| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 888.1 | problem one | BLAKE::DERAMO | Daniel V. D'Eramo, VAX LISP developer | Sun Jun 19 1988 21:44 | 61 |
| Problem 1:
>> 1. Suppose f is differentiable on [a, b], f(a) = 0, and there is
>> a real number A such that |f'(x)| <= A|f(x)| on [a, b]. Prove that
>> f(x) = 0 for all x in [a, b].
Assume A is positive.
f is differentiable on [a, b], so it is continuous on
[a, b]. Let c and d be arbitrary reals such that
1) a <= c <= d <= b
2) f(c) = 0
3) d - c < 1/A
Since the closed interval [c, d] is compact and f is
continuous, the image of [c, d] under f is bounded, and
there is a point x of [c, d] such that |f(x)| = M
= the least upper bound of |f(t)| for t in [c, d].
I will show that M = 0, i.e. f is identically zero in
[c, d].
Now, I forget what it is called, Rolle's theorem maybe,
but there is a point e of [c, x] such that
f(x) = f(c) + (x - c)f'(e)
Now play with the symbols for a while
M = |f(x)|
= |f(c) + (x - c)f'(e)|
= |(x - c)f'(e)| since f(c) = 0
= (x - c)|f'(e)| c <= x <= d
<= (x - c) A |f(e)| property of f'
<= (x - c) A M |f(e)| <= M
If M is nonzero, divide throught by M to get
1 <= (x - c) A
But x - c <= d - c < 1/A, so the right hand side is < 1.
The contradiction refutes the assumption that M is non-zero,
so M must be zero, and f is identically zero over [c, d].
Now, choose positive integeral n large enough so that
(b - a)/n < 1/A.
We can choose a and a + (b - a)/n for c and d in the above,
[c = a so f(c) = f(a) = 0, and d - c = (b - a)1/n < 1/A],
thus showing that f is identically zero over [a, a +
(b - a)/n]. Then choose a + (b - a)/n and a + 2(b - a)/n
for c and d, to show that f is identically zero over
that subinterval. Repeat this n times until you have
"covered" all of [a, b], showing that f is identically
zero on all of [a, b].
Dan
P.S. Lots of line noise, but I've tried to remove all of
the garbage it added to my reply.
|
| 888.2 | | CLT::GILBERT | | Mon Jun 20 1988 11:46 | 8 |
| I'm uncomfortable with the proof of problem 1, because of functions
such as:
f(x) = x - x^3 (0 <= x <= 1) { Rolle's thm doesn't hold }
-1/x
f(x) = e , for 0 < x { f(x) is infinitely differentiable,
= 0 for x <= 0 and at x=0, all derivatives are 0 }
|
| 888.3 | You left off one of the conditions. | AITG::DERAMO | I am, therefore I'll think. | Mon Jun 20 1988 12:36 | 17 |
| Re .2
Are these counterexamples to Rolle's theorem or to the statement
of problem 1? For the first problem, f'(0) = 1, and so there can
be no real A such that |f'(x)| <= A |f(x)| for all x in [0, 1].
For problem two,
-1/x
e
f'(x) = -----
2
x
2
and so |f'(x)| / |f(x)| = 1/(x ) and again no real A can be chosen
(in a domain that includes 0) to satisfy the statement of problem
1.
Dan
|
| 888.4 | | CLT::GILBERT | | Mon Jun 20 1988 17:34 | 2 |
| The proof is good, of course (BTW, it's easy to understand graphically).
I was just worrying about pathological cases.
|
| 888.5 | more on what I called Rolle's theorem | ZFC::DERAMO | Daniel V. D'Eramo, VAX LISP developer | Mon Jun 20 1988 18:22 | 54 |
| I typed up this proof of Rolle's theorem before reading
.-1, but I'll post it anyway. There may be pathological
cases, but they will still obey Rolle's theorem.
Let f be differentiable [and therefore continuous] on [a, b].
Let g be the linear function equal to f(a) at a and equal to
f(b) at b, i.e.,
f(b) - f(a)
g(x) = ----------- (x - a) + f(a)
b - a
Finally, let h be the function on [a, b] given by
h(x) = f(x) - g(x)
Note that h is continuous on [a, b], differentiable on [a, b],
and that h(a) = h(b) = 0.
If h is not identically zero, then there is a point e in
[a, b] such that h(e) takes on that maximum or minimum, and
furthermore such that h'(e) = 0. Think about what the ratio
h(x + delta_x) - h(x)
---------------------
delta_x
must be as delta_x approaches zero near an x such that h(x)
takes on its largest or smallest value; for instance, near a
maximum the numerator is >= 0 whether delta_x is positive or
negative, so that the ratio is nonnegative for positive
delta_x and nonpositive for negative delta_x, and since the
limit h' exists it must be zero.
