Uc is closed. Let


and define f on d by


whose support is


Finally define ϕ on d by


Because V is bounded and f is C, the function ϕ is C. The support of ϕ is

suppϕ=supp(1V*f)supp 1V+suppf¯=V+Bδ/3¯¯=K+B2δ/3¯U.

Because 1V and f are nonnegative, so is their convolution ϕ. For any x,


so 0ϕ1. For xK, if yVc then |x-y|δ/3. But f(u)=0 for |u|δ/3, so in this case f(x-y)=0. This implies that for xK the functions y1V(y)f(x-y) and yf(x-y) are equal, hence


This shows that ϕ=1 on K, verifying all the assertions made about ϕ.

The function ψ is radial and so f is too. If V is invariant under SO(d), then the indicator function 1V is radial. Thus, if K is invariant under SO(d) then 1V is radial, and the convolution of two radial functions is also radial, which means that ϕ 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 δ=d(K,Uc)=1 and V=B4/3. In Figure 1 we plot the bump function ϕ constructed in the above theorem.

Figure 1: The bump function ϕ, for d=1, K=[-1,1], U=(-2,2); δ=1 and V=(-4/3,4/3)