Show that if $f$ is continuously differentiable on $[0,1]$, then $$\lim\limits_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)-\int_{0}^{1}f(x)dx\right)=\frac{f(1)-f(0)}{2}$$
Observe that
\begin{align*}
\lim\limits_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)-\int_{0}^{1}f(x)dx\right)&=\lim\limits_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)-\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}f(x)dx\right)\\
&=\lim\limits_{n\rightarrow\infty}n\left(\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}\left[f\left(\frac{i}{n}\right)-f(x)\right]dx\right)\\
&=\lim\limits_{n\rightarrow\infty}n\left(\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}f'(c_i)\left[\left(\frac{i}{n}\right)-x\right]dx\right)
\end{align*}
where the last equality follows from the Mean Value Theorem.
Let $m_i=\inf\{f'(x):x\in[(i-1)/n,i/n]\}$ and $M_i=\sup\{f'(x):x\in[(i-1)/n,i/n]\}$, then we have the follow inequality: $$m_i\int_{(i-1)/n}^{i/n}\left[\left(\frac{i}{n}\right)-x\right]dx\leq\int_{(i-1)/n}^{i/n}f'(c_i)\left[\left(\frac{i}{n}\right)-x\right]dx\leq M_i\int_{(i-1)/n}^{i/n}\left[\left(\frac{i}{n}\right)-x\right]dx$$
Consequently $$\frac{1}{2n}\sum_{i=1}^{n}m_i\leq n\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}f'(c_i)\left[\left(\frac{i}{n}\right)-x\right]dx\leq\frac{1}{2n}\sum_{i=1}^{n} M_i$$ where $\int_{(i-1)/n}^{i/n}\left[\left(\frac{i}{n}\right)-x\right]dx=\frac{1}{2n^2}$.
I stuck at this step. And it seems not right because when I take the limit both sides, I have $0$. Can someone give me a hint or suggestion. Thanks in advanced.
Best Answer
Your bounds
$$ \frac{1}{2n} \sum_{i=1}^{n} m_i \quad \text{and} \quad \frac{1}{2n} \sum_{i=1}^{n} M_i $$
converge to the same quantity, namely $\frac{1}{2}(f(1) - f(0))$. This is essentially because they are Riemann sums for $\frac{1}{2}f'(x)$.
Proof using Taylor Theorem. Let $x_i = i/n$ for brevity and consider $F(x) = \int_{0}^{x} f(t) \, dt$. Then we may write
$$ n \left( \frac{1}{n} \sum_{i=1}^{n} f(x_i) - \int_{0}^{1} f(x) \, dx \right) = n \sum_{i=1}^{n} (F(x_{i-1}) - F(x_i) + \tfrac{1}{n}F'(x_i)). $$
By Taylor Theorem, we can pick $c_i \in [x_{i-1}, x_i]$ such that
$$ F(x_{i-1}) = F(x_i - \tfrac{1}{n}) = F(x_i) - \tfrac{1}{n}F'(x_i) + \tfrac{1}{2n^2}F''(c_i). $$
Plugging this back, we have
$$ n \left( \frac{1}{n} \sum_{i=1}^{n} f(x_i) - \int_{0}^{1} f(x) \, dx \right) = \frac{1}{2n} \sum_{i=1}^{n} f'(c_i). $$
Taking $n \to \infty$, this converges to $\frac{1}{2}(f(1) - f(0))$ as desired.