Let $f: I \to \mathbb{R}$ be a function, where $I \subset \mathbb{R}^+.$ Let $u>0,E=\{x \in I,f(x)>u\}$ . Define $y \in \overline{\mathbb{R}}^+:y=\inf E$ if $E \neq \emptyset$ and $y=+\infty$ otherwise.
i. If $I=\mathbb{N},$ can you declare that: $\forall x \in I,f(\min(x,y)) \leq u$ ?
ii. Repeat question i. when $I=\mathbb{R}^+.$
Attempt: i. Let $v= \min(x,y) \leq y,$ this implies from the definiton of the infimum $\forall k \in I,$ such that $f(k) >u,f(k) \leq y,$ but it seems this is not very helpful.
Any ideas are aprreciated!
No, this is not true in either case without additional assumptions on $f$. For example, if we take $f(x) = 2$ for all $x \in I$ and $u = 1$ then and $f(\min(x,y)) = 2 > u$ for all $x \in \mathbb{N}$.
In the first case, where $I = \mathbb{N}$, the only way this could be true is if $f(x) \le u$ for all $x \in I$. Otherwise we will have $f(y) > u$, so $f(\min(x,y)) > u$ when $x \ge y$.
In the second case, where $I = [0,\infty)$, this is true if we assume that $f$ is increasing, continuous, and $u \ge f(0)$.
Proof: Because $f$ is increasing, $f(\min(x,y)) \le f(y)$ for all $x \in I$, so we just need to show $f(y) \le u$. If $y = 0$, this is true because we assume $f(0) \le u$, so suppose without loss of generality $y > 0$. Now assume working towards a contradiction $f(y) > u$. Then there exists $\varepsilon > 0$ such that $f(y) > u + \varepsilon$, but since $f$ is continuous this implies there exists $x < y$ with $f(x) > u + \frac{\varepsilon}{2} > u,$ which contradicts $y = \inf E$. Hence we conclude $f(y) \le u$ and therefore $f(\min(x,y)) \le f(y) \le u$ as well.