Re: Range of the averages convex?
- From: "TCL" <tlim1@xxxxxxxxxxx>
- Date: Mon, 09 Oct 2006 12:17:22 GMT
"The World Wide Wade" <waderameyxiii@xxxxxxxxxxxxxxxxxxxx> wrote in message
news:waderameyxiii-22DD27.13150008102006@xxxxxxxxxxxxxxxxxxxxxxxxxxx
In article <iY9Wg.1802$ji5.603@trnddc04>,
"johnson" <johnson246@xxxxxxxxxxx> wrote:
"johnson" <johnson246@xxxxxxxxxxx> wrote in message
news:hq9Wg.2204$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.
If f is real valued, let a = ess inf f, b = ess sup f. Let eps >
0. By the Lebesgue differentiation theorem, we can find intervals
I and J, of equal length h, such that 1/m(I)*int_I f < a + eps,
1/m(J)*int_J f > b - eps. The function x -> (1/h)int_(x,x+h) f is
continuous, hence its range is an interval, which must contain
[a+eps, b-eps]. This shows S contains (a,b), and since S is
contained in [a,b], S is an interval, hence is convex.
What if f is complex valued? The above argument does not seem to work.
.
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