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SEEMOUS 2011 Problem 4

Let f :[0,1]\to \Bbb{R} be a twice continuously differentiable increasing function. Define the sequences given by
L_n=\displaystyle \frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k}{n}\right) and
U_n=\displaystyle \frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right) for n \geq 1.
The interval [L_n,U_n] is divided into three equal segments. Prove that for large enough n, the number I=\displaystyle \int_0^1 f(x)dx belongs to the middle one of these segments.
SEEMOUS 2011 Problem 4

Solution: Define F(x)=\displaystyle \int_0^x f(t)dt. Then we can write I=F(1)-F(0)=F(1)-F(\frac{n-1}{n})+F(\frac{n-1}{n})-F(\frac{n-2}{n})+...+F(1/n)-F(0). We apply Taylor’s expansion formula of order 2 for F in points 0,1/n,...,\frac{n-1}{n}.

We have F(x)=F(k/n)+(x-k/n)f(k/n)+\frac{(x-k/n)^2}{2}f^\prime(\theta),\ \theta \in [k/n,x],\ k=0...n-1. For x=1/n,...,1 we get F(\frac{k+1}{n})-F(\frac{k}{n})=\frac{1}{n}f(\frac{k}{n})+\frac{1}{2n^2} f^\prime(\theta_k),\ \theta_k \in [\frac{k}{n},\frac{k+1}{n}]. Summing this for k=0,1,...,n-1 we get I=L_n+\frac{1}{2n}\sigma_n, where \sigma_n is a Riemann sum for \int_0^1 f^\prime(t)dt=f(1)-f(0).

The middle intervals are [\frac{2}{3}L_n+\frac{1}{3}U_n,\frac{1}{3}L_n+\frac{2}{3}U_n]=[u_n,v_n]. If f(1)=f(0) the function is constant, and we know that f is increasing and therefore, non-constant (increasing means strictly increasing). Therefore f(1)>f(0).

Now, we calculate n(I-u_n)= \frac{1}{3}(L_n-U_n)+\frac{1}{2}\sigma_n=\frac{1}{3}(f(0)-f(1))+\frac{1}{2}\sigma_n, and taking the limit for n \to \infty we get that n(I-u_n)\to \frac{1}{6}(f(1)-f(0))>0, which means that for n great enough we have I>u_n. On the other hand, n(v_n-I)=\frac{2}{3}(U_n-L_n)-\frac{1}{2}\sigma_n\to\frac{1}{6}(f(1)-f(0))>0.

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