4

I want to show that the square of the supremum of any set of non-negative numbers is the supremum of the squares. I.e. of $A\subseteq \mathbb{R}^+$ then \begin{equation*} \sup_{a\in A} a^2 = \left( \sup_{a\in A}a\right)^2 \end{equation*} This is obvious if the supremum is actually attained (so it is the maximum). However, I am struggling to prove it rigorously for the case that the maximum is actually not attained. Could show this carefully?

Thanks

User086688
  • 653
  • https://math.stackexchange.com/questions/81721/is-the-supremum-of-a-function-squared-the-square-of-its-supremum?rq=1 – user29418 Sep 21 '17 at 05:15

2 Answers2

9

In case you want to show this rigorously, you will need to show that $\sup \{a^2\} \leq (\sup \{a\})^2$ and vice-versa.

To show one direction, note that $\sup \{a\} \geq a$ for all $a$, hence $(\sup \{a\})^2 \geq a^2$ for all $a$, so $(\sup \{a\})^2$ is an upper bound for $\{a^2\}$, hence it is greater than the least upper bound, i.e. $(\sup\{a\})^2 \geq \sup\{a^2\}$.

To show the other direction, let $N$ be large enough such that $\sup \{a\} - \frac 1N > 0$ (if $\sup \{a\} = 0$ and all $a$ are non-negative, then it is clear what the set is and the result itself). Now for any $n>N$, the quantity $\sup \{a\} - \frac 1n$ is not an upper bound of $\{a\}$, Hence, there is some $a_n$ such that $a_n > \sup\{a\} - \frac 1n$. Square both sides(note that by non-negativity of both sides, this preserves sign) to see that $a_n^2 > \frac 1{n^2} + (\sup \{a\})^2 - \frac{2\sup \{a\}}{n}$. Now, by definition of supremum, we have $$\sup \{a^2\} \geq a_n^2 > \frac 1{n^2} + (\sup \{a\})^2 - \frac{2\sup \{a\}}{n}$$

This applies for all $n>N$. Since $(\sup \{a\})^2$ is a bounded quantity, letting $n \to \infty$, we see that $\sup \{a^2\} \geq (\sup \{a\})^2$. Hence, equality follows.

Note the trick played in the second half of the proof. You will need to use it repeatedly and become comfortable with it.

Sarvesh Ravichandran Iyer
  • 76,105
  • Ah thank you very much! That was exactly what I was looking for! – User086688 Sep 21 '17 at 06:00
  • You are welcome! – Sarvesh Ravichandran Iyer Sep 21 '17 at 06:06
  • @астонвіллаолофмэллбэрг I do not quite understand the trick you use in the second part of the proof. It seems that letting $n\rightarrow \infty$ makes the claim holds only under condition that $n$ is very large. – Mr.Robot Feb 14 '19 at 22:59
  • 1
    @Mr.Robot The claim is $\sup{a^2} \geq \sup({a})^2$, which does not depend on $n$. The trick in the second part of the proof is roughly this : for any $n$, the number $\sup{a^2} - \frac 1n$ is not an upper bound of ${a^2}$, so we find an $a_n \in A$ such that the above is less than $a_n^2$. After squaring and comparing with $\sup {a^2}$ we get some inequality which holds for all $n$. Therefore, I may choose the $n$ which maximizes the right hand side, so that I get the best possible inequality. But as I increase $n$, the RHS goes to $(\sup {a})^2$, the best bound. – Sarvesh Ravichandran Iyer Feb 15 '19 at 05:23
  • 1
    @TeresaLisbon , the proof is flawed. When you square the inequality $a_n>\sup{a}-\frac{1}{n}$ it is not guaranteed that it still holds. Hence, the resulting inequality doesn't hold for all $n$. You must assume that $n$ is sufficiently large, e.g. $0\leq\sup{a}-\frac{1}{n}$. – Philipp Aug 26 '20 at 15:53
  • @Philipp Kindly see the edits, I would like you to be satisfied before I leave this answer, I don't want to mislead others. – Sarvesh Ravichandran Iyer Aug 26 '20 at 16:07
  • @TeresaLisbon the index $n$ of $a$ in the expression $\sup a_n -\frac{1}{n}>0$ should be omitted. It should be $\sup a -\frac{1}{n}>0$ – Philipp Aug 26 '20 at 16:26
  • @Philipp I've checked things after your recommendation, and it seems ok (once you agree). Thank you for the corrections. – Sarvesh Ravichandran Iyer Aug 26 '20 at 16:29
  • @TeresaLisbon, looks good now :) – Philipp Aug 26 '20 at 16:33
  • @Philipp No point of up voting your comment, so I've up voted your two questions I liked best. Once again, thank you for the help. – Sarvesh Ravichandran Iyer Aug 26 '20 at 16:35
0

A swifter approach would be if you use the fact that an upper bound $S$ of $A:=\{a_1,a_2, \cdots\}$ is a supremum iff:

"For all $\epsilon'>0$ there exists an element $a\in A$ such that: $ S-\epsilon' < a$".

We now set $\epsilon':= \frac{\epsilon}{2S}$. Assuming that $S$ is a supremum of $A$ it is clear that $S^2$ is an upper bound of $A^2:=\{a_1^2,a_2^2, \cdots\}$ so it remains to show that:

"For all $\epsilon>0$ there exists an element $a^2\in A^2$ such that: $ S^2-\epsilon < a^2$".

If $\epsilon> S^2$ the statement is obviously true. If $0<\epsilon<S^2$, then there exists an $a\in A$ and an $a^2 \in A^2$ such that: $$ S-\epsilon' < a\\ \Rightarrow S^2-2S\epsilon'+\epsilon'^2= (S-\epsilon')^2 < a^2 \\ \Rightarrow S^2-\epsilon=S^2-2S\frac{\epsilon}{2S}< a^2 .$$ (Note that: $S^2-\epsilon>0\implies S-\frac{\epsilon}{2S}>0$, so squaring the first inequality doesn't flip the inequality sign.)

Philipp
  • 4,483
  • 2
  • 10
  • 22