If $f\in L^1(\mathbb{R}^d)$ is uniformly continuous, do its integrals over spheres of increasing radius decay to zero?

Question

To be more precise, if $f\in L^1(\mathbb{R}^d), d>1$ and is uniformly continuous, is it true that $$\lim_{R\to\infty} \int_{|x|=R} |f(x)| \ dS(x) = 0 \ ?$$

where $dS$ is the surface measure on the sphere of radius $R$.

If $f\in L^1(\mathbb{R})$ and uniformly continuous, it's well-known that $|f(x)|\to 0$ as $|x|\to\infty$. But this problem seems to require more precise decay, and I have a feeling it's false. One idea for a counterexample: pick $f$ to be a radial series of bump functions centered at $|x|=n$ with slowly decaying heights and whose supports are disjoint and shrink fast enough to ensure integrability. This feels like main scenario for a counterexample (if one exists), but I wasn't able to get the details to work out.

Answer 1

You're right, it's false.

First, consider the function $$\operatorname{bump}(x;x_0,a)=\begin{cases}\exp\left(\frac{-1}{1-(x-x_0)^2/a^2}\right)&|x-x_0|<a \\ 0 & |x-x_0|\geq a\end{cases}$$ Where $x,x_0,a\in\Bbb{R}_+$. This is a "bump function" centered at $x_0$, with width $2a$.

Some remarks -

• This function is everywhere nonnegative (importantly, equal to its own absolute value)

• This function is uniformly continuous, since it is continuous on a compact set.

We first consider an integral which will later be useful (trust me!) $$\int_{\Bbb R_+}x\operatorname{bump}(x,x_0,a)\mathrm dx=\int_{x_0-a}^{x_0+a} x\exp\left(\frac{-1}{1-((x-x_0)/a)^2}\right)\mathrm dx \\= \int_{-a}^a (z+x_0)\exp\left(\frac{-1}{1-(z/a)^2}\right)\mathrm dz \\ = \underbrace{\int_{-a}^a z\exp\left(\frac{-1}{1-(z/a)^2}\right)\mathrm dz}_{=0 ~\text{b.c odd func. on symm. interv.}}+x_0\int_{-a}^a \exp\left(\frac{-1}{1-(z/a)^2}\right)\mathrm dz \\ =x_0\int_{-1}^1 \exp\left(\frac{-1}{1-s^2}\right)a\mathrm ds \\ =c~x_0a$$ Where $$c=\int_{-1}^1 \exp\left(\frac{-1}{1-x^2}\right)\mathrm dx \\ \text{and obviously,}~~0<c<1$$ Finding a closed form for $c$ is of no importance (and I suspect could be impossible). The point is that the integral of the bump function scales linearly with its width.

Now, consider the function $$F(x)=\sum_{n=1}^\infty \frac{1}{n}\operatorname{bump}\left(x,n,\frac{1}{2n^2}\right)$$

(This is uniformly continuous - should be easy enough to check.)

Finally, consider the function

$$\Phi : \Bbb{R}^2\to \Bbb{R} \\ \Phi(\boldsymbol x)=F(|\boldsymbol x|)$$

We can see that $\Phi$ is $L^1$.

$$\int_{\Bbb{R}^2}|\Phi(\boldsymbol x)|\mathrm d^2 \boldsymbol x = \int_0 ^ \infty F(r)~2\pi r~\mathrm dr \\ =2\pi\sum_{n=1}^\infty \int_0^\infty r\frac{1}{n}\operatorname{bump}\left(r,n,\frac{1}{2n^2}\right)\mathrm dr$$ Which, using our previous result, is $$2\pi\sum_{n=1}^\infty \int_0^\infty r\frac{1}{n}\operatorname{bump}\left(r,n,\frac{1}{2n^2}\right)\mathrm dr =2\pi \sum_{n=1}^\infty c n \frac{1}{n}\frac{1}{2n^2}=\pi c\sum_{n=1}^\infty \frac{1}{n^2}=c\pi \zeta(2)<\infty$$

However, if we consider the sequence of integrals of $\Phi$ on concentric circles around the origin,

$$A_n=\int\limits_{\partial \mathbb{B}(0,n) }\Phi(\boldsymbol x)\mathrm d^1 \boldsymbol x$$ One can easily check that $$A_n=2\pi \\ \forall n$$ And so $$\lim_{R\to\infty} \int\limits_{\partial \mathbb{B}(0,R)}|\Phi(\boldsymbol x)|\mathrm d^1 \boldsymbol x=2\pi \neq 0$$ I wonder if it's possible to make this limit infinite. Anyway, this serves fine as a counterexample.