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Limit of $L^p$ norm

On the $L_p$ spaces, when is $$\lim_{p\to \infty}\| f\|_{p}=\| f\|_{\infty}$$ true? Or what condition is necessary for this?


marked as duplicate by Davide Giraudo, Henry T. Horton, Stefan Hansen, Asaf Karagila, rschwieb Feb 5 '13 at 15:01

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  • $\begingroup$ This is always true, however, if the space is not $\sigma$-finite you need a funky definition of $||\cdot ||_\infty$ $\endgroup$ – user45150 Feb 5 '13 at 4:02
  • $\begingroup$ @user45150: Why is that so? What is wrong in the non $\sigma$-finite case? $\endgroup$ – Jack Mar 26 '17 at 19:26

It's a relatively easy proof that this holds in $L_p(X)$ for $f \in L^\infty(X)$ iff $\mu (X) < \infty$. The following was an exercise in Rudin's Real and Complex Analysis:

Suppose $\mu$ is a positive measure on $X, \mu(X) < \infty, f \in L^\infty(\mu),||f||_\infty > 0, \text{and}$ $$a_n = \int_X|f|^n\,d\mu~~~~~(n=1,2,3,...).$$ prove that$$\lim_{n\to \infty}\frac{a_{n+1}}{a_n} = ||f||_\infty$$

That you might want to try to develop a further understanding.

  • $\begingroup$ I can see how $$\lim_{n\to \infty}\frac{a_{n+1}}{a_n} \leq ||f||_\infty$$, but how do you show the other direction? $\endgroup$ – PrimeRibeyeDeal Jul 22 '13 at 22:46
  • $\begingroup$ I could not see it, how this limit is equal to $\Vert f \Vert_\infty$, thanks $\endgroup$ – seriously divergent Jan 26 '14 at 9:56
  • $\begingroup$ I have no idea where is the part $\mu(X)<\infty$ from. $\endgroup$ – Jack Mar 26 '17 at 19:21

If $\|f\|_\infty<\infty$, then for the property to hold it is necessary and sufficient that there exists $p<\infty$ such that $\|f\|_p<\infty$. The only way for this to always hold is for the measure space to have finite total measure.

If $\|f\|_\infty=\infty$, then it holds regardless.


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