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I'm having difficulty with a proof in the paper "Approximation capabilities of multilayer feedforward networks" by Kurt Hornik. By definition $\psi$ is supposed to be bounded and non-constant, on page 256 in the proof of theorem 5 it says:

the integral $\int_\mathbb{R}\psi(t)f(t)dt$ vanishes for all integrable functions $f$ which have zero integral. It is easily seen that this implies that $\psi$ is constant which was ruled out by assumption.

Well, I'm not easily seeing the implication, so far I have that for every $A,B\subseteq\mathbb{R}$ with $\int_\mathbb{A}1dt=\int_\mathbb{B}1dt$ we get $\int_\mathbb{A}\psi(t)dt=\int_\mathbb{B}\psi(t)dt$ but of course that can also be the case for non-constant $\psi$.

Any help would be appreciated, thanks.

Sumanta
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Jack Larsson
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1 Answers1

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For $\psi\in L^{\infty}(\Bbb R)$ suppose $\int_{\Bbb R}f\psi=0$ for all $f\in L^1(\Bbb R)$ with $\int_{\Bbb R}f=0$. Hence for the function $g=\chi_{(-\epsilon,\epsilon)}-\chi_{(a-\epsilon,a+\epsilon)}$ we have $\int_{\Bbb R}g=0$ for all $a\in \Bbb R$ and $\epsilon>0$. So that, $\int_{\Bbb R}g\psi=0$. Therefore, $\int_{(-\epsilon,\epsilon)}\psi=\int_{(a-\epsilon,a+\epsilon)}\psi$ for all $a\in \Bbb R$ and $\epsilon>0$.

Hence, for almost every $x\in \Bbb R$ we have, $\psi(x)=\lim_{r\to 0+}\frac{1}{2r}\int_{(x-r,x+r)}\psi=\lim_{r\to 0+}\frac{1}{2r}\int_{(0-r,0+r)}\psi=\psi(0)$. So $\psi$ is constant almost everywhere.

Here I have used a lemma related Hardy-Littlewood maximal function which says that, for a locally integrable function $\phi$ defined over $\Bbb R^n$ we have $$\lim_{r\to 0+}\frac{1}{\lambda(B_r(x))}\int_{B_r(x)}\phi(y)dy=\phi(x)\text{ a.e. }x\in \Bbb R^n.$$

Sumanta
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  • Thank you very much! I actually thought about using this technique but dismissed it because I was sure it only holds for continuous functions and didn't know the name. – Jack Larsson Nov 17 '19 at 12:49