Let $f:[a,b] \to \mathbb{R}$ be an integrable function and $(a_n) \subset \mathbb{R}$ such that $a_n \to 0.$ $\textbf{a) }$ If $A= \{m \cdot a_n \mid m,n \in \mathbb{N}^* \},$ prove that every open interval of strictly positive real numbers contains elements from $A.$ $\textbf{b) }$ If, for any $n \in \mathbb{N}^*$ and for any $x,y \in [a,b]$ with $|x-y|=a_n,$ the inequality $\left| \int_x^yf(t)dt \right| \leq |x-y|$ is true, prove that $$\left| \int_x^y f(t)dt \right| \leq |x-y|, \: \forall x,y \in [a,b]$$ Nicolae Bourbacut
Problem
Source: Romania NMO - 2018
Tags: function, inequalities, calculus, integrals, real analysis
28.03.2019 13:19
I assume the sequence $(a_n)$ consists of infinitely positive numbers (else the conclusion might be false). Here is the solution for part (a):
For part $\textbf(b)$, we can use the triangle inequality and part $\textbf(a)$ multiple times.
22.01.2022 22:11
The sequence must consists of infinitely many positive numbers. (a)Let $(x,y)$ be an open, with $0< x<y$.Since the sequence converges to 0 with infinitely many positive numbers, hence there is $n$ such that, $$2\le \frac{y-x}{a_n}$$Hence, there is some integer $m$ between $\frac{x}{a_n}< m<\frac{y}{a_n}$.We are done.$\blacksquare$ (b)Let $x,y\in [a,b]$ with $x<y$.Pick any $\varepsilon$ such that $y-x>\varepsilon>0$.By (a) we can find integer $i$ and sime $a_n$ such that $$y-x-\varepsilon <ia_n\le y-x$$Hence $x+ia_n\in(y-\varepsilon,y)$.Since $f$ is integrable there is $M\ge |f(x)|,\forall x\in [a,b]$.Now observe that, $$|\int_x^y f(t) dt|= |\int_x^{x+ia_n} f(t) dt+\int_{i+a_n}^y f(t) dt|\le |\int_x^{x+ia_n} f(t) dt|+|\int_{i+a_n}^y f(t) dt|$$Now partitioning the interval $[x,x+ia_n]$ into $i$ intervals of length $a_n$ we get $$|\int_x^{x+ia_n} f(t) dt|\le ia_n\le y-x$$Again, $|\int_{i+a_n}^y f(t) dt|\le M(y-ia_n)\le M\varepsilon$ Since $\varepsilon $ is arbitrary,we are done.$\blacksquare$