Analysis with Thomae's function.
mina_world
f : [0,1] -> R,
f(x) = 1/p if x = q/p is a rational number. (p, q)=1
= 0 if x is irrational or 0.
1) f(x) is discontinuous at all rational.
2) f(x) is cotinuous at all irrational.
3) f(x) is not differentiable at all rational.
4) f(x) is not differentiable at all irrational.
5) f(x) is Riemann integrable by (1),(2).
6) f(x) is Lebesgue integrable by (1),(2).
--------------------------------------------------
Ok, I will show (1)~(4).
Let's go...
(1) f(x) is discontinuous at all rational.
There is a irrational sequence (x_n) such that (x_n) -> r for r in Q.
f(x_n) = 0, f(r) =/= 0.
so, f(x_n) -/-> f(r).
so, f(x) is discontinuous at all rational.
(2) f(x) is cotinuous at all irrational.
I must show that |x-c| < d ==> |f(x)-f(c)| < e for c in I(irrational).
Of course, f(c) = 0.
so, |x-c| < d ==> |f(x)| < e.
If x in I /\ (c-d, c+d), |f(x)| = 0 < e.
If x in Q /\ (c-d, c+d), I must show that |f(x)| < e.
There is N >0 such that 1/N < e.
Let A = {1}U{1/2}U{1/3, 2/3}U{1/4, 2/4, 3/4}U ...
... U{1/N, 2/N, ... , (N-1)/N} on x-axis.
so, for each r in A, f(r) in {1, 1/2, 1/3, ... , 1/N} ==> f(r) >= 1/N.
so, for each r' in Q-A, f(r') < 1/N < e.
[Q-A = {1/(N+1), 2/(N+1), ... , N/(N+1)}U{1/(N+2), 2/(N+2), ....]
Since A is finite, there is d >0 such that |x-c| < d ==> x not in A.
so, x in Q /\ (c-d, c+d), |f(x)| < 1/N < e.
(3) f(x) is not differentiable at all rational.
Since f(x) is discontinuous at all rational, trivial.
(Example)
If c = 0,
f'(0) = lim{x->0} [f(x) - f(0)] / [x - 0] = lim{x->0} f(x) / x
If (x_n) is a irrational sequence with x_n -> 0,
lim{x_n ->0} f(x_n) / (x_n) = 0
If (y_n) is a rational sequence with y_n = 1/n,
lim{y_n ->0} f(y_n) / (y_n) = 1
so, f(x) is not differentiable at x=0.
(4) f(x) is not differentiable at all irrational.
If c in I(irrational),
f'(c) = lim{x->c} [f(x) - f(c)] / [x - c] = lim{x->0} f(x) / (x - c)
If (x_n) is a irrational sequence with x_n -> c,
lim{x_n ->c} f(x_n) / (x_n - c) = 0.
I must find the sequence (z_n) such that lim{z_n ->c} f(z_n) / (z_n - c) =/=
0.
and then f(x) is not differentiable at c.
Let c = 0.(d_1)(d_2)(d_3)....
Let z_n = 0.(d_1)(d_2)...(d_n)
so, z_n -> c
|z_n - c| = 0.0....0(d_(n+1))(d_(n+2)).... < 1/(10^n)
|f(z_n) - f(c)| = |f(z_n)| >= 1/(10^k) with k = min{n | d_n =/= 0}.
If n >= k,
lim{z_n ->c} |f(z_n) / (z_n - c)| > [1/(10^k)] / [1/(10^n)] = 10^(n-k) >= 1
=/= 0.
so, f(x) is not differentiable at all irrational.
END.
M Ashok Kumar
Can you prove the given function has simple continuity at all rationals, i.e. left limit and the right limit exist at all rationals but not equal to f(x).
Dave L. Renfro
Is this a trick question?
Dave L. Renfro
David C. Ullrich
wrote:
>Hello teacher,
>Can you prove the given function has simple continuity at all
>rationals, i.e. left limit and the right limit exist at all rationals but not equal to f(x).
