User:IssaRice/Taking inf and sup separately

This page describes a trick that is sometimes helpful in analysis.

Satement

Let $A$ and $B$ be bounded subsets of the real line. Suppose that for every $a\in A$ and $b\in B$ we have $a\ge b$. Then $inf(A)\ge sup(B)$.

Actually, do $A$ and $B$ have to be bounded? I think they can even be empty!

Proof

Let $a\in A$ and $b\in B$ be arbitrary. We have by hypothesis $a\ge b$. Since $b$ is arbitrary, we have that $a$ is an upper bound of the set $B$, so taking the superemum over $b$ we have $a\ge sup(B)$ (remember, $sup(B)$ is the least upper bound, whereas $a$ is just another upper bound). Since $a$ was arbitrary, we see that $sup(B)$ is a lower bound of the set $A$. Taking the infimum over $a$, we have $inf(A)\ge sup(B)$, as required.

Applications

liminf vs limsup

(Notation from Tao's Analysis I.)

Let $(a_n{)}_{n=m}^{\mathrm{\infty }}$ be a sequence of real numbers. Let $L^{-}:={lim inf}_{n\to \mathrm{\infty }}{a}_n$ and let $L^{+}:={lim sup}_{n\to \mathrm{\infty }}{a}_n$. Then we have $L^{-}\le L^{+}$.

Consider the sequences $(a_N^{-}{)}_{N=m}^{\mathrm{\infty }}$ and $(a_N^{+}{)}_{N=m}^{\mathrm{\infty }}$ defined by $a_N^{-}:=inf(a_n{)}_{n=N}^{\mathrm{\infty }}$ and $a_N^{+}:=sup(a_n{)}_{n=N}^{\mathrm{\infty }}$.

Now consider the sets $A:=\{a_N^{+}:N\ge m\}$ and $B:=\{a_N^{-}:N\ge m\}$. If we can show that $a_j^{+}\ge a_k^{-}$ for arbitrary $j,k\ge m$, then we can apply the trick to these sets to conclude that $L^{+}=inf(a_N^{+}{)}_{N=m}=inf(A)\ge sup(B)=sup(a_N^{-}{)}_{N=m}=L^{-}$.

Lower and upper Riemann integral

(Notation from Tao's Analysis I.)

We have

${\overline{\int }}_{I}f:=inf\{p.c.{\int }_{I}g:g\text{ is a p.c. function on }I\text{ that majorizes }f\}$

${\underset{\_}{\int }}_{I}f:=sup\{p.c.{\int }_{I}g:g\text{ is a p.c. function on }I\text{ that minorizes }f\}$

We want to show ${\underset{\_}{\int }}_{I}f\le {\overline{\int }}_{I}f$.

Define

$A:=\{p.c.{\int }_{I}g:g\text{ is a p.c. function on }I\text{ that majorizes }f\}$

$B:=\{p.c.{\int }_{I}g:g\text{ is a p.c. function on }I\text{ that minorizes }f\}$

Then we have ${\overline{\int }}_{I}f=inf(A)$ and ${\underset{\_}{\int }}_{I}f=sup(B)$. To apply the trick all we need to do is to let $g$ be a p.c. function on $I$ that majorizes $f$, and let $h$ be a p.c. function on $I$ that minorizes $f$, and show that $p.c.{\int }_{I}g\ge p.c.{\int }_{I}h$.