Is Phi Upper Semicontinuous in Rudin's Analysis Problem?

[benorin]

upper semicontinuity problem (Papa Rudin)

By Papa Rudin I do mean Real and Complex Variables by Walter Rudin. This is part of my grad analysis homework...
Let f be an arbitrary complex function on $$\mathbb{R}^1$$, and define
$$\phi(x,\delta)=\sup\{|f(s)-f(t)|:s,t\in(x-\delta,x+\delta)\}$$,
$$\phi(x)=\inf\{\phi(x,\delta):\delta > 0\}$$.
Prove that $$\phi$$ is upper semicontinuous, that f is continuous at a point x if and only if $$\phi(x)=0$$.
I can get the rest (I hope) from there.
The working definition of upper semicontinuous (ucs) is: a function f:X-->R^1 is usc if {x:f(x)<a} is an open set for every a in R^1.

 

[benorin]

OK, so I read the FAQ: got to show some work...

I don't have much done but:
So $$\delta_{1}<\delta_{2}\Rightarrow \phi(x,\delta_{1})\leq\phi(x,\delta_{2})$$ since sup-ing over less domain means less possibilities for big values in the co-domain. If knew f was bounded... but no.

 

[CarlB]

None of the 21st century math whizzes have stepped up to the plate so far.

My instincts on this are a quarter century rusty but it seems to me that you can do this by doing an epsilon and delta type argument where you set epsilon = (a-phi(x))/2 > 0. That is, first you use epsilon to back out the inf to get an open neighborhood U of x that satisfies phi(U) < phi(x) + epsilon. This gives you the usc requirement. I would guess the remainder to be similar.

Carl

 

[mathwonk]

because most people do not like to do other peoples hw.

this kind of thing is a perfect il;lustration of why I dislike rudin's book, i.e. a simple idea is made obscure and difficult.this appears much more cleasrly in riemann e.g.

the first notion defined above is an attempt to make precise the notion of "local oscilaltion", i.e. how much does the function jump up and down on a delta neighborhood of x?the second notion is an attempt to make rpecise the concept of infinitesimal oscillation AT x, i.e. how much (at least) does the function jump up and down on EVERY neighborhood of x?a function is continuos if and only if as delta goes to zero, the oscillation of f on a delta neighborhood of x goes to zero.

An upper semi continuous function is one that only "jumps UP" at points, not down. i.e. it behaves like the dimension of the kernel of a family of linear maps, which can jump up at x if the determinant of the map vanishes at x.

so let's see, if a function oscillates by at least K on every nbhd of x, where x is a variable point converging to a, then any nbhd of a will contain a point where f oscillates by at least K, hence f will also oscillate by at elast K on every nbhd of a.

thus if x converges to a, the value of phi at a will be at least as great as the value at poimnts near a.

so the set where phi is at elast K will be closed. i hope that is the right definition of upper semi blah blah blah...

the point is to understand what you are trying to do first intuitively, then write it up rigorously.

good luck.