Re: Green Rudin Ch. 3 #26 (quoted inside)

In article <aderamey.addw-22EEA7.18420801022008@xxxxxxxxxxxxxxxxxxxxxx>,
The World Wide Wade <aderamey.addw@xxxxxxxxxxx> wrote:
In article <b_qdnWezkcvXJj7anZ2dnUVZ_o6knZ2d@xxxxxxxxxxxxx>,
"NotP" <spam@xxxxxxxx> wrote:

"The World Wide Wade" <aderamey.addw@xxxxxxxxxxx> wrote in message
news:aderamey.addw-C67D68.14011801022008@xxxxxxxxxxxxxxxxxxxxxxxxx
In article <auGdndvZ4eFkGj7anZ2dnUVZ_oGjnZ2d@xxxxxxxxxxxxx>,
"NotP" <spam@xxxxxxxx> wrote:

Rudin asks which of the following integrals is larger for positive
measurable functions f:

int_{0 to 1} f(x) log f(x) dx

or

[int_{0 to 1} f(x) dx]*[int_{0 to 1} log f(x) dx]

It seems to me that the top one is larger, but I can't see how to prove
that.

Hint: f(x)log f(x) = (g o f)(x).

Perfect hint! I tried everything BUT that.

There is a slight problem here. One of these integrals may not exist.
Which one?

Actually, none need to exist, but if we define

f(x) = 1/{x log(x)^2} for x in (0,1/2)

= exp(1/{x-1}) for x in (1/2,1)

= 1 for x in { 0, 1/2, 1 }

Then f is integrable on [0,1], but neither f log(f) nor log(f) is.

As to your previous hint; it appears that you might be thinking of
applying Jensen's inequality in some way. I gave a different proof,
but I am curious as to the proof at which you were hinting. I did
not see a simple way to apply Jensen's inequality to this problem.

Rob Johnson <rob@xxxxxxxxxxxxxx>
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Relevant Pages

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