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How would you formally justify this? Or is it just notationally obvious? (As opposed to 'conceptually' obvious, which is never an excuse in mathematics.)

Edit: For some $c \geq 0$

$\text{sup} \ \{ c \cdot f(x): \text{some domain of $x$} \}$ = $c \cdot \text{sup} \ \{f(x): \text{same domain of $x$} \}$

To illustrate what I mean, consider $\text{sup}\ \{ 1, 5\} = 5$ which however conceptually obvious can still be proven.

Suppose otherwise, if $\text{sup}\ \{ 1, 5\} < 5$, then the supremum is less than an element of set. A contradiction. If $\text{sup}\ \{1, 5\} > 5$, then there exists the number 5 less than the supremum which nonetheless is an upper bound of the set. A contradiction.

But I'm not sure how I would 'prove' my original assertion.

user_hello1
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2 Answers2

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Note that you need the condition $c>0$ in order to prove your point.

The proof needs to address two things.

First you show that upon multiplying by $c$ the upper bound property is preserved

Then you have to show that you have the least upper bound.

Both statements are done by definitions and the fact that multiplying by the positive number $c$ preserve the orientation of inequalities.

Mohammad Riazi-Kermani
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If $g:\mathbb{R}\to \mathbb{R}$ is increasing and continuous, then $g(\sup A)=\sup \{g(x): x\in A\}$. Here $g(\infty)=\lim_{x\to \infty}g(x)$ and $g(-\infty)= \lim_{x\to -\infty} g(x)$.

Fix $x\in A$ and note that, since $x\leqslant \sup A$, $g(x)\leqslant g(\sup A)$. Therefore $g(\sup A)$ is an upper bound for $\{g(x): x\in A\}.$ In order to finish, we need to show that $g(\sup A)$ is the least upper bound, which means we need to show that values in $\{g(x): x\in A\}$ can be arbitrarily close to $g(\sup A)$. By definition of $\sup A$, we can find sequence $(x_n)_{n=1}^\infty$ in $A$ with $\lim_n x_n=\sup A$. By continuity, $\lim_n g(x_n)=g(\sup A)$, so among the points $g(x_n)$, we can indeed get arbitrarily close to $g(\sup A)$.

Now apply this general fact to the specific case in which $g(x)=cx$ for some $c\geqslant 0$ and $A=\{f(x): x\in D\}$, where $D$ is your domain of $x$.

  • I do not see the name $f$ in the question at all, nor did I say it was there. I see the function $f(x)=cx$ in the question, without the name $f$. But I do not understand the point of your question. –  May 21 '19 at 16:28
  • There is an $f$ in “$\text{sup} \ { c \cdot f(x): \text{some domain of $x$} }$ = $c \cdot \text{sup} \ {f(x): \text{same domain of $x$} }$”, and I erroneously thought your $f$ is the same function. – Martin R May 21 '19 at 16:31
  • Ahh, now I see the point. I introduced a new function named $f$ without realizing there was already a function named $f$ in the problem. I will edit my answer accordingly. Thank you for pointing it out. –  May 21 '19 at 16:32