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Suppose $\mu (X) = 1$ & $||f||_{r} < \infty$ for some $r>0$ . Show that: $lim_{p \to 0} ||f||_{p} =$ $exp. [\int_{X} {log|f|} d\mu ]$ .

Now, there are arising lot of questions:

1) How the existence of limit is assured??

2) After assuming the existence of limit is assured, to prove the equality.

My Thoughts:

1) $||f||_{p} ^{p} = \int|f|^{p} d\mu = \int(|f|^{r})^{p/r} d\mu \le [\int |f|^{r}d\mu]^{p/r} < \infty$ for $0<p<r$ ; by Jensen's Inequality.

But how does it imply the limit exist??

2) to prove the equality, again by Jensen's Inequality ,

$log ||f||_{p} = log [ (\int_{X}|f|^{p} d\mu)^{1/p}]=\frac{1}{p}log[ \int_{X}|f|^{p} d\mu] \ge \frac{1}{p} \int_{X}(log |f|^{p}) d\mu = \int_{X} log|f| d\mu$ .But then???

Please help in this regard...Thank You!!

user92360
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    See Ayman Hourieh's answer in (http://math.stackexchange.com/questions/282271/scaled-lp-norm-and-geometric-mean). – user78270 Apr 20 '14 at 14:11
  • But can you please explain... how D.C.T was used in the last step?? Because to use it the sequence of function : $\frac{f^{1/n}-1}{1/n}$ is to be bounded by a dominating function.... & I cannot see such a dominating function... so, please help! – user92360 Apr 21 '14 at 08:27
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    @user92360 I edited my answer to clarify this. In your question, you have the assumption that $|f|_r < \infty$ for some $r$. This gives you the dominating function. – Ayman Hourieh Apr 21 '14 at 11:10
  • If I am not right... please feel free to rectify, what I have understood is : for large enough $n$ , $\frac {1}{n} \le q$ .... but in that case the denominator will have opposite inequality sign... isn't it?? .. – user92360 Apr 21 '14 at 12:41

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