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Dec 28, 2002, 12:46:19 AM12/28/02

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Let f: (A,B) \subset R --> R. Define f to be _convex_ when, for any

closed subinterval [a,b] of (A,B) and for every point x in [a,b],

closed subinterval [a,b] of (A,B) and for every point x in [a,b],

(x - a) (f(b) - f(a)) <= (b - a) (f(x) - f(a)).

I would like to see a proof or counterexample to the following:

If f is convex on (A,B), then f is continuous on (A,B).

This is *not* true if the domain is not open. (A counterexample is

the sign function on [0,1].)

Thanks

Geoffrey

--

"You have the right to remain silent. Anything you say on Usenet will

be forged, taken out of context, misquoted, and used against you."

Geoffrey T. Falk, BSc MA; SCJ2P, SCSadm7, FreeBSD <gtf(@)cirp.org>

Dec 28, 2002, 2:15:47 AM12/28/02

to

In article <LUaP9.177093$Qr.45...@news3.calgary.shaw.ca>

gtfN...@NOSPAM.cirp.org (Geoffrey T. Falk) writes:

>Let f: (A,B) \subset R --> R. Define f to be _convex_ when, for any

>closed subinterval [a,b] of (A,B) and for every point x in [a,b],

>

> (x - a) (f(b) - f(a)) <= (b - a) (f(x) - f(a)).

>

>I would like to see a proof or counterexample to the following:

>If f is convex on (A,B), then f is continuous on (A,B).

>

>This is *not* true if the domain is not open. (A counterexample is

>the sign function on [0,1].)

>

It is easy to transform this definition into the standard definition>Let f: (A,B) \subset R --> R. Define f to be _convex_ when, for any

>closed subinterval [a,b] of (A,B) and for every point x in [a,b],

>

> (x - a) (f(b) - f(a)) <= (b - a) (f(x) - f(a)).

>

>I would like to see a proof or counterexample to the following:

>If f is convex on (A,B), then f is continuous on (A,B).

>

>This is *not* true if the domain is not open. (A counterexample is

>the sign function on [0,1].)

>

f(alpha*a + (1-alpha)*b) >= alpha*f(a) + (1 - alpha)*f(b)

when a, b in (A, B) and 0 <= alpha <= 1. Geometrically,

if (a, f(a)) and (b, f(b)) are two points on the graph of f,

then (x, f(x)) is on or above the line segment passing through

(a, f(a)) and (b, f(b)) whenever x is between a and b.

Look for a function with graph resembling (in fixed font)

--------------------

/

/

/

with a discontinuity where the diagonal line ends

and the horizontal line begins.

--

A local drug store selling wine boasts a drug and alcohol free workplace.

A local grocery store advertises Hot Buys even on frozen foods.

Peter-Lawren...@cwi.nl Home: San Rafael, California

Microsoft Research and CWI

Dec 28, 2002, 3:16:33 AM12/28/02

to

In article <LUaP9.177093$Qr.45...@news3.calgary.shaw.ca>,

gtfN...@NOSPAM.cirp.org (Geoffrey T. Falk) wrote:

gtfN...@NOSPAM.cirp.org (Geoffrey T. Falk) wrote:

> I would like to see a proof or counterexample to the following:

> If f is convex on (A,B), then f is continuous on (A,B).

Hint: Given b in (A,B), choose a and c in (A, B) with a < b < c. Now

consider the line through (a,f(a)) and (b,f(b)) and the line through

(b,f(b)) and (c,f(c)). To the right of b, f has to stay on or above the

first line and on or below the second line. To the left of b, an analogous

statement holds.

Dec 28, 2002, 3:25:43 AM12/28/02

to

>

>

>>Let f: (A,B) \subset R --> R. Define f to be _convex_ when, for any

>>closed subinterval [a,b] of (A,B) and for every point x in [a,b],

>>

>> (x - a) (f(b) - f(a)) <= (b - a) (f(x) - f(a)).

>>

>>I would like to see a proof or counterexample to the following:

>>If f is convex on (A,B), then f is continuous on (A,B).

>>

>>This is *not* true if the domain is not open. (A counterexample is

>>the sign function on [0,1].)

