$|F^* - \tilde F^*| \to 0$ as $\epsilon \to 0$

Question

Define $F^* := \max\limits_{x\in [0, \alpha]} F(x),\quad \tilde F^* := \max\limits_{x\in [0, \tilde \alpha]} F(x)$. Assume that $F(x)$ is continous function and $|\alpha -\tilde \alpha| \leq \epsilon$. How can get the upper bound of $|F^*-\tilde F^*|$ such that this upper bound will imply $|F^*-\tilde F^*| \to 0$ as $\epsilon \to 0$?

I have no idea to solve it. Please give me some help. Thank you in advance!

you know that the max function is continuous as well if F is continuous? — Simonsays, Jun 19 '17 at 10:23

Simonsays · Answer 1 · 2017-06-19T14:13:33.070

1

Define $F^*(\alpha):=max_{x \in [0,\alpha]}F(x)$.
This is an continuous function (see also here), because $F$ is continuous. It also holds $F^*(\bar{\alpha})= \bar{F}^*(\bar{\alpha})$, so to be precise, both functions are the same, they are just calculated at different points on the intervall.
The proof simply follows by continuity of $F^*$:
Now say without loss of generality $\alpha = \bar{\alpha}- \epsilon$:
$|F^*(\alpha)-\bar{F}^*(\bar{\alpha})|=|F^*(\alpha)-F^*(\alpha + \epsilon)| \rightarrow 0$, as $\epsilon \rightarrow 0$

edited Jun 19 '17 at 14:13

answered Jun 19 '17 at 10:41

Simonsays

1,823

how can you prove that $F^*(\alpha)$ is continous? – Thomas Edison Jun 19 '17 at 12:37
look here – Simonsays Jun 19 '17 at 14:12
If function $F$ is Lipchitz with constant $L$. Can we show that $|F^(\alpha) - F^(\bar \alpha) | \leq L\delta$? – Thomas Edison Jun 19 '17 at 14:49
well I guess you mean $\epsilon$ instead of $\delta$, that would be the definition of Lipschitz continuity. But this functions $F^*$ is not necessarily Lipschitz ( and you dont need it to be). – Simonsays Jun 19 '17 at 14:53

$|F^* - \tilde F^*| \to 0$ as $\epsilon \to 0$

1 Answers1