Intro Real Analysis, (Most of) Lec 14: Uniform Continuity Non-Examples, Variation of a Function

Lecture 14, Part 1.

(0:00) Remarks about last lecture.
(0:33) Goals for lecture.
(1:08) Reminder that derivatives satisfy the intermediate value property.
(2:59) Definition of uniform continuity.
(5:34) Uniform continuity theorem and miscellaneous facts.
(6:46) Sketch of proof of 1/x not being uniformly continuous on (0,1).
(14:33) Sketch of sin(x^2) not being uniformly continuous on R.
(26:15) Summary of facts about monotone functions.
(32:58) Start discussion of the variation of a function.
(33:58) Partition of a closed and bounded interval.
(37:58) Definition of the variation of a function over an interval.
(40:34) The variation of a monotone function.

Bill Kinney, Bethel University Department of Mathematics and Computer Science. St. Paul, MN.

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[jonaskoelker]

23:40 "pi over [sqrt(a + b*n) + sqrt(c + d*n)] goes to 0 as n goes to infinity, not proven".

It involves some boring legwork, but one can show that the (non-negative) square root function is increasing and grows without bound. But then one term in the denominator can be made arbitrarily large and the other at the very least positive (and in fact also arbitrarily large). But then this converges like pi/N which goes to 0. [Add rigor according to taste.]

[muhammadaafaquerajput]

Thanks for the video lectures Prof. Kinney. One question regarding this lecture for Case-I when \delta >=1, you took the x=0.5 and y=0.25, so |x-y|=0.25=\delta, which is less than 1 not >=1. Can you please explain this? Is my understanding correct? Thanks in advance.

[jonaskoelker]

I was arguing with Zeno about Achilles and the turtle the other day, and I managed to persuade Zeno that if Achilles is at a particular place at a particular time, he must have also been at every other place closer to the starting line at some other (earlier) point in time. Furthermore, since motion and stillness are opposites, Achilles is never in any given place at two different points in time.

Theorem: let f: [a, b] -> R be strictly increasing and surjective. Then f is continuous.
Proof:
Let some x in [a, b] be given.
Let y_l = f(x) - epsilon and y_h = f(x) + epsilon.
Let x_l and x_h satisfy f(x_l) = y_l and f(x_h) = y_h.
Since f is surjective the x_l and x_h exist.
[Since the increasingness of f is strict they are unique, but we don't depend on this fact.]
Let delta_1 = x - x_l and delta_2 = x_h - x.
Let delta = min(delta_1, delta_2).
Consider now some y.
If x < y < x + delta <= x_h, that is, y is to the right of x and close, then f(x) < f(y) < f(x_h) = f(x) + epsilon.
If x > y > x - delta >= x_l, that is, y is to the left of x and close, then f(x) > f(y) > f(x_l) = f(x) - epsilon.
So f(y) is between f(x) - epsilon and f(x) + epsilon, i.e. |f(y) - f(x)| < epsilon.
[Note: if f is the sign function, f(x) = -1 when x < 0 and =1 when x > 0 and =0 when x = 0, and we are looking at f exactly at x=0, then we cannot confine f to a small neighborhood around f(0) by confining the inputs to a small interval (-delta, delta). So we need the monotonicity to be strict, not weak.]
Corollary: if f is strictly decreasing and otherwise as required above, then -f is strictly increasing and thus continuous, but then -(-f) = f is also continuous.

Surjective essentially means "no jump points". So this theorem suggests that monotone functions are distinguished from continuous ones exactly by the presence or absence of jump points.

Note that at some jump point c, we can let a = sup { f(x) | x < c } and b = inf { f(x) | x > c }.
By monotonicity (assume WLOG increasing) we have a <= f(c) <= b.
By definition of a jump point we cannot have a = f(c) = b.
But then we must have a < b.
So there must be some open interval (a, b) that is disjoint from the range of f.
Or maybe there are two of them, (a, f(c)) and (f(c), b), if f(c) is not in {a, b}.
[If f is decreasing then -f is increasing and the argument can be repeated.]

[chandankar]

Say f(x)=x^2 in [1, 2] and [0, infinity)
Then the first is uniformly cont. and the second is not !

I mean how to show this like :
Say for any \epsilon s.t.
|x+y|

Now till this, everything is same why in one case it is UC and in othercase its not ?

[beback_]

You guys should do a noise gate, the background hum is really bad.