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Let $f : \small \mathbb{R} → \mathbb{R}$ be a continuous nondecreasing function. Let $A$ be a nonempty, bounded set.

(a) Show that $f(\sup A) = \sup(f(A)).$

(b) If we drop the assumption that $f$ is continuous, what can you say?

Attempt:

$x\le \sup(A)\ \ \forall x \in A$

$\Rightarrow f(x)\le f(\sup(A))\ \ \forall x \in A$

$\Rightarrow \sup(f(A))\le f(\sup(A))\ \ $

also, $f(x)\le \sup(f(A))\ \ \forall x \in A$

$\Rightarrow x\le f^{-1}(\sup(f(A)))\ \ \forall x \in A$

$\Rightarrow \sup(A)\le f^{-1}(\sup(f(A)))\ \ $

$\Rightarrow f(\sup(A))\le \sup(f(A))\ \ $

Thus $f(\sup(A))= \sup(f(A))$

But in the Highlighted line I have used that $f$ is injective, but $f$ may not be injective as $f$ is given to be non decreasing. So how to correct that part.

Community
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User
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1 Answers1

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You are right about your argument: you can't use $f^{-1}$, so you need something else. Let $y = \sup(A)$. Using the definition of the supremum, for each $n \ge 1$ we can pick $x_n \in A$ such that $y-1/n < x_n \le y$. The squeeze lemma tells us that $x_n \to y$ as $n \to \infty$. Since $f$ is continuous we know that $f(x_n) \to f(y)$ as $n\to \infty$. Let $\epsilon >0$ and pick $N$ such that $n \ge N$ implies that $\vert f(x_n) - f(y) \vert < \epsilon$. Then $n \ge N$ implies that $$ f(y) = f(y) - f(x_n) + f(x_n) \le \vert f(y) - f(x_n) \vert + f(x_n) < \epsilon + f(x_n) \\ \le \epsilon + \sup f(A). $$ Since $\epsilon >0$ was arbitrary, we deduce that $$ f(\sup A) = f(y) \le \sup f(A). $$

Glitch
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  • Thank you so much for your help sir. – User Dec 04 '16 at 22:03
  • Here you didn't use the fact that $f$ is non decreasing, did you? – Motaka Dec 31 '21 at 15:59
  • Not in this part, no, but this part also doesn't establish equality, just one side of the matching inequalities. The other side was established in the post between "attempt" and "also," and it is there that $f$ being non-decreasing is used. – Glitch Dec 31 '21 at 16:31