Evidently you meant simple _discontinuity_.
What is f?
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
ashok
> On Fri, 17 Oct 2008 05:34:57 EDT, M Ashok Kumar <ashokma...@yahoo.com>
> wrote:
>
> >Hello teacher,
> >Can you prove the given function has simple continuity at all
> >rationals, i.e. left limit and the right limit exist at all rationals but not equal to f(x).
>
> Evidently you meant simple _discontinuity_.
>
> What is f?
> David C. Ullrich
>
> "Understanding Godel isn't about following his formal proof.
> That would make a mockery of everything Godel was up to."
> (John Jones, "My talk about Godel to the post-grads."
> in sci.logic.)
The function is the so-called Thamae's function. i.e.
f(x)=0 if x is irrational, and 1/n if x is rational and x=m/n where
(m,n)=1 and n>0.
Thank you.
Dave L. Renfro
>>> Can you prove the given function has simple
>>> continuity at all rationals, i.e. left limit
>>> and the right limit exist at all rationals but
>>> not equal to f(x).
David C. Ullrich wrote:
>> Evidently you meant simple _discontinuity_.
>>
>> What is f?
M Ashok Kumar wrote:
> The function is the so-called Thamae's function.
> i.e. f(x)=0 if x is irrational, and 1/n if x is
> rational and x=m/n where (m,n)=1 and n>0.
Isn't it fairly clear that the same reasoning which
shows f is continuous at each irrational number can
be used to show that both unilateral limits exist and
are equal to 0 at each nonzero [*] rational number?
Note this is stronger than what you asked for, since
the unilateral limits are equal.
[*] It seems to me that your definition of f does not
define a value of f(0).
Dave L. Renfro
mina_world
>> >Can you prove the given function has simple continuity at all
>> >rationals, i.e. left limit and the right limit exist at all rationals
>> >but not equal to f(x).
>>
>> Evidently you meant simple _discontinuity_.
>>
>> What is f?
>> David C. Ullrich
>>
>> "Understanding Godel isn't about following his formal proof.
>> That would make a mockery of everything Godel was up to."
>> (John Jones, "My talk about Godel to the post-grads."
>> in sci.logic.)
>
> The function is the so-called Thamae's function. i.e.
> f(x)=0 if x is irrational, and 1/n if x is rational and x=m/n where
> (m,n)=1 and n>0.
> Thank you.
New member ?
Welcome to mina_world.
M Ashok Kumar
For irrational x, for given c>0, I choose n large enough so that 1/n<c. And I define 'delta' to be the distance form x to the set {k/j: k is in Z and j=1,2, ..,n}. Since x is irrational 'delta' is always >0. And I can show that f is continuous at x. But in case of rational x, how to get the delta to show that the left hand and right hand limits of f at x are 0?
Dave L. Renfro
> Sorry Dave L. Renfro. Ya.. I did not define the function at 0.
> It is, actually, f(0)=1.
I assumed this was the case and was simply pointing out
an oversight in your definition.
> For irrational x, for given c>0, I choose n large enough so
> that 1/n<c. And I define 'delta' to be the distance form x
> to the set {k/j: k is in Z and j=1,2, ..,n}. Since x is
> irrational 'delta' is always >0. And I can show that f is
> continuous at x. But in case of rational x, how to get
> the delta to show that the left hand and right hand limits
> of f at x are 0?
So how did you decide to use the value of 'delta' that
you used? You must have had some underlying insight about
what was going on, and it's this insight that will also
allow you to show that at each real number (thus, at each
rational number) both unilateral limits exist and are
equal to 0. Hint: Given any positive integer q and any
bounded interval J, there are at most a finite number
of rational numbers in J whose reduced fractional expression
has a denominator (assumed positive) that is not greater
than q. This implies that given any positive value c,
the set {x in R: f(x) >= c} is "locally finite", and
from this the result easily follows.
Dave L. Renfro