Inequality Involving Integrals

Question

You are given that $M:I\rightarrow \mathbb{R}$ and $N:I\rightarrow \mathbb{R}$ are continuous, where $I=[x,y]$.

a) Suppose that $\forall u \in [x,y]$, $\int_{x}^{u} N \leq \int_{x}^{u}M$. Prove that $\forall u \in [x,y]$, $N(u)≤M(u)$ or give a contradictory example.

b)Now suppose that $\forall u ,v \in [x,y]$, $\int_{v}^{u} N \leq \int_{v}^{u}M$. Prove that $\forall u \in [x,y]$, $N(u)≤M(u)$ or give a contradictory example.

I was able to find a counter-example for the first part ($f(x) = 16 - x^{2}$ and $g(x) = 0$ for $I = [0,5]$) but how would we tackle the second part?

Answer 1

Let $f(t)=M(t)-N(t) \ \forall \in [x,y]$.

If $f(a)<0$ for some $a\in [x,y]$ then by continuity for $\epsilon=-f(a)/2>0$ we can find $\delta>0$ such that $|f(t)-f(a)| <\epsilon=-f(a)/2 $ whenever $|t-a|<\delta $ and $ t\in [x,y]$.

Equivalently, $f(t)\in (3/2f(a),1/2f(a)) $, hence $f(t)<0$ for some neighborhood $U$ of $a$.

Now, take $u,v \in U$ then $\int_u^v f(t) dt<0$.

Answer 2

(a): The counter example looks good
(b): Suppose that it is not true. Then there are some $x^*\in[x,y]$ such that $N(x^*)\gt M(x^*)$. Since $N$ and $M$ are continuous, there are some interval $[x_1,x_2]$ (why?) such that for all $x\in[x_1,x_2]$, $N(x^*)\gt M(x^*)$. Then let $u=x_1$ and $v=x_2$, it can be shown that $\int _u^v N(x)-M(x)\gt0$, which is a contradiction.