Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 23I (Analysis of Functions)

For $f\in L^{\infty }({\mathbb{R}}^n)$ and $g\in L^1({\mathbb{R}}^n)$, define their convolution byIt is bounded becauseIf ${\tau }_{h}g(z)=g(z+h)$, thenContinuity of translation in $L^1$ makes the right-hand side tend to zero with $h$, uniformly in $x$. Thus the convolution of L infinity and L1 functions is bounded and continuous.

Now let $f=1_A$ and $g(x)=1_A(-x)$. ThenAt zero this equals $|A|>0$. By continuity it remains positive throughout some open neighbourhood $U$ of zero. Positivity at $z$ means that some $y$ satisfies $y\in A$ and $y-z\in A$, so $z\in A-A$. HenceThis is the Steinhaus theorem.

The conclusion also holds when $|A|=\infty$. Since ${\mathbb{R}}^n$ is the union of bounded balls, some measurable subset $A\cap B(0,R)$ has finite positive measure. Its difference set lies inside $A-A$ and already contains a neighbourhood of zero.

If $u,v\in f^{-1}(B_{\epsilon })$, additivity gives $f(-v)=-f(v)$ andTherefore

For every positive integer $k$, additivity gives $f(kx)=kf(x)$. HenceConversely, if $z\in f^{-1}(B_{k\epsilon })$, write $z=k(z/k)$. Thenso $z/k\in f^{-1}(B_{\epsilon })$. Thus

To prove continuity, fix $\epsilon >0$ and putThese sets are measurable and ${\mathbb{R}}^n={\bigcup }_{k\ge 1}{E}_k$, so at least one $E_k$ has positive Lebesgue measure. Part (ii) gives $E_k=k{E}_1$, and therefore $E_1$ also has positive measure. By the Steinhaus theorem, $E_1-E_1$ contains an open neighbourhood $U$ of zero. Part (i) givesThis proves continuity at zero. Finally,as $h\to 0$, so $f$ is continuous at every $x$. Thus every such map is an instance of the theorem that a measurable additive function is continuous.