Now plug these in:
0 = h'(e) we picked e that way
= f'(e) - g'(e) because h = f - g
f(b) - f(a)
= f'(e) - ----------- from definition of g
b - a
and so f(b) = f'(e) (b - a) + f(a), i.e., Rolle's theorem.
Note that all we used to derive this was [lots of facts
about the topolgy of the real line, about limits,
arithmetic, calculus, etc., and] that f:[a, b] -> R was
differentiable over all of [a, b].
Dan
Oh by the way, if h is identically zero then f is a linear
function and e = b will satisfy Rolle's theorem.
|
| 888.6 | First problem | HPSRAD::KUNDU | | Mon Jun 20 1988 20:54 | 45 |
| >> 1. Suppose f is differentiable on [a, b], f(a) = 0, and there is
>> a real number A such that |f'(x)| <= A|f(x)| on [a, b]. Prove that
>> f(x) = 0 for all x in [a, b].
>> f is differentiable on [a, b], so it is continuous on
>> [a, b]. Let c and d be arbitrary reals such that
>>
>> 1) a <= c <= d <= b
>> 2) f(c) = 0
>> 3) d - c < 1/A
I feel the approach in 888.1 is in the right direction. However,
the proof appears to be incomplete/incorrect because of the
assumption f(c) = 0. One needs to show that there can be a c, a <
c, where f(c) = 0. Below, I attempt to supply a cleaner steps of
reasoning.
We (linear) transform the problem onto [0,1], where f(0) = 0, and
|f'(x)| <= B|f(x)| on [0,1]. B is a constant depending on a, b,
and A. B is not 0, otherwise there is nothing to be solved.
Now, f is continuous (being differentiable) on a compact set [0,1].
Thus |f(x)| is also a continuous function on the compact set [0,1].
So |f(x)| has a maximum, and let
M = maximum of |f(x)| over [0,1].
Let p is the point where a maximum of |f(x)| occurs. Note that for
any x, -M <= f(x) <= M according to our definition of M. If p = 0,
then it implies, |f(x)| <= M = |f(0)| = 0 which means f(x) = 0 for
all x.
Now suppose p > 0.
By Rolles theorem, f(p) - f(0) = pf'(q), for some q, 0 <= q <= p.
So, |f(p)| = |pf'(q)| (use f(0) = 0)
= p|f'(q)| (p is positive)
<= pB|f(q)| (use |f'(x)| <= B|f(x)| at q)
<= pbM.
Therefore, M = |f(p)| <= pBM. Since, p and B are not 0, the
only non-negative M that satisfies above inequality is M = 0.
This completes the proof.
Snehamay K.
|
| 888.7 | typo in .6 | HPSRAD::KUNDU | | Mon Jun 20 1988 21:20 | 1 |
| "pbM" should read "pBM".
|
| 888.8 | re: f(c) = 0 | ZFC::DERAMO | Daniel V. D'Eramo, VAX LISP developer | Mon Jun 20 1988 21:20 | 24 |
| Re .6:
Well, the purpose of the conditions on c and d, including
f(c) = 0, was that the result that f is identically zero
on [c, d] could "slide" along the "potentially longer"
interval [a, b]. The first use I made of it was to show
that we could take a and a+((b-a)/n) for c and d and
by doing so show that f is always zero in that subinterval
of [a, b]. So there c = a, and f(c) = f(a) = 0. Also,
having just proved that f(d) = f(a + (b - a)/n) = 0,
I could again invoke the result about c and d, this time
with c = a+((b-a)/n) and d = a+2((b-a)/n), and again
with this choice f(c) = 0.
So I wasn't assuming that there was some random point
c in [a, b] with f(c) = 0. I was assuming that *IF*
there was a sufficiently small [i.e., d - c < 1/A]
subinterval [c, d] of [a, b] and *IF* f(c) = 0 then f
is zero over all of [c, d].
But each application of that "lemma" used a value of
c for which it was known that f(c) = 0.
Dan
|
| 888.9 | ans in .6 is incorrect | HPSRAD::KUNDU | | Tue Jun 21 1988 10:36 | 7 |
| Thank you Dan. You are right. In my reply .6, M <= pBM fails to
imply M = 0 when pB >= 1. And I can't reason why pB should be < 1.
So, my answer in .6 is incorrect.
I now see what you have done in .1 and its good.
/Snehamay K.
|
| 888.10 | Solutions | HPSTEK::XIA | | Sun Jun 26 1988 18:07 | 11 |
| Problem 1:
The solution Dan gave in .1 is correct, so I am not going to
do it again here.