>>

>

>

> It is easy to transform this definition into the standard definition

>

> f(alpha*a + (1-alpha)*b) >= alpha*f(a) + (1 - alpha)*f(b)

>

>when a, b in (A, B) and 0 <= alpha <= 1. Geometrically,

>if (a, f(a)) and (b, f(b)) are two points on the graph of f,

>then (x, f(x)) is on or above the line segment passing through

>(a, f(a)) and (b, f(b)) whenever x is between a and b.

>

Actually, this is the definition of a *concave* function. (Here we go >

>>Let f: (A,B) \subset R --> R. Define f to be _convex_ when, for any

>>closed subinterval [a,b] of (A,B) and for every point x in [a,b],

>>

>> (x - a) (f(b) - f(a)) <= (b - a) (f(x) - f(a)).

>>

>>I would like to see a proof or counterexample to the following:

>>If f is convex on (A,B), then f is continuous on (A,B).

>>

>>This is *not* true if the domain is not open. (A counterexample is

>>the sign function on [0,1].)

>>

>

>

> It is easy to transform this definition into the standard definition

>

> f(alpha*a + (1-alpha)*b) >= alpha*f(a) + (1 - alpha)*f(b)

>

>when a, b in (A, B) and 0 <= alpha <= 1. Geometrically,

>if (a, f(a)) and (b, f(b)) are two points on the graph of f,

>then (x, f(x)) is on or above the line segment passing through

>(a, f(a)) and (b, f(b)) whenever x is between a and b.

>

again...)

No matter: both concave and convex functions on open sets are

continuous. (Indeed, one of f or -f is convex iff the other is

concave.)

--

Stephen J. Herschkorn hers...@rutcor.rutgers.edu

Dec 28, 2002, 8:29:30 AM12/28/02

to

Geoffrey T. Falk <gtfN...@NOSPAM.cirp.org>

[sci.math Dec 28 2002 12:50:29:000AM]

http://mathforum.org/discuss/sci.math/m/468956/468956

[sci.math Dec 28 2002 12:50:29:000AM]

http://mathforum.org/discuss/sci.math/m/468956/468956

wrote (in part):

> I would like to see a proof or counterexample to the following:

> If f is convex on (A,B), then f is continuous on (A,B).

I'm sure that others will answer your specific question. Below

is part of an old post of mine that gives a summary of a few key

results about convex functions in case you or others are interested.

Dave L. Renfro

------------------------------------------------------------------

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http://mathforum.org/epigone/ap-calc/khendgeequer

http://mathforum.org/epigone/ap-calc/stelzonzhel

Subject: [ap-calculus] Convex Functions

Author: Dave L. Renfro <dlre...@gateway.net>

Date: Thu, 14 Jun 2001 20:38:52 -0500

DEFINITION: Let f be defined on an interval I. We say that f

is convex on I if whenever x1, x2 belong to I, then

the line segment whose endpoints are (x1,f(x1)) and

(x2,f(x2)) lies on or above {(x,f(x)): x in [x1, x2]}.

If "on or above" is strengthened to "strictly above", we get a

geometric condition for concave up. Many of the results given

below continue to hold when f is concave up. Some of these will

be automatic (e.g. when the hypothesis includes "convex", since

concave up implies convex) and some of these will continue to hold

for other reasons. However, I don't really have the time or desire

right now to try and sort out which continue to hold when "convex"

is replaced with "concave up". <snip>

Here is another characterization of convex functions.

THEOREM : A function f is convex on an interval I if and only if

the following condition holds:

Whenever x1 < x2 < x3 belong to I, then

[f(x2) - f(x1)] / (x2 - x1) < or = [f(x3) - f(x2)] / (x3 - x2).

In other words, the average rate of change of f on

[x1, x2] does not exceed the average rate of change

of f on [x2, x3].

---------- SOME RESULTS ----------

Convex functions have many applications, both in pure mathematics

and in applied mathematics -- Many useful inequalities, including

the arithmetic and geometric mean inequality, can be obtained very

easily using convex functions. In recent years there has been a lot

of interest in convex functions defined on Banach spaces, especially

in their differentiability properties. Finally, convex functions play

a crucial role in linear programming and in optimization theory. For

more information, go to <http://www.google.com/> and search using

"differentiability +of convex functions" (quotes included),

"convex functions optimization" (quotes NOT included),

and "convex functions linear programming" (quotes NOT included).

1. If f is convex on an open interval I, then f is continuous at

each point in I.

