How to prove the convergence of the multiplication of a periodic function and a continuous function on [0, 1]

Question

I'm stuck in this problem:

Let $\lambda \in(0,1),$ and $a, b \in \mathbf{R},$ and $h: \mathbf{R} \rightarrow \mathbf{R}$ be the periodic extension of the function $$ h(x)=\left\{\begin{array}{ll} a & \text { if } 0 \leq x<\lambda \\ b & \text { if } \lambda \leq x<1 \end{array}\right. $$ and define $g_{n}:[0,1] \rightarrow \mathbf{R}$ by $g_{n}(x)=h(n x) .$ For each $\varphi(x) \in C^{1}[0,1],$ prove the convergence of $$ \lim _{n \rightarrow \infty} \int_{0}^{1} g_{n}(x) \varphi(x) d x $$ I don't even know where to start.

Answer 1

Let $A=\int_0^{1} h(t)dt$. The required limit is $\phi(1) A-\int_0^{1} x\phi'(x)dx$. First integrate by parts: $\int g_n(x)\phi(x) dx=G_n(x)\phi (x)dx|_0^{1}-\int_0^{1} G_n(x) \phi '(x)dx$ where $G_n(x)=\int_0^{x} g_n(x)dx =\frac 1n \int_0^{nx} h(t)dt$. Next check that $G_n(x) \to x A$ for each $x$ and that $\frac 1n \int_0^{n} h(t)dt$ is uniformly bounded. By DCT we see that the limit is $\lim G_n(1)\phi (1)-\int_0^{1} x \phi '(x)dx$. But $G_n(1)=\frac 1n\int_0^{n} h(t)dt=A$ for each $n$

Answer 2

Let $c= \lambda a + (1-\lambda ) b$ and note that on any interval $I \subset (0,1)$ we have $\int_I g_n = c \cdot mI$.

Let $s_k $ be a sequence of step functions that converge uniformly to $\phi$.

Then $|\int_0^1 (g_n-c)\phi| \le |\int_0^1 (g_n-c)s_k| + |\int_0^1 (g_n-c)(\phi-s_k)| $.

Choose $\epsilon>0$ and $k$ such that $|\phi(x)-s_k(x)| < \epsilon $ for all $x$, then $|\int_0^1 (g_n-c)\phi| \le |\int_0^1 (g_n-c)s_k| + \max(|a|,|b|) \epsilon$. Now choose $n$ large enough so that $|\int_0^1 (g_n-c)s_k| \le \max(|a|,|b|) \epsilon$, then $|\int_0^1 (g_n-c)\phi| \le 2 \max(|a|,|b|) \epsilon$.

Hence $\lim_n \int_0^1 g_n \phi = c \int_0^1 \phi$.

Note:

Since the step functions are dense in $L^1[0,1]$, the above argument can be adapted to show that it holds for $L^1[0,1]$ as well.

Answer 3

Another answer, which, I think, doesn't require $\phi$ to be $\mathcal C^1$, only $\mathcal C^0$.

Note that for all $k\in\{0,\dots,n-1\}$ and all $x\in\left[\frac kn,\frac{k+1}{n}\right[$, $$g_n(x)=\begin{cases} a & \text{if }\frac kn\le x< \frac{k+\lambda}{n} \\ b & \text{if }\frac{k+\lambda}{n}\le x < \frac{k+1}{n} \end{cases}$$ so $$I_n = \int_{0}^{1} g_{n}(x) \varphi(x) {\mathrm d} x = \sum_{k=0}^{n-1} \int_{\frac{k}{n}}^{\frac{k+1}{n}} g_{n}(x) \varphi(x) {\mathrm d} x = \sum_{k=0}^{n-1} \left(\int_{\frac{k}{n}}^{\frac{k+\lambda}{n}} a\varphi(x) {\mathrm d} x + \int_{\frac{k+\lambda}{n}}^{\frac{k+1}{n}} b\varphi(x) {\mathrm d} x \right)$$ For each $k\in\{0,\dots,n-1\}$, there exist $\mu_k$ and $\nu_k$ in $\left[\frac kn,\frac{k+1}{n}\right[$ such that $$\int_{\frac{k}{n}}^{\frac{k+\lambda}{n}} a\varphi(x) {\mathrm d}x = a\frac\lambda n\phi(\mu_k)\text{ and } \int_{\frac{k+\lambda}{n}}^{\frac{k+1}{n}} b\varphi(x) {\mathrm d} x = b\frac{1-\lambda}{n}\phi(\nu_k)$$ (mean theroem, assuming continuity for $\phi$), so that $$I_n = a\lambda \frac1n\sum_{k=0}^{n-1} \phi(\mu_k) + b(1-\lambda)\frac1n\sum_{k=0}^{n-1} \phi(\nu_k)$$ By Riemann theorem (still assuming just continuity for $\phi$), both sums converge to $I=\int_0^1 \phi$, so $$\lim_{n\to\infty} I_n = (\lambda a+(1-\lambda)b)\int_0^1\phi$$