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[hops]

I've been at this one single question for 5 hours now. I hate myself and nobody would help me.

[MagmaMcFry]

You've been way too generous with the inequalities. You have to choose your approximations wisely, or else they won't be bounded anymore. That's the hard part in this kind of question: proving an inequality in two variables. Proving inequalities isn't something you can just do by the book, it's different every time and usually done with wizardry and eldritch skills. There's absolutely no general guideline on how to do that, but here's some general tricks you should try:

- Use both sides of your inequality, don't stuff everything onto one side.

- Break the symmetry of symmetrical inequalities: If you have an inequality that is symmetrical in x and y, then you only need to consider the case that x >= y, the other case works identically.

- Occasionally it helps to substitute variables.

For example, let's have a look at the first function, and show that it's Lipschitz continuous with a factor of 0.5 (we could have chosen any larger number too but we're doing 0.5 for the thrill):

We need to show

|sqrt(x+1)-sqrt(y+1)| <= 0.5|(x-y)|

We can assume x>=y because of symmetry, and since the square root function is increasing, we need to show

2(sqrt(x+1)-sqrt(y+1)) <= x-y.

Let's substitute x' = sqrt(x+1), y' = sqrt(y+1). We get

2(x' - y') <= x'^2 - y'^2.

But hold up, we can factor the right side as (x'+y')(x'-y')! Since both sides are positive and x'-y' is positive too, we can divide by x' - y' and it remains to show that
2 <= x' + y'.
This is true, because x' >= 1 and y' >= 1. We're done!

[Shadowlord]

Shit, I don't even know what a Lipschitz anything is.

[hops]

Aaaaaa thanks MagmaMcFry, I couldn't understand any of the other explanations online ;-;

[Dozebôm Lolumzalìs]

Shit, I don't even know what a Lipschitz anything is.

Me neither. Now if we were talking matrices, I'd be all over these problems...

[MagmaMcFry]

A function f is called Lipschitz continuous (or just Lipschitz) with factor v such that for all x, y the inequality |f(x)-f(y)| <= v|x-y| holds.

To visualize: If f(x) describes the position of a dog at time x, then f is Lipschitz with factor v if the dog has a speed limit of v.

[Dozebôm Lolumzalìs]

Ah, so as long as a function obeys some speed limit it must be continuous, right?

(assuming that f(x) even exists)

[Shadowlord]

That seems like a very confusing way to write |(f(x)-f(y))/(x-y)| <= v

[MagmaMcFry]

Ah, so as long as a function obeys some speed limit it must be continuous, right?

(assuming that f(x) even exists)

Yes. Any Lipschitz function is also continuous. However not every continuous function is also Lipschitz, for example the function f with f(x) = x^2 is not. However, this f is still locally Lipschitz, which means that for any time x there is *some* speed limit being followed.

Yet there are still continuous functions that are neither Lipschitz nor locally Lipschitz. An example is the cube root function f with f(x) = x^(1/3). Around x = 0 the dog breaks all speed limits.

That seems like a very confusing way to write |(f(x)-f(y))/(x-y)| <= v

I find my notation a) more intuitive and b) way more well-defined in case x = y.

[hops]

Sorry about relying so much on this thread but I don't even know how to approach i. I know that that is a method to prove a value exists in an interval, but how am I supposed to prove that it's a minimum when I don't know what the minimum is?

[da_nang]

Sorry about relying so much on this thread but I don't even know how to approach i. I know that that is a method to prove a value exists in an interval, but how am I supposed to prove that it's a minimum when I don't know what the minimum is?

Check the end points of the interval for p(x). What conclusions can you draw? What do those conclusions mean for f(x)?

[hops]

Sorry about relying so much on this thread but I don't even know how to approach i. I know that that is a method to prove a value exists in an interval, but how am I supposed to prove that it's a minimum when I don't know what the minimum is?

Check the end points of the interval for p(x). What conclusions can you draw? What do those conclusions mean for f(x)?

...I mean, I still don't know what the minimum is? If they told me 'The minimum f(c)=2 exists' then it would be another story.

[MagmaMcFry]

Any continuous function defined on a compact set has a minimum. This is a theorem you should have heard about by now.

To show that this minimum is less than or equal to 4, just find some x in [-2, 2] with f(x) <= 4. This proves that f(x_) <= f(x) <= 4.

[hops]

Also, I understand intermediate value theorem, but how do I prove that a value doesn't exist within an interval?

[hops]

Or, you know, any ways whatsoever to solve ii

Sigh, I really need to revise