Problem 2:
The answer is false. Counterexample:
Let f: R^2-->R such that x = (0,0) ==> f(x) = 0. if x = (x1, x2)
!= (0,0) then f(x) = x1^3 / (x1^2 + x2^2).
Comments: The first problem is very intuitive, but the second problem
is totally counter intuitive (at least to me).
|
| 888.11 | request for more on the counterexample | BLAKE::DERAMO | Daniel V. D'Eramo, VAX LISP developer | Sun Jun 26 1988 22:06 | 7 |
| re .-1, problem 2.
Could you please give an example of the C function that shows
this f to be a counterexample? [I haven't translated
this problem to someone I could visualize yet.]
Dan
|
| 888.12 | | HPSTEK::XIA | | Mon Jun 27 1988 10:06 | 3 |
| C is supposed to be any arbitrary continuously differentiable function
that is not vanishing at 0. (The problem states "for all such C").
Eugene
|
| 888.13 | | HPSTEK::XIA | | Mon Jun 27 1988 12:53 | 5 |
| I guess I did not explain quite clearly about what I meant in .12
The point is if f is defined as in .10, then any continuously
differentiable function C will make g(t) = f(C(t)) continuous at
0. Yet f is not differentiable at 0.
Eugene
|
| 888.14 | | ZFC::DERAMO | Daniel V. D'Eramo, VAX LISP developer | Mon Jun 27 1988 18:16 | 7 |
| Ok, thanks. I understand it now. Showing that f is
not differentiable at 0 is easy enough, so to complete
the proof one just has to show that f(C(t)) is continuous
at 0 for all of the relevant C. I can't read .0 from
this buffer so I'll think about that later. :-)
Dan
|
| 888.15 | | HPSTEK::XIA | | Mon Jun 27 1988 18:30 | 7 |
| It is not quite that simple to show that f is not differentiable
at 0 though. I personally think it is the hard part of the problem
(Again there might be some simple and clever ways of doing it. It
is just that I am not aware of it).
Have fun!
Eugene
|
| 888.16 | | ZFC::DERAMO | Daniel V. D'Eramo, VAX LISP developer | Mon Jun 27 1988 19:21 | 43 |
| Doesn't this suffice to show that f is not differentiable
at (0,0)?
Consider the function f from .10
3 2 2
f(x,y) = x / (x + y )
The partial derivatives are
4 2
- 2x 3x
f = ---------- + ----------
x 2 2 2 2 2 2
(x + y ) (x + y )
3
- 2x y
f = ----------
y 2 2 2
(x + y )
In polar coordinates [using r and o (instead of theta)]
these are
4 4 2 2
- 2r cos o 3 r cos o 4 3 2
f = -------------- + --------- = cos o - - cos o
x 4 4 r
r r
3
f = - 2 cos o sin o
y
These do not approach a single limiting value as (x,y) ->
(0,0) along different rays through the origin; for example,
along the positive x axis f decreases without bound as x -> 0.
x
So f and f do not exist at (0,0).
x y
Dan
|
| 888.17 | Perhaps Not | HPSTEK::XIA | | Tue Jun 28 1988 10:03 | 3 |
| re .16
This only shows that the derivitive is not continuous at (0,0).
Eugene
|
| 888.18 | Correction | HPSTEK::XIA | | Tue Jun 28 1988 12:54 | 5 |
| re .12
Correction:
It should be "whose derivitive is not vanishing at 0" not "that
is not vanishing at 0". The function C is vanishing at 0.
Eugene
|
| 888.19 | another try | ZFC::DERAMO | To err is human; to moo, bovine. | Tue Jun 28 1988 23:07 | 31 |
| To show: f(x,y) = x^3 / (x^2 + y^2) is not differentiable
at (0,0) [where f(0,0) = 0].
f (0,0) = lim (f(h,0) - f(0,0)) / h
x h -> 0
1 h^3
= lim - (-------------- - 0)
h -> 0 h h^2 + 0^2
= lim 1
h -> 0
= 1
So f (0,0) = 1
x
f (0,0) = lim (f(0,k) - f(0,0)) / k
y k -> 0
1 0^3
= lim - (------------- - 0)
k -> 0 k 0^2 + k^2
= lim 1 / (k^3)
k -> 0
... the limit does not exist, so f is not defined at (0,0).
y
Dan
|
| 888.20 | I am afraid that is not right either. | HPSTEK::XIA | | Wed Jun 29 1988 10:47 | 28 |
| re .19
There is a problem in your computation of f (0,0).
y
f (0,0) = lim (f(0,k) - f(0,0)) / k
y k -> 0
1 0^3
= lim - (------------- - 0) = lim 1/k (0) = 0.
k -> 0 k 0^2 + k^2 k->0
= lim 1 / (k^3)
k -> 0
^^^^^^^^^^^^^^^^^^
This is not right.
