Contraction mapping Puns

Hey I need your help. I'm currently learning convex optimization in university and all that's thrown at me is lemmas and definitions but I don't really get the things behind the scenes. For example: What is the goal of contraction mapping in relation to convex optimization? It must has to do something with the fact that a contraction mapping has one unique fixed point (if Banach can be applied)

I would like to get a brief understanding first and then deeper dives into it but those mathematics publications are not the right start for me I feel like.

Thanks in advance. Hope this is a good subreddit for it otherwise please refer me to others.

Hey guys, currently learning the contraction mapping and I was wondering whether someone could me an example for this definition?

$f: \Bbb R \mapsto \Bbb R$

$g: \Bbb R \mapsto \Bbb R$

$h: \Bbb R \mapsto \Bbb R$

$h:=\max\{f(x), g(x)\}$

Is $h$ a contraction on $ \Bbb R$ if $f$ and $g$ are both so?

First attempts of mine -although they are not very fruitful- is the following:

since every contraction mapping is Lipschitz cont's and hence uniformly cont's, and I just proved that $h$ is uniformly cont's if $f$ and $g$ are so, $h$ could be a contraction if $f$ and $g$ are so.

http://math.stackexchange.com/questions/1826987/example-of-contraction-mapping-theorem-failing-for-strict-metric-map

Is there an example of f:[0,1]→[0,1] s.t. |f(x)−f(y)|<|x−y|

but a sequence x0,f(x0),f2(x0)... doesn't converge to its fixed point?

We can easily show that there is a unique fixed point, but I don't know what to do from here. I feel like any counterexample would be non-differentiable, but I'm not sure.

I need some help with this proof.

Let (X,d) be a compact metric space. Let f: X -> X be continuous. Let f be a contraction mapping. (then d(f(x),f(y)) <= d(x,y) for all x,y in X). Show there is a unique fixed point, f(x) =x.

Attempt at proof: First since X is compact d(x,y) is bounded. This is because otherwise you could take the open cover (B(x,1),B(x,2),B(x,3),....) for some x in X, and this open cover has no finite cover.

Now we define f^{1}: X ->f^{1} (x) and in general f^{n}: f^{n-1} (x) ->f^{n}(x)

Now if f^{n} has more than one point, then f^{n+1} is strictly contained in f^{n}.

Pf: f^{n} is compact, as X is compact and f is continuous and images of compact spaces are compact. Now take x,y in f^{n} x not equal to y. and d(x,y) is maximized. (this is possible since d(x,y) is bounded in X by above)

At least one of x,y does not belong in f^{n+1}. This is because f is a contraction mapping, so d(f(z),f(w)) <= d(z,w) for all z,w in f^{n}, therefore d(x,y) is never attained in f^{n+1}. Therefore at least one of x,y does not belong in f^{n+1}

At this point I'm not sure how to proceed... I'd like to say that eventually f^{k} has only one point in it and that point is the fixed point but I don't believe that is possible.

Any help is welcome.

Kind of a cool theorem in real analysis: A contraction mapping is a mapping of a metric space X into X such that d(f(x),f(y))< c*d(x,y) where c<1 for all x,y in X. In complete metric spaces, contraction mappings leave exactly one point fixed. I'm trying to think of a real life situation that can be viewed as a contraction mapping in which this theorem would apply (for instance in R^3)

How can I use the contraction mapping principle (sometimes known as successive approximations?) to solve differential equations? If I know differential equation, say f' = f + 1, and an initial condition such as f(0) = 1, how could I solve it with CMP?

I’m sure there are technical limitations and other items to consider, but thought I’d share the idea with the community!

Edit: Thanks folks! Turned out I just didn't use the right chest. Ah well, can't unsend the hateful carrier pigeon messages I sent to R*.

I plan to have a mapping variable that stores address => uint256. Every time a user interacts with the contract my contract will add to the mapping if the address doesn't exist, check the count if it does exist, update the count at the end of the tx.

This mapping is expected to grow to around 5k. Will this logic add insanely high gas costs to the user as the mapping grows?

Yesterday: Alpine 4 Holdings (ALPP) Subsidiary, Vayu Aerospace Corporation, Advances to Phase 3 of our US Air Force SBIR Allowing for Sole Source Procurement

https://finance.yahoo.com/news/alpine-4-holdings-alpp-subsidiary-165200329.html

Today: Alpine 4 Holdings (ALPP) Subsidiary, Identified Technologies, has Been Awarded a Multi-Year Drone-Mapping Contract with The US Army Corps of Engineers

https://finance.yahoo.com/news/alpine-4-holdings-alpp-subsidiary-154500435.html

https://reddit.com/link/s2liiz/video/u75835e6ncb81/player

Like we got new maps now so eventually we can get contracts for the smismass maps right?

No fud I really do want to know. I'm not new in crypto so I don't have the same fearlessness as newer people do when investing in brand new projects here in the crypto space. Looking for reassurance that the contract is clean and safe. Thank you for any basic info on future audit plans.

I just returned to playing after several months. I noticed that when my teammates ping a contract, vehicle, etc. It shows up but when I pull up the map it doesnt show.

I know that it use to.

Is this a bug, change or a setting I need to fix? On PS5.

I've just started working on map contracts (cp_mercenarypark is the first one) but as of writing there are no casual matches, and no community servers using that map. Now I'm aware that this won't always be the case, but what exactly do I do? I tried running my own server for a while although no one joined causing contracts to stay inactive. Just confused as all hell over here after an hour of no hits.

I completed the contract legitimately, quite fun actually. However, the two huge trailers are now back on the map after I loaded the game later.

Is there a way to get rid of them?

PC Steam

Polygon Contract: 0x57194feaca970A4E98A19C365FE144fB54F657DB

It would be MUCH cheaper gas fees to acquire tokens on Polygon, however I am unsure about how that works exactly, my understanding is the token is an exact mirror and should have the same value, when you wish to move it from Matic back to ETH simply use the bridge (obviously at a later date when the value goes to moon)

I did the same with ELON and so far the token price is exactly the same, but ELON on Coingecko actually states their Polygon address as well as ETH.

Can anyone confirm that it is the exact mirrored value/same token (maybe even a developer??) this would help A TON of people acquire UFO cheap!!!!

$f: R \mapsto R$

$g: R \mapsto R$

$h: R \mapsto R$

$h$:= $max${$f(x)$, $g(x)$}

Is $h$ a contraction on $R$ if $f$ and $g$ are both so?

First attemps of mine -although they are not very fruitful- is the following:

since every contraction mapping is Lipschitz cont's and hence uniformly cont's, and i just proved that h is uniformly cont's if f and g are so, h could be a contraction if f and g are so.

If g = f compose f, ie: g(x) = f(f(x)) on metric space (X,d) and g is a contraction mapping, how can one show f has a fixed point?

I can ascertain that if g is a contraction, then g is continuous on the metric space it is defined upon, this implied it has a unique fixed point by Banach, ie: there exists z such that g(z) = z, but how can one calculate that f has a fixed point.

If g is continuous, then is f continuous by definition on metric space (X,d)?

How can I use the contraction mapping principle (sometimes known as successive approximations?) to solve differential equations?

If I know differential equation, say f' = f + 1, and an initial condition such as f(0) = 1, how could I solve it with CMP?