Super Mario 64 120 Stars Cheat
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Description: How to get 120 stars, without any glitches, glitches, cheats, tricks, hacks or. If this isn't what you want, you can use our cheat device to get all.
Super Mario 64 120 Stars Trick
My 120 Stars Super Mario 64 hack will give you all 120 stars in the game. though I don't know how it works, since it's the first hack i've.
Star War 64 cheat tool (Hacks, Cheat Codes, Info) - NewSuperCheat.Q:
Is $f(x,y) = \ln x$ Lipschitz continuous on $\mathbb{R}^2$?
Let $f:\mathbb{R}^2\to\mathbb{R}$ be defined by $f(x,y) = \ln x$.
Is $f$ Lipschitz continuous?
My intuition is that the answer is no, since
$|f(x,y) - f(x',y')| = \left| \ln\frac{x}{x'}\right| \le \frac{1}{x'} + \frac{1}{x}$.
(I think that in fact $f$ is not Lipschitz continuous, since it fails to be so if $x'=x$.)
I have tried to prove this by using the Mean Value Theorem. Let $M$ be a bound for $f$: $|f(x,y) - f(x',y')| \le |(x,y) - (x',y')| \cdot M$.
However, since $f(x,y) = \ln x$ depends on two variables, it is unclear how to find such $M$ bounding $f$ with respect to the $L^1$ norm.
I am not sure whether there is a nice proof for this (or an easy counterexample), but it might have something to do with $\mathbb{R}$ being infinite.
A:
You are right, there is no such bound, but we can get a similar bound to your one.
If we write $f(x,y) = \log(x)$, then we have $f(x,y) \le \log(x)$. This is clear, as $x \ge e^y$, and this is 0b46394aab
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