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
Seicchi28
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):
We can argue by contradiction. Assume there is an open interval $(\alpha, \beta)$ which doesn't contain elements of $A$. Since rational number is dense in $\mathbb{R}^+$, we can find two distinct rational numbers $\frac{p}{q}$ and $\frac{r}{s}$ for which $\alpha < \frac{p}{q} < \frac{r}{s} < \beta$. Let $k,l,m$ be positive integers such that $\frac{p}{q}=\frac{l}{m}$ and $\frac{r}{s}=\frac{k}{m}$. (Thus $l<k$.)
Now consider the intervals $[\frac{1}{2m}, \frac{1}{m}] , [\frac{1}{3m}, \frac{1}{2m}], \dots, [\frac{1}{(n+1)m}, \frac{1}{nm}], \dots$. Since $a_n \rightarrow 0$, we can found a big enough $N$ (Specifically, $N > \frac{l}{k-l} \implies Nk > (N+1)l$ ), such that there exists a positive term of this sequence, call it $a_M$, lying in $[\frac{1}{(N+1)m}, \frac{1}{Nm}]$. Therefore:
\[ \alpha < \frac{l}{m} < (N+1)la_M < Nka_M < \frac{k}{m} < \beta \]
so, $(N+1)la_M$ and $Nka_M$ are inside $(\alpha, \beta)$, a contradiction.
For part $\textbf(b)$, we can use the triangle inequality and part $\textbf(a)$ multiple times.
Mathematicsislovely
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.$\blacksquare$