Weak convergence to the average

Question

Let $\varphi\in L_\infty(\mathbb R)$ be a function for which an average value exists in the following sense: $$ \bar\varphi=\lim_{r\to+\infty}\frac1{2r}\int_{-r}^r\varphi(x)\,dx\in\mathbb R. $$ Is it true that for any $f\in L_1(\mathbb R)$ $$ \lim_{r\to+\infty}\int_{-\infty}^\infty f(x)\varphi(rx)\,dx= \bar \varphi\int_{-\infty}^\infty f(x)\,dx? $$ In other words functions $\varphi(rx)$ converge weakly to the average $\bar \varphi$ in $L_\infty(\mathbb R)$.

Answer 1

Res. Take $\varphi(x)={\text{sign}}(x)$. Then $\overline{\varphi}=0$, and then take $f$ to be any $L^1$ function that is $0$ on $(-\infty,0)$ and for which $\int f \neq 0$.

However. If we consider $\varphi(x)$ even (i.e. $\varphi(x)=\varphi(-x)$ for all $x$), the result holds.

Proof.

We show the result by measure-theoretic "induction".

Consider a bounded interval $A = [a,b]$, and let $f(x)={\bf{1}}_A(x)$ bethe characteristic function of the interval $A$.

Then $$ \int_{-\infty}^\infty f(x)\varphi(r x)dx= \int_a^b \varphi(r x) dx= \frac{1}{r}\int_{r a}^{r b} \varphi(x) dx\,. $$ Write $$ \frac{1}{r}\int_{r a}^{r b} \varphi(x) dx = b\cdot \left(\frac{1}{r b}\int_{0}^{r b} \varphi(x) dx\right) - a\cdot \left(\frac{1}{r a}\int_{0}^{r a} \varphi(x) dx\right) \to (b-a) \cdot\overline{\varphi} \,, $$ because now $\lim\limits_{r\to\infty} \frac{1}{r}\int_0^r \varphi(x)dx=\overline{\varphi}$ as $\varphi$ is even (even if we consider negative $r$).

Thus we get on with the induction: the result holds for characteristic functions bounded intervals, hence for linear combinations of characteristic functions of bounded intervals.

Now, linear combinations of characteristic functions of bounded intervals are dense in $L^1(\mathbb{R})$ and you may prove that this implies the result because $\varphi\in L^\infty(\mathbb{R})$ (and so in particular it is bounded! which is crucial to making this step work).

Answer 2

Let $\varphi$ be the sign function, that is, $\varphi(x)=-1$ if $x$ is negative and $1$ if $x$ is positive. Then $\bar\varphi=0$ and for any positive $r$, $$\int_{-\infty}^\infty f(x)\varphi(rx)\,dx=\int_{0}^\infty f(x) \,dx-\int_{-\infty}^0f(x) \,dx.$$