I am stuck on this problem for a while now, any help would be appreciated. I am working on the proof of a Network Calculus theorem, and I would like to show that the infimum is reached in the horizontal deviation of two functions. I simplified the problem, and left only what is essential I think :
Let $f$ : $\mathbb{R}^+ \rightarrow \mathbb{R}^+ \:;\: a\in\mathbb{R}^+ \:;\: t\in\mathbb{R}^+ \:;\: d\in\mathbb{R}^+$. Here are a few assumptions that I have :
- $f$ is non-decreasing: $x\leq y \rightarrow f(x) \leq f(y)$.
- $f(0)=0$.
- $\inf \{\tau: \tau\ge0 \textrm{ and } a \leq f(t+\tau)\} \leq d$.
- $\exists \theta\geq 0,\ a \leq f(t+\theta)$.
- Also, I am pretty sure that we need : $f$ is continuous.
Let $A = \{\tau: \tau\ge0 \textrm{ and } a \leq f(t+\tau)\}$
I want to prove that: $\inf A \in A$.
Thank you
I did some work and came up with the following. But there are still some gaps left.... $0 \le \inf A \textrm{ and } a = f(t+ \inf A) \longrightarrow \inf A \in A$
Proving $0 \leq \inf A$ is trivial, since all elements of A are positive.
Next, let us show that $a = f(t+ \inf A)$ by contradiction:
Let us assume that $a \neq f(t+ \inf A) \, ,$
hence: $0 < |f(t + \inf A) - a| \, ,$
with the density of reals we can obtain an $ε$ where $0 < ε \text{ and } ε < |f(t + \inf A) - a| \, . \qquad (1)$
Moreover, $a \in \mathbb{R}^+ \text{ and } f(0) = 0 \longrightarrow f(0) \leq a$
With this and the assumption about the existence of $\theta$, we can use the theorem of intermediate values to obtain $c$ where $0 \leq c \text{ and }c \leq t + \theta \text{ and } f(c) = a \, .$
With (1), we have: $ε < |f(t + \inf A) - f(c)| \, . \qquad (2)$
And from the continuity of f, we can obtain an $η$ where : $∀x. |(t + \inf A) - x| < η ⟶ |f (t + \inf A) - f(x)| < ε$
Now, I am not sure if euclidean division can be used with real, so I used this construction which is an equivalent of the reminder of an euclidean division: $x_η = c - \lfloor {c - (t+\inf A) \over η} \rfloor \times η \, ,$
We then obtain: $t + \inf A \leq x_η\, \qquad (3.1)$ $x_η \leq c \, \qquad (3.2)$ $|(t + \inf A) - x_η| < η \, \qquad (3.3).$
With the definition of $η$, we have: $|f (t + \inf A) - f(x_η)| < ε$
I need to prove $f(x_η) = f(c)$, which would allow us to conclude, with (2).
From 3.2 and f incresing, we easily obtain $f(x_η) \leq f(c)$.
But from 3.1 and f increasing, all I can get, is that $a \leq f(t + x_η)$. I need to prove that $t + \inf A = \inf \{\tau: t \leq \tau \text{ and } a \leq f(\tau)\}$ But I have no clue on how to prove this...