Question

@KingGeorge

Answer 1

Its my weekly question :)

Answer 2

If you dont like this one I can give you another one

Answer 3

It's been a little while since I've done actual analysis :P

Answer 4

Ya so let me get you smth easier

Answer 5

Well, the first part is pretty easy. It's basically just the definition of "not uniformly continuous"

Answer 6

BTW Congratulations. Where is ur offer from?

Answer 7

U of Rochester. I'm still waiting to hear back from most schools I applied to. Unfortunately, I also just recently got a rejection from Princeton :(

Answer 8

awwwwww :( That is sad
They are stupiidddddd

Answer 9

Ah. This new one I can do. It's just the IVP (intermediate value theorem).

To be honest, I wasn't expecting them to accept me.

Answer 10

Let's see how much I remember. Last time proved this, we started with the assumption that \(f(a)f(b)\le 0\). In this particular case, if \(f(a)=0\) or \(f(b)=0\), we're done. If \(f(a)f(b)<0\), then one of \(f(a),f(b)\) is negative, and the other positive. At this point, we had another theorem that proved there was then some \(c\in[a,b]\) s.t. \(f(c)=0\).

Then define \(g(x)=f(x)+y\). Then, from what we just proved, it immediately follows that \(g(c)=y\).

If we started with \(g(b)\le y\le g(a)\), take \(f(x)=g(x)-y\), so \(f(b)\le 0 \le g(a)\). From this we get that \(f(b)f(a)\le 0\) and we're done.

Answer 11

However, I cannot immediately remember how we proved that if \(f(a)f(b)\le0\), then there was some \(c\in[a,b]\) s.t. \(f(c)=0\).

Answer 12

hmmm gonna check my notes

Answer 13

hmmm I found a proof in my book using the supremum

Answer 14

Well, I'm sorry, but I've got to go now. I need to get up early tomorrow, so I need to head to bed. Have a good night, and good luck with this!

Answer 15

Thanks @KG :)

Answer 16

Good night