Let \(f,g\) be continuous \(1\)-periodic functions on \(\mathbb R\). Prove that \[ \lim_{n\to\infty}\int_0^1 f(x)g(nx)\,dx =\left(\int_0^1f\right)\left(\int_0^1g\right). \]
Solution
Partition \([0,1]\) into the intervals \[ \left[\frac jn,\frac{j+1}{n}\right], \qquad j=0,\ldots,n-1. \]
1. Rewrite the integral using periodicity of \(g\).
On the \(j\)th interval set \[ x=\frac{j+t}{n}, \qquad 0\le t\le1. \] Then \(dx=dt/n\) and, because \(g\) is \(1\)-periodic, \[ g(nx)=g(j+t)=g(t). \] Therefore \[ \begin{aligned} \int_0^1 f(x)g(nx)\,dx &=\sum_{j=0}^{n-1}\frac1n \int_0^1 f\left(\frac{j+t}{n}\right)g(t)\,dt\\ &=\int_0^1 g(t)A_n(t)\,dt, \end{aligned} \] where \[ A_n(t)=\frac1n\sum_{j=0}^{n-1} f\left(\frac{j+t}{n}\right). \]
2. The shifted Riemann sums \(A_n(t)\) converge uniformly in \(t\) to \(\int_0^1f\).
Because \(f\) is continuous and \(1\)-periodic, it is uniformly continuous on \(\mathbb R\). Let \[ \omega_f(\delta) =\sup\{|f(x)-f(y)|:|x-y|\le\delta\}. \] Then \[ \omega_f(\delta)\to0 \] as \(\delta\to0\).
For \(0\le t\le1\), \[ \left|\frac{j+t}{n}-\frac jn\right|\le\frac1n, \] so \[ \left|A_n(t)-\frac1n\sum_{j=0}^{n-1}f(j/n)\right| \le\omega_f(1/n). \] The ordinary left-endpoint Riemann sums satisfy \[ \frac1n\sum_{j=0}^{n-1}f(j/n) \longrightarrow\int_0^1f(x)\,dx. \] The displayed bound is independent of \(t\), hence \[ A_n(t)\longrightarrow\int_0^1f(x)\,dx \] uniformly for \(t\in[0,1]\).
3. Pass to the limit in the integral.
Since \(g\) is continuous on \([0,1]\), it is bounded. Uniform convergence from step 2 gives \[ \begin{aligned} \lim_{n\to\infty}\int_0^1 f(x)g(nx)\,dx &=\int_0^1g(t) \left(\int_0^1f(x)\,dx\right)dt\\ &=\left(\int_0^1f(x)\,dx\right) \left(\int_0^1g(t)\,dt\right). \end{aligned} \] This is the desired formula.