Show that $\lim_{n\to\infty} \int_{0}^{1}f(x)g(nx)dx = 0$ $g$ periodic on $\mathbb{R}$

Question

Let $g$ be a continuous periodic function on $\mathbb{R}$ with $g(x + 1) = g(x)$. Assume that $\int_{0}^{1}g(x)dx = 0$.

(a) Let $f$ be continuous on all $\mathbb{R}$

Show that $\lim_{n\to\infty} \int_{0}^{1}f(x)g(nx)dx = 0$

(b) Let $f$ be Lebesgue integrable on all $\mathbb{R}$

Show that $\lim_{n\to\infty} \int_{0}^{1}f(x)g(nx)dx = 0$

My attempt:

(a)

For part (a) I dont know how to proceed. I have a solution that uses the following:

$$\lim_{n\to\infty} \int_{0}^{1}f(x)g(nx)dx = \int_{0}^{1}f(x)\int_{0}^{1}g(x)dx$$ and by hypothesis the results follow. There is another approach ?

(b) (This needs verification)

$g(x)$ is continuous and periodic on $\mathbb{R}$, then $\exists B > 0$ such that $|g(x)|< B$.

$f$ Lebesgue integrable on $[0,1]$ $\rightarrow \forall \epsilon > 0, \exists h$ step function such that $\int_{0}^{1}|f - h| < \frac{\epsilon}{B}, \forall x \in [0,1]$

then

$$| \int_{0}^{1}f(x)g(nx)dx | \leq \int_{0}^{1}|f(x) - h(x)| |g(nx)|dx + |\int_{0}^{1} h(x)g(nx)dx| \leq \epsilon + |\int_{0}^{1} h(x)g(nx)dx|$$

We must show that $\lim_{n\to \infty}\int_{0}^{1} h(x)g(nx)dx|$

Since $h$ is a step function,

$$h(x) = \sum_{i=1}^{k}\lambda_i\chi_{O_i}(x), \mbox{ where } \bigcup_{i=1}^{k}O_i = [0,1]$$

Hence, for $M = \max_{i = 1,...,k}|\lambda_i|$

$$\int_{0}^{1} h(x)g(nx)dx = \sum_{i=1}^{k}\lambda_i\int_{O_i}g(nx)dx$$ $$\leq M\sum_{i=1}^{k}\int_{O_i}g(nx)dx$$ $$= M\int_{0}^{1}g(nx)dx| = \frac{M}{n}\int_{0}^{n}g(x)dx = M\int_{0}^{n}g(x) = 0$$

Then

$$|\int_{0}^{1}f(x)g(nx)dx| < \epsilon$$

Answer 1

For part (a): Let $I=\int_{0}^{1}|g(x)|dx$. Let $\epsilon>0$ be given. Note that $f$ is uniformly continuous on $[0,1]$, so there exists $\delta>0$ such that $|f(x)-f(y)|<\frac{\epsilon}{I+1}$ whenever $x,y\in[0,1]$ with $|x-y|<\delta$. Choose $N\in\mathbb{N}$ such that $\frac{1}{N}<\delta$. By considering the substitution $y=nx$, we may rewrite \begin{eqnarray*} \int_{0}^{1}f(x)g(nx)dx & = & \frac{1}{n}\int_{0}^{n}f(\frac{y}{n})g(y)dy\\ & = & \frac{1}{n}\sum_{k=1}^{n}\int_{k-1}^{k}f(\frac{y}{n})g(y)dy. \end{eqnarray*} Let $n\geq N$ be arbitrary. Denote $f_{k}=f(\frac{k}{n})$, for $k=1,2,\ldots n$. By periodicity of $g$, we have $\int_{k-1}^{k}g(y)dy=\int_{0}^{1}g(y)dy=0$. Therefore \begin{eqnarray*} & & |\int_{k-1}^{k}f(\frac{y}{n})g(y)dy|\\ & = & |\int_{k-1}^{k}f(\frac{y}{n})-f_{k}g(y)dy|\\ & \leq & \int_{k-1}^{k}|f(\frac{y}{n})-f_{k}||g(y)|dy\\ & \leq & \int_{k-1}^{k}\frac{\epsilon}{I+1}|g(y)|dy\\ & = & \frac{\epsilon}{I+1}\int_{0}^{1}|g(y)|dy\\ & = & \epsilon\cdot\frac{I}{I+1}\\ & \leq & \epsilon. \end{eqnarray*} Now it is clear that \begin{eqnarray*} & & |\int_{0}^{1}f(x)g(nx)dx|\\ & \leq & \frac{1}{n}\sum_{k=1}^{n}|\int_{k-1}^{k}f(\frac{y}{n})g(y)dy|\\ & \leq & \frac{1}{n}\sum_{k=1}^{n}\epsilon\\ & = & \epsilon. \end{eqnarray*}

Answer 2

For (b): Since $g$ is continuous, there exists $M>0$ such that $|g(x)|\leq M$ for all $x\in[0,1]$. By periodicity of $g$, it follows that $|g(x)|\leq M$ for all $x\in\mathbb{R}$. Recall that Lebesgue integrable function can be approximated by integrable continuous function in $||\cdot||_{1}$-norm. More precisely, given $\epsilon>0$, there exists a continuous function $h:\mathbb{R}\rightarrow\mathbb{R}$ such that $\int_{0}^{1}|f-h|<\epsilon$.

Let $\epsilon>0$ be given. Choose continuous function $h:\mathbb{R}\rightarrow\mathbb{R}$ such that $\int_{0}^{1}|f-h|<\frac{\epsilon}{2M}.$ By part (a), there exists $N$ such that $|\int_{0}^{1}h(x)g(nx)dx|<\frac{\epsilon}{2}$ whenever $n\geq N$. Now let $n\geq N$ be aribitrary, then \begin{eqnarray*} & & |\int_{0}^{1}f(x)g(nx)dx|\\ & \leq & |\int_{0}^{1}[f(x)-h(x)]g(nx)dx|+|\int_{0}^{1}h(x)g(nx)dx|\\ & \leq & M\int_{0}^{1}|f(x)-h(x)|dx+\frac{\epsilon}{2}\\ & < & \epsilon. \end{eqnarray*}