jensen for a derivative
Li Yi
f( (x+y)/ 2 ) <= ( f(x) + f(y) ) / 2 (*)
Show that f is continuous.
======================================
I know some similar statements. (assume that I is an open interval)
(1) If g(x+), g(x-) exist for all x, and g satisfies (*) then it is
continuous
(2) If g is bounded and satisfies (*), then it is continous.
And I know derivative has the following two properties:
(1) It doesn't have a jump discontinuity;
(2) It has the intermediate value property.
But I don't know how to do the problem. Maybe we can try to show that
f(x+),f(x-) exist? Or something else?
José Carlos Santos
> Let f be a derivative of a function on interval I, and f satisfies
> f( (x+y)/ 2 ) <= ( f(x) + f(y) ) / 2 (*)
>
> Show that f is continuous.
The assertion
f((x + y)/2) <= (f(x) + f(y))/2 for all x and all y in I
is equivalent to the assertion "f is convex"; this follows from the
fact that f is a derivative and therefore that it has the intermediate
value property (i. e. if x and y belong to I and if c is somewhere
betwenn f(x) and f(y), then there's some z between x and y such that
f(z) = c).
Suppose that you want to prove that f is continuous at some point _a_
of I. Take some b > a. You can assume that f(b) = f(a) (if it isn't,
then replace f(x) by f(x) - c.x for some appropriate c). If f were not
right-continuous at _a_, there would be two possibilities:
1) lim_{x -> a^+} f(x) exists and it is different from f(a); this is
impossible, by the intermediate value property;
2) lim_{x -> a^+} f(x) does not exist. This is equivalent to the
assertion:
(*) there's some r > 0 such that for each d > 0 there are elements x
and y of I such that |x - a|,|y - a| < d and that |f(x) - f(y)| >= r.
Now, since f(a) = f(b) and since f is convex, you have f(c) <= f(a) for
each c in [a,b]. It follows from this fact and from (*) that there's
some decreasing sequence (a_n)_n of elements of [a,b] such that lim_n
a_n = a and that f(a_n) <= f(a) - r, for each natural n. Using again the
convexity of f, it follows that f(x) <= f(a) - r for each x in ]a,a_1].
This is impossible, by the intermediate value property.
> I know some similar statements. (assume that I is an open interval)
If you assume that, your problem becomes easy and you don't need the
hypothesis that f a derivative; see theorem 3.2 in Rudin's "Real and
Complex Analysis" or problem 11 from the appendix to chapter 11 of
Spivak's "Calculus" (3rd edition).
Best regards,
Jose Carlos Santos
Li Yi
José Carlos Santos wrote:
> The assertion
>
> f((x + y)/2) <= (f(x) + f(y))/2 for all x and all y in I
>
> is equivalent to the assertion "f is convex"; this follows from the
> fact that f is a derivative and therefore that it has the intermediate
> value property (i. e. if x and y belong to I and if c is somewhere
> betwenn f(x) and f(y), then there's some z between x and y such that
> f(z) = c).
I don't follow that why f is convex.
What role does intermediate value property play here?
José Carlos Santos
Oops, I guess that I forgot something here! :-(
In what follows, I will "forget" to write "for all x and y in I". It is
easy to deduce from
f((x + y)/2) <= (f(x) + f(y))/2
that
(1) for all t in [0,1] of the forma p/2^n (with p non-negative integer),
f(tx + (1 - t)y) <= tf(x) + (1 - t)f(y).
To say that f is convex is to say that
(2) for all t in [0,1], f(tx + (1 - t)y) <= tf(x) + (1 - t)f(y).
It's clear that (2) => (1), but you want to prove is that (1) => (2). Of
course, the points of [0,1] that can be written as p/2^n with p a
non-negative integer form a dense subset of [0,1], but, in general, this
is not enough to deduce that (2) => (1). I was under the illusion that
the intermediate value property would be enough to establish the result,
but right now I am unable even to deduce it from the fact that f has a
primitive.
I hope that this helps. Sorry for wasting your time.
David C. Ullrich
<jcsa...@fc.up.pt> wrote:
>On 25-07-2005 1:51, Li Yi wrote:
>
>>>The assertion
>>>
>>> f((x + y)/2) <= (f(x) + f(y))/2 for all x and all y in I
>>>
>>>is equivalent to the assertion "f is convex"; this follows from the
>>>fact that f is a derivative and therefore that it has the intermediate
>>>value property (i. e. if x and y belong to I and if c is somewhere
>>>betwenn f(x) and f(y), then there's some z between x and y such that
>>>f(z) = c).
>>
>>
>> I don't follow that why f is convex.
>> What role does intermediate value property play here?
>
>Oops, I guess that I forgot something here! :-(
Yeah, I was wondering about that as well - imagine my
disappointment at your reply<g>.
(It seems "clear" to me that the result must be true, because
a discontinuous "midpoint-convex" function must be _very_
discontinuous, and a derivative can't be all that discontinuous.
But that's just a little too vague to count as a proof...)
>[...]I was under the illusion that
>the intermediate value property would be enough to establish the result,
>but right now I am unable even to deduce it from the fact that f has a
>primitive.
>
>I hope that this helps. Sorry for wasting your time.
>
>Best regards,
>
>Jose Carlos Santos
************************
David C. Ullrich
Robert Israel
Li Yi <liy...@gmail.com> wrote:
>Let f be a derivative of a function on interval I, and f satisfies
>f( (x+y)/ 2 ) <= ( f(x) + f(y) ) / 2 (*)
>Show that f is continuous.
I'll assume I is open.
Any derivative is the pointwise limit of continuous functions, and
therefore is measurable.
Let B_n = {x in I: f(x) < n}. Then I = union {B_n: n=1,2,...}
so some B_n has positive measure. But then
(B_n + B_n)/2 = {(x+y)/2: x, y in B_n} contains a nonempty open
interval, and by (*) so does B_n. Now if (a,b) is contained in
B_n and b < c < 2b-a with c in I, then [b,(b+c)/2) is contained
in B_{(n+f(c))/2}. Repeating this, we find that f is bounded
above on every compact subinterval of I.
Now we can rewrite (*) as f(x-d) >= 2 f(x) - f(x+d), and then
by induction f(x-kd) >= f(x) + k(f(x)-f(x+d)) for all positive
integers k for which x - kd is in I.
Similarly, f(x+kd) >= f(x) + k(f(x+d)-f(x)) if x + kd is in I.
If x+kd and x-kd are in B_n, these inequalities imply
(f(x)-n)/k <= f(x+d)-f(x) <= (n-f(x))/k.
Given x in I and epsilon > 0, x is in some open interval J contained
in some B_n. Take k large enough that 2n/k < epsilon, and delta
small enough that x + k delta and x - k delta are in J, and
then for |d| < delta we have |f(x+d)-f(x)| < epsilon.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada