Re: Range of the averages convex?

On Sun, 08 Oct 2006 16:47:10 GMT, "johnson" <johnson246@xxxxxxxxxxx>
wrote:


"johnson" <johnson246@xxxxxxxxxxx> wrote in message
news:hq9Wg.2204$e65.413@xxxxxxxxxxx

"johnson" <johnson246@xxxxxxxxxxx> wrote in message
news:sm9Wg.2203$e65.1080@xxxxxxxxxxx
Let f be an essentially bounded Lebesgue measurable function on [0,1].
Consider the set
\int_E f d\mu
where \mu is the Lebesgue measure and E are Lebesgue measurable subsets
of [0,1].
Must f be convex?

Correction:

Let f be an essentially bounded Lebesgue measurable function on [0,1].
Consider the set S=
\int_E f d\mu
where \mu is the Lebesgue measure and E are Lebesgue measurable subsets of
[0,1].
Must S be convex?


Sorry. I meant 1/mu(E) times \int_E f d\mu
for E such that mu(E)>0.

After two corrections I still have a question: Are you assuming
that f is real-valued? Seems to me that the default assumption
based on what you wrote would be that f is complex-valued, but
the solutions given are for real-valued f.

I suspect the answer is yes for complex-valued f but the
proof is not quite so simple. I can show that the _closure_
of S is convex:

Say K is the essential range of f; note that K is compact.
Say H is the convex hull of K. Then the closure of S is
equal to H:

The fact that S is contained in H is clear. Now say
z_1, z_2 are elements of K and z = t z_1 + (1-t) z_2
is a convex combination of z_1 and z_2 (the argument
for more than two elements of K is the same, just
harder to type).

Let's say f_E is the average of f on E. Note that
if A and B are disjoint and C is the union of A and
B then

f_C = mu(A)/mu(C) f_A + mu(B)/mu(C) f_B,

a convex combination of f_A and f_B.

Now let eps > 0. There exist A and B such that
|f - z_1| < eps at every point of A and
|f - z_2| < eps at every point of B. Since the
same holds for any subsets of A and B we can
assume that A and B are disjoint.

Now there exists A' contained in A and B' contained
in B such that if C' is the union of A' and B' then
mu(A')/mu(C') = t (and hence mu(B')/mu(C') = 1-t.
It follows that |f_C' - z| < eps. QED.

Unfortunately the set of all functions of the form
chi_E/mu(E) (where chi_E is the characteristic function
of E) is not compact in any relevant topology so I
don't see how to refine this into a proof that S
is actually equal to H (or the interior of H or
whatever). If S has an interior point it's not hard
to show that everything works...


************************

David C. Ullrich
.

Relevant Pages

  • Re: Range of the averages convex?
    ... Must S be convex? ... interior points, making it not convex. ... Now let eps> 0. ... in B such that if C' is the union of A' and B' then ...
    (sci.math)
  • Re: Range of the averages convex?
    ... Must S be convex? ... interior points, making it not convex. ... Now let eps> 0. ... in B such that if C' is the union of A' and B' then ...
    (sci.math)
  • Re: Range of the averages convex?
    ... where \mu is the Lebesgue measure and E are Lebesgue measurable subsets ... Must S be convex? ... David C. Ullrich ...
    (sci.math)
  • Re: Range of the averages convex?
    ... where \mu is the Lebesgue measure and E are Lebesgue measurable subsets ... Must S be convex? ... Now let eps> 0. ... If S has an interior point it's not hard ...
    (sci.math)
  • Re: Range of the averages convex?
    ... Must S be convex? ... interior points, making it not convex. ... Now let eps> 0. ... David C. Ullrich ...
    (sci.math)