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Let $\Omega = \mathopen]0,1\mathclose[$ and let a function $A_n: \Omega \to \mathbb R$ defined as: $$A_n(x) = \begin{cases}\alpha &\text{if } k \epsilon \leq x < (k+\tfrac{1}{2}) \epsilon \\ \beta &\text{if } \big(k+\tfrac{1}{2}\big) \epsilon \leq x < (k+1) \epsilon \end{cases} $$ for $k=0$, $1,\ldots,n-1$ where $\alpha$, $\beta > 0$ and $\epsilon=1/n$. Let $f: [0,1] \rightarrow \mathbb{R}$ be a continuous function. How can we prove that $$\lim_{n \to \infty} \int_0^1 A_n(x) f(x) dx = \lim_{n \to\infty} \sum_{k=0}^{n-1} \left(\alpha \int_{k \epsilon}^{(k+1/2)\epsilon} f(x) dx + \beta \int_{(k+1/2)\epsilon}^{(k+1)\epsilon} f(x) dx\right) \\= \dfrac{\alpha + \beta}{2} \int_0^1 f(x) dx$$ Thanks for your help.

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Hint: Use that $f$ is in fact uniform continuous on the closed interval $[0,1]$. –  gerw Apr 5 '13 at 12:19
    
Yes, $f$ is uniform continuous on $[0,1],$ so for all $\eta > 0,$ there exist $\epsilon > 0: |f(x) - f(k \epsilon)| < \eta$ for all $x \in [k \epsilon,(k+1)\epsilon]$ but how we can use thise to compute the limite? –  jijiii Apr 5 '13 at 12:23
    
It might be worth pointing out that this question is about clarifying my answer to this question. I could not find the time to go into the necessary detail, unfortunately. –  Harald Hanche-Olsen Apr 5 '13 at 12:24

1 Answer 1

Hints:

  1. All you need to prove is that $$ \lim_{n\to\infty} \sum_{k=0}^{n-1}\int_{k/n}^{(k+1/2)/n} f(x)\,dx = \frac{1}{2}\int_0^1 f(x)\,dx $$

  2. Remember that (Riemann's sums theorem) $$ \lim_{n\to\infty} \sum_{k=0}^{n-1} \frac{1}{n}f(k/n)\,dx = \int_0^1 f(x)\,dx $$

  3. By using the uniform continuity of $f$, you can justify an estimate of the type $$ \int_{k/n}^{(k+1/2)/n} f(x)\,dx \simeq \frac{f(k/n)}{2n} $$ for large values of $n$, uniformly in $k$.


Extension to the case $f\in L^1[0,1]$

Use the density of continuous functions in $L^1[0,1]$ and the fact that for all $n$ we have \begin{equation} \left|\int_0^1A_n(x)f(x)\,dx - \int_0^1A_n(x)g(x)\,dx\right| \leq \max\{|\alpha|,|\beta|\} \|f-g\|_{L^1} \end{equation}

hence $$ \left|\langle A_n,f\rangle - \frac{\alpha+\beta}{2}\int_0^1 f\right| \leq C_{\alpha,\beta} \|f-g\|_{L^1} + \left|\langle A_n,g\rangle - \frac{\alpha+\beta}{2}\int_0^1 g\right| $$ where $C_{\alpha,\beta}=\max\{|\alpha|,|\beta|\}+\frac{\alpha+\beta}{2}$.

Thus, for every $g$ continuous we have (according to what you did in the first part) $$ \limsup_{n\to\infty} \left|\langle A_n,f\rangle - \frac{\alpha+\beta}{2}\int_0^1 f\right| \leq C_{\alpha,\beta} \|f-g\|_{L^1} $$ and $\|f-g\|_{L^1}$ can be made as small as necessary, so finally $$ \limsup_{n\to\infty} \left|\langle A_n,f\rangle - \frac{\alpha+\beta}{2}\int_0^1 f\right| = 0 $$

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Okay! I think that i'm understand. $f$ is continuous on a compact so she is uniformly continuous so for all $\eta > 0$ there exist $\epsilon > 0$ : $|f(x) - f(y)| < \epsilon$ –  jijiii Apr 5 '13 at 13:35
    
for $y = k \epsilon,$ we have $|f(x) - f(k \epsilon)| < \eta$ i.e., $f(x) \simeq f(k \epsilon)$ so $\int_{k \epsilon}^{(k+1/2)\epsilon} f(x) dx \simeq \dfrac{f(k/n)}{2 n}.$ and for the Riemann's sums theorem, $\lim_{n \rightarrow + \infty} \sum_{k=0}^{n-1} \int f(x) dx = \dfrac{1}{2} \int_0^1 f(x) dx.$ And we could de conclude. –  jijiii Apr 5 '13 at 13:35
    
My last question is: how we can prouve this relation for all $f \in L^1(0,1)$? –  jijiii Apr 5 '13 at 13:36
    
Let $g \in L^1,$ then, there exist a sequence $(f_m)_m $ of continuous functions such as $f_m \rightarrow g$ dans $L^1.$ So $\lim_{m \rightarrow + \infty} A_n(x) f_m(x) dx = \int_0^1 A_n(x) g(x) dx.$ Then, $$\lim_{n \rightarrow + \infty} \lim_{m \rightarrow + \infty} \int_0^1 A_n(x) f_m(x)dx = \dfrac{\alpha + \beta}{2} \lim_{m \rightarrow + \infty} \int_0^1 f_m(x) dx = \dfrac{\alpha + \beta}{2} \int_0^1 g(x) dx = \lim_n \int_0^1 A_n(x) f_n(x) dx.$$ correct? –  jijiii Apr 5 '13 at 16:29
    
It's true, we must justify the inversion of limits.What's the uniformity with respect to $n$ please? –  jijiii Apr 5 '13 at 17:18

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