2. If f is convex on an open interval I, then f is Lipschitz

continuous on each closed subinterval of I. [Lipschitz continuous

means the difference quotients are bounded.] This strengthens #1.

3. If f is convex on an open interval I, then there are at most

countably many points at which f is not differentiable.

Any countable set can be the set of non-differentiability points

for some convex function. Curiously, I couldn't find this in any

of the real analysis texts I looked at. However, this is a special

case of theorem 4.20 on p. 93 of [1].

4. Assume that f is convex on an open interval I. Then, at each point

of I, both the left derivative of f and the right derivative of

f exist. This strengthens #3. [It can be shown that this property

implies the property given in #3, but not conversely.]

5. If f is convex on an open interval I, then the left derivative

of f is a non-decreasing function, the right derivative of f

is a non-decreasing function, and at each point the left

derivative is less than or equal to the right derivative.

(See [7], p. 109.) This strengthens #4.

6. If f is convex on an open interval I, then the second derivative

of f exists at every point of I except for a set of measure zero.

[This follows from #3, #5, and the fact that monotone functions

are differentiable almost everywhere.]

7. If f is convex on an open interval I, and g is either the left

derivative of f or the right derivative of f (it doesn't matter

which one you let g be), then

f(b) - f(a) = integral of g on the interval [a,b]

for all a,b in I. [Monotone functions are Riemann-integrable,

so this is the usual calculus integral.]

8. Suppose f'' exists at each point of an open interval I. Then f is

convex on I if and only if f''(x) is nonnegative for each x in I.

9. Suppose f is continuous. Then f is convex on an open

interval I if and only if

limit as h --> 0 of [ f(x+h) + f(x-h) - 2*f(x) ] / (h^2)

is nonnegative for each x in I. [This strengthens #8, since

the existence of f'' at a point implies the limit above exists

at that point, and the converse fails.]

Riemann introduced and used this "second order symmetric

derivative" in an 1854 memoir on trigonometric series. It

was this memoir, incidentally, that Riemann introduced what

we now call the Riemann integral. LaTeX, .dvi, .ps, and .pdf

files of Riemann's 1854 memoir are available at

<http://www.maths.tcd.ie/pub/HistMath/People/Riemann/Trig/>.

10. Suppose f' exists at each point of an open interval I. Then f is

convex on I if and only if f' is non-decreasing on I. This

strengthens #8 and neither implies nor is implied by #9.

11. If h is non-decreasing on an open interval I and 'a' belongs

to I, then the function f defined on I by

f(x) = integral of h on the interval [a,x]

is convex on I. [This refines a result that arises by putting

#5 and #8 together.]

---------- SOME REFERENCES ----------

[1] Yoav Benyamini and Joram Lindenstrauss, "Geometric Nonlinear

Functional Analysis", Volume 1, Colloquium Publications #48,

American Mathematical Society, 2000. [chapter 4: "Differentiation

of Convex Functions", pp. 83-98]

[2] Ralph P. Boas, "A Primer of Real Functions", 4'th edition

(revised and updated by Harold P. Boas), Carus Mathematical

Monographs 13, Mathematical Association of America, 1996.

[pages 175-186]

[3] Andrew M. Bruckner, "Differentiation of Real Functions",

CRM Monograph Series #5, American Mathematical Society, 1994.

[pages 131-134 (advanced)]

[4] Krishna M. Garg, "Theory of Differentiation", Canadian

Mathematical Society Series of Monographs and Advanced Texts

#24, John Wiley and Sons, 1998. [pages 195-198 (very advanced)]

[5] R. Kannan and Carole King Krueger, "Advanced Analysis on the

Real Line", Springer-Verlag, 1996. [pages 74-76]

[6] A.C.M. van Rooij and W.H. Schikhof, "A Second Course on Real

Functions", Cambridge University Press, 1982. [pages 14-18]

[7] H. L. Royden, "Real Analysis", 2'nd edition, MacMillan, 1968.

[pages 108-110]

[8] Brian S. Thomson, "Symmetric Properties of Real Functions",

Pure and Applied Mathematics #183, Marcel Dekker, 1994.

[pages 202-209 (advanced)]

[9] Richard L. Wheeden and Antoni Zygmund, "Measure and Integral",

Pure and Applied Mathematics #43, Marcel Dekker, 1977.

[pages 118-124]

Dave L. Renfro

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