
Re: Some derivative/continuity questions
Posted:
Aug 4, 2013 1:58 PM


On 08/04/2013 11:22 AM, G Patel wrote: > On Saturday, August 3, 2013 6:51:47 PM UTC4, phol...@gmail.com wrote: >> On Saturday, August 3, 2013 1:37:09 PM UTC7, G Patel wrote: >> >>> 1) If f is differentiable at x, is it continuous on an interval about x? >> >>> >> >>> It seems intuitively like it should be. >> >>> >> >>> >> >>> >> >>> 2) Do you know a function that is differentiable at an isolated point? >> >>> >> >>> >> >>> >> >>> Thanks >> >> >> >> Well, what is normally taught in Calc I is the relationship between a limit, continuity and differentiability. >> >> A function is differentiable at a point if it is ..... >> >> 1/ Defined at that point in the domain of the function. >> >> 2/ Continuous at that point (both sides, left and right). This precludes the end points of a line segment. >> >> 3/ The left determined derivative equals the right derivative (using the limit definition of a derivative). This precludes a corner point or cusp. >> >> >> >> So if you apply those requirements to an isolated point.......? >> >> Phil H > > Ooops. I now realize my problem in stating my question. > I meant, the function is differentiable at a point, but not anywhere near it. > That is, it causes f' to be defined at an isolated point. > > Does such a function exit? >
Suppose we want the function to be differentiable at 0, but not near 0. Then, we can fix the interval K of definition to be [a, b], with a = 1/2 and b = 1/2.
Suppose W: R> R is a Weierstrasstype continuous, nowhere differentiable function.
Then, the proposed function will be f(x) := h(x) W(x), for x in [a, b], with h(0) = 0 and h(x)> 0 if a <= x < 0 or 0 < x <= 1/2 .
Further, h should be smooth (infinitely differentiable) on [a, b].
There are things called "bump functions", cf. http://en.wikipedia.org/wiki/Bump_function
The Examples section there gives a standard 1D bump function, Psi: R > R.
Psi(1) = Psi(1) = 0, and Psi is supported on [1, 1].
Psi is positive in [1, 1].
Suppose h(x) := Psi(1x^2), for 1/2 <= x <= 1/2 .
Then h(.) is very smooth and very small near x = 0, but is still strictly positive if x =/= 0 and 1/2 <= x <= 1/2.
So the proposed function is f: [1/2, 1/2] > R,
f(x) := Psi(1x^2) W(x) .
I expect that f will be differentiable at x=0.
If x ~= 0, but x =/= 0, x in [1/2, 1/2], then Psi(1x^2) > 0 and it is "wellbehaved" near x. So possibly in this case with x in [1/2, 1/2] and x =/= 0 we have that f(x) = Psi(1x^2) W(x) is nondifferentiable at x.
david
 abc?

