Formal proof of $\sup \limits_{0<r} f(r) = \sup \limits_{0<R} \sup \limits_{0<r<R} f(r)$

Question

This question by Dal is in fact a question about the behavior of the supremum. In particular, he asks that if we have a function $f:\mathbb{R} \rightarrow \mathbb{R}$, then do we necessarily have:

$$ \sup \limits_{0<r} f(r) = \sup \limits_{0<R} \sup \limits_{0<r<R} f(r) $$

I gave a rigorous proof of this fact (and restated it as an answer below), but it seems very wordy to me. Does anyone have a more concise proof of this fact?

Answer 1

Clearly $\sup_{r >0} f(r) \ge f(x)$ for all $x >0$ hence $\sup_{r >0} f(r) \ge \sup_{R>x>0} f(x)$ for all $R$ and hence $\sup_{r >0} f(r) \ge \sup_{R>0} \sup_{R>x>0} f(x)$.

Similarly, $f(r) \le \sup_{R>0} \sup_{R>x>0} f(x)$ for all $r >0$. Hence $\sup_{r>0} f(r) \le \sup_{R>0} \sup_{R>x>0} f(x)$.

Answer 2

Let $f:(0,\infty) \rightarrow \mathbb{R}$. We want to show that the following two quantities are equal:

\begin{align} A &= \sup \{ f(r) : 0 < r < \infty \} \\ B &= \sup \{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \} \end{align}

First remark that by definition, $A$ is an upper bound of $\{ f(r) : 0 < r < \infty \}$. Since this set includes $ \{ f(r) : 0 < r < R \}$ as a subset, $A$ is also an upper bound for this set for any $R$. But since the supremum of a set is less or equal than any upper bound, we have

$$ \sup \{ f(r) : 0 < r < R \} \leq A $$

for every $0 < R < \infty$. Hence $A$ is an upper bound on $\{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \}$.

To show that $A$ is the supremum of this set, we must show that it is the least upper bound. Suppose that $D$ is another upper bound on $\{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \}$. Let $0 < r < \infty$, and pick $R=r+1$. Then we know that we must have $D \geq \sup \{ f(s) : 0 < s < R \}$. We know $f(r)$ is in this set, since $0 < r < r+1$, and since the supremum is an upper bound, it is greater than $f(r)$. Therefore, $D \geq f(r)$, and this holds for every $0 < r < \infty$.

Thus we conclude that $D$ is an upper bound on $\{ f(r) : 0 < r < \infty \}$. But since $A$ is the supremum of that set, we have that $A \leq D$. Hence we have shown that $A$ is less or equal to any upper bound of $\{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \}$, and since we have also shown that $A$ itself is an upper bound of this set, we get, by definition:

$$ A = \sup \{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \} = B $$

And this is what we wanted to show.

Answer 3

Here is a slightly less wordy proof. Let $$ E=\{f(r):r>0\}\\ F=\{\sup_{0<s<R} f(s):R>0\} $$ Notice that for every element $y$ of $E$, there is a corresponding element $y'$ of $F$ for which $y\le y'$. Namely, given $y=f(r)\in E$, then you can take $y'= \sup_{0<s<r+1} f(s)\in F$. This shows any upper bound for $F$ is also an upper bound for $E$. In particular, $\sup F$ is an upper bound for $E$, so $\sup E\le \sup F$.

On the other hand, every element $y'=\sup_{0<s<R}f(s)$ of $F$ satisfies $$ y'=\sup_{0<s<R}f(s)\le^* \sup_{0<s} f(s)=\sup E $$ so $\sup E$ is an upper bound for $F$, proving $\sup E\ge \sup F$.

---

The inequality $\le^*$ follows from the easy lemma that when $G\subseteq H$, we have $\sup G\le \sup H$.