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 $a\geq b$. Then $\inf(A)\geq \sup(B)$.

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 $a\geq b$. Since $b$ is arbitrary, we have that $a$ is an upper bound of the set $B$, so taking the superemum over $b$ we have $a\geq \sup(B)$ (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 $\inf(A)\geq \sup(B)$, as required.

Applications

liminf vs limsup

(Notation from Tao's Analysis I.)

Let $(a_{n})_{n=m}^{\infty }$ be a sequence of real numbers. Let $L^{-}:=\liminf _{n\to \infty }a_{n}$ and let $L^{+}:=\limsup _{n\to \infty }a_{n}$. Then we have $L^{-}\leq L^{+}$.

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

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

Lower and upper Riemann integral

(Notation from Tao's Analysis I.)

We have

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

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

We want to show ${\underline {\int }}_{I}f\leq {\overline {\int }}_{I}f$.

Define

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

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

Then we have ${\overline {\int }}_{I}f=\inf(A)$ and ${\underline {\int }}_{I}f=\sup(B)$. 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 $p.c.\int _{I}g\geq p.c.\int _{I}h$.