Hence, f (0,0) does exist and is 0.
y
Eugene
Hint: g is continuously differentiable for all C, so you can expect
all the directional derivitive to exist. That is why this
is the hard part of the problem.
|
| 888.21 | oops ^ 17 | ZFC::DERAMO | To err is human; to moo, bovine. | Wed Jun 29 1988 12:38 | 7 |
| Silly me, that's what I get for being stubborn. When
I cancelled the zeros in the numerator and denominator
I left the product the same on top [oops] and the sum
the same on the bottom [okay]. What node is conference
ARITHMETIC on? :-)
Dan
|
| 888.22 | shall I try a third time? :-) | ZFC::DERAMO | To err is human; to moo, bovine. | Wed Jun 29 1988 12:43 | 10 |
| Re .20
>> Hint: g is continuously differentiable for all C, so you can expect
>> all the directional derivitive to exist. That is why this
>> is the hard part of the problem.
Can I just take f as a function of a single complex variable
and show that it isn't analytic at zero?
Dan
|
| 888.23 | C does not mean the complex plane here | HPSTEK::XIA | | Wed Jun 29 1988 13:36 | 16 |
| re .22
The C in my hint is the function defined in the base note. It
does not mean the complex plane.
As to whether it is sufficient to prove that f is nonanalytic a
0. That trivializes the problem since we know that analytic functions
are infinitely differentiable. And by your previous work, we know
that the derivitive is not continuous at 0. Hence, the second
derivitive does not exist at 0.
Well, since I never had a course in Complex Analysis, I am not sure
my above handwaving proof is right. For one thing, I am not 100%
certain that f' is not differentiable in R^2 => f' is not diferentiable
in C. Maybe you can verify this for me.
Eugene
|
| 888.24 | Solutions! Solutions! | HPSTEK::XIA | | Thu Jun 30 1988 11:12 | 47 |
| The notation in this proof is consistant with the ones in the base note.
#1. f is not differentiable at (0,0).
Proof: Let v = (3/5, 4/5). Hence v is a unit vector. Now we compute
the diretional derivitive in the direction of v.
Let C(t) = (C1(t), C2(t)). Let h: R-->R^2
such that h(t) = f(v*t). Then:
h(t) = (3/5 * t)^3 / ((3/5 * t)^2 + 4/5 * t)^2)
= (3/5 * t)^3 / t^2 = 27/125 * t.
Hence, h'(t) = 27/125. ==> h'(0) = 27/125. Hence, by definition,
the directional derivitive of f at (0,0) in the direction of v
is 27/125.
By Dan's previous work, we know that f (0) = 1 and f (0) = 0.
x y
let F = (f (0), f (0))^T = (1, 0)^T
x y
Then v * F = (3/5, 4/5) * (1, 0)^T = 3/5.
Hence, v * F != h'(0). Since h'(0) is a directional derivative,
this means F is not linear. Hence, f is not differentiable at
(0, 0).
#2. show that g is continuously differentiable for all C at 0.
Proof: Again let C(t) = (C1(t), C2(t)). Then g(t) = f(C1(t), C2(t))
= (C1(t))^3 / (C1(t)^2 + C2(t)^2).
g'(0) = lim (g(0+h) - 0)/h where h -> 0.
= lim g(h)/h
= lim C1(h)^3 / (h * (C1(h)^2 + C2(h)^2))
= lim (C1(h)/h)^3 / ((C1(h)^2 + C2(h)^2) / h^2)
= C1'(0)^3 / ||C'(0)||^2 (This step is allowed since
we know that ||C'(0)|| > 0.)
Hence, g is differentiable for all C. I will leave the proof of
g being continuously differentiable to the readers.
Have fun.
Eugene
|
| 888.25 | by the way ... :-) | ZFC::DERAMO | To err is human; to moo, bovine. | Thu Jun 30 1988 19:20 | 11 |
| Re .20, .24
>> Hint: g is continuously differentiable for all C, so you can expect
>> all the directional derivitive to exist. That is why this
>> is the hard part of the problem.
How exactly is differentiability of f defined (I guess I
mean in the general case of f:R^m -> R^n)? Obviously having
the partial derivatives all exist is not enough.
Dan
|
| 888.26 | Definition of Derivitive. | HPSTEK::XIA | | Fri Jul 01 1988 10:05 | 13 |
| re .25
Let h be in R^m.
Let T be a linear operator (map) from R^m-->R^n such that
lim ||(f(x + h) - f(x) - Th)|| / ||h|| exist and is equal to 0.
h->0
Then T is call the derivitive of f at x. Incidentally, derivitive
is not restricted to R^n. You can replace R^m, R^n in the above
definition by any normed vector space.
Eugene
|