$f$ is 1-periodic prove $\lim_{n \to \infty} \int_{0}^{1} \sin^2(πx)f(nx)\,dx = \frac{1}{2} \int_{0}^{1} f(x)\,dx$

Question

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Let $f(x)$ be a continuous function $f:R \to R$ with period $1$. Prove that $\displaystyle \lim_{n \to \infty} \int_{0}^{1} \sin^2(\pi x)f(nx)\,\mathrm{d}x = \frac{1}{2} \int_{0}^{1} f(x)\,\mathrm{d}x$.

Added the solution as an answer.

Answer 1

An alternative solution: Let $I = \int_0^1 f(t) \, dt$. Then $$ \int_0^1 \sin^2(\pi x)f(nx)\,dx = I \int_0^1 \sin^2(\pi x) \, dx + \int_0^1 \sin^2(\pi x)(f(nx)-I)\,dx \\ = \frac 12 I + \int_0^1 \sin^2(\pi x)(f(nx)-I)\,dx \, . $$ Define $G(x) = \int_0^x (f(t) - I) \, dt$. Then $G'(x) = f(x) - I$ and $G(0) = G(1) = 0$. In particular, $G$ is continuous and $1$-periodic and therefore bounded on $\Bbb R$. Now do integration by parts: $$ \int_0^1 \sin^2(\pi x)(f(nx)-I)\,dx = \int_0^1 \sin^2(\pi x)G'(nx) \, dx \\ = - \frac \pi n \int_0^1 \sin(2 \pi x) G(nx) \, dx $$ and that converges to zero for $n \to \infty$ because the integral is bounded independently of $n$.

Answer 2

$$\lim_{n \to \infty} \int_{0}^{1} \sin^2(πx)f(nx)dx$$

$$=\lim_{n \to \infty} \sum_{r=0}^{n-1} \int_{r/n}^{r/n+1/n} \sin^2(πx)f(nx)dx $$

$$=\lim_{n \to \infty} \sum_{r=0}^{n-1} \sin^2(πr/n) \int_{r/n}^{r/n+1/n} f(nx)dx$$ $nx=t$.

$$=\sum_{r=0}^{n-1}(\sin^2(πr/n))\frac{1}{n} \int_{r}^{r+1} f(t)dt$$

$$=\int_{0}^{1}f(t)dt \int_{0}^{1} \sin^2(πx)dx = \frac{1}{2} \int_{0}^{1} f(x)dx$$

Answer 3

Here is another solution based on this post and this solution which states that if $\phi\in \mathcal{L}_1(\mathbb{R})$, $f$ is a bounded $T$ periodic function and $a_n$ is any real numeric sequence, then

$$ \lim_n\int \phi(x)f(nx+a_n)\,dx=\Big(\frac{1}{T}\int^T_0f\Big)\int \phi \tag{1}\label{one} $$

In terms of the OP, $T=1$, $a_n=0$, and $\phi(x)=\mathbb{1}_{[0,1]}(x)\sin^2(\pi x)$. The rest follows by substituting $\phi$ on $\eqref{one}$.