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Consider the $S_{n} = \int_{a}^{b}f(x)dx - \frac{b-a}{n}\sum f(a+k \frac{b-a}{n})$. Prove that $nS_{n} \to \frac{(b-a)(f(b)-f(a))}{2}$. Of course $f \in C[a,b]$

My attempt: we could get some partition in integral and say that there is a point $\xi \in (x_{0} , x_{0}+x) : \int_{x_{0}}^{x_{0}+x}f(t)dt = f(\xi) \triangle_{[x_{0},x+x_{0}]}$ for any segment. So will have in limit: $ \displaystyle\lim_{n \to \infty} nS_{n} = (b-a) \lim_{n \to \infty} \sum_{k = 0}^{n} (f(\xi_{i}) - f(a+k \frac{b-a}{n}))$

But how can we estimate difference of functions in sum?

openspace
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1 Answers1

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If $f$ is continuously differentiable, apply the mean value theorem to obtain

$$\begin{align}nS_n &= n\sum_{k=1}^n\int_{x_{k-1}}^{x_k}\left[f(x) - f(x_k)\right] \, dx \\ &= n\sum_{k=1}^n\int_{x_{k-1}}^{x_k}f'(\xi_{k,n})(x - x_k) \, dx \end{align}$$

where $x_k = a + (b-a)k/n$, and $\xi_{k,n} \in (x, x_k) \subset [x_{k-1}, x_k].$

Defining

$$M_{k} = \sup \{f'(x): x_{k-1} \leqslant x \leqslant x_k \}, \\ m_{k} = \inf \{f'(x): x_{k-1} \leqslant x \leqslant x_k \}. $$

we get the bounds

$$n \sum_{k=1}^nm_{k}\int_{x_{k-1}}^{x_k}(x - x_k) \, dx \leqslant nS_n \leqslant n \sum_{i=1}^nM_{k}\int_{x_{k-1}}^{x_k}(x - x_k) \, dx. $$

Note that

$$\int_{x_{k-1}}^{x_k}(x - x_k) \, dx = -\frac{1}{2}\left(x_{k-1} - x_k \right)^2 = - \frac{(b-a)^2}{2n^2},$$

and

$$-\frac{b-a}{2}\underbrace{\frac{b-a}{n}\sum_{k=1}^nm_{k}}_{\text{ lower sum}} \leqslant nS_n \leqslant -\frac{b-a}{2}\underbrace{\frac{b-a}{n}\sum_{k=1}^nM_{k}}_{\text{upper sum}}. $$

Now take the limit as $n \to \infty$. The upper and lower Riemann (or Darboux) sums indicated above converge to the integral of $f'$ and by the squeeze theorem

$$\lim_{n \to \infty} nS_n = - \frac{b-a}{2} \int_a^bf'(x) \, dx = \frac{(b-a)(f(a)-f(b))}{2}$$

RRL
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  • You mean that $f$ should be differentiable? – openspace Mar 22 '17 at 06:36
  • At least differentiable with a Riemann integrable derivative for this proof to work. – RRL Mar 22 '17 at 07:23
  • last sum is equivalent to $\frac{-(b-a)^{2}}{2n} \sum f'(\xi_{k})$ ? – openspace Mar 22 '17 at 10:16
  • @openspace: That is correct. Now that sum can be squeezed between lower and upper sums for $\int_0^1f'(x) , dx$ giving the desired result. This of course assumes that $f'$ is bounded and integrable. For more general continuous $f$, this is a subtle problem Notice that it works for $f(x) = \sqrt{x}$ even though the derivative in this case is unbounded on $[0,1]$. There is apparently a class of continuous functions (perhaps nowhere differentiable) where the result does not hold. – RRL Mar 22 '17 at 17:25
  • But why lower and upper sums tends to $\frac{(b-a)(f(a)-f(b))}{2}$ – openspace Mar 22 '17 at 21:05
  • @openspace: I added to my answer to clarify this for you. – RRL Mar 22 '17 at 21:30