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I encountered the next problem and I'm having problems solving it. It says:

Let $(X,\mathcal{S},\mu)$ be a measure space and let $\{A_n\}$ be a sequence of measurable sets satisfying $0< \mu(A_n)<\infty$ for each $n$ and $\lim \mu(A_n)=0$. Fix $1<p<\infty$ and let $g_n=[\mu(A_n)]^{-\frac{1}{q}}\chi_{A_n}$ where $\frac{1}{p}+\frac{1}{q}=1$ and $\chi_E$ denotes the characteristic function of the set $E$. Show that $\lim \int fg_n\, d\mu=0$ for each $f\in L_p(\mu)$.

At a first glance, I thought it was just an easy application of Holder's inequality applied to the function $fg_n$ but then I realized that it only gives $\int |fg_n|\, d\mu\leq \Vert f\Vert _p$ since $\Vert g_n\Vert_q=1$ for each $n$ and that doesn't help very much.

I tried to use the classic argument of simple functions: showing that the result is valid for that kind functions, extend it to non-negative measurable functions and finally for every integrable function. It is pretty easy to see that $\lim \int \phi g_n\, d\mu=0$ if $\phi$ is simple and integrable. However, when I try to prove this for a nonnegative measurable function, I arrive to a double limit of the kind $$\lim\limits_{n\to\infty}\lim\limits_{m\to\infty}\int \phi_m g_n\, d\mu$$ (where $\{\phi_m\}$ is a sequence of simple functions converging to $f$ monotonically) and I can't justify the change in the order of the limits. Also, I don't think this would be the way to go since I haven't use that $f\in L_p(\mu)$.

Could you give some hints to solve the problem? Thanks in advance.

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  • $\begingroup$ Is that $[\mu(A_n)]^{1-\frac{1}{q}}$ or $[\mu(A_n)]^{1-\frac{1}{p}}$? $\endgroup$ – user99914 Mar 17 '14 at 8:18
  • $\begingroup$ It is a $q$, the conjugate exponent of $p$. $\endgroup$ – Brandon Mar 17 '14 at 8:28
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    $\begingroup$ $(\,\int |g_n|^q \,)^{1/q} = [\mu (A_n)]^{1-1/q} (\,\int_{A_n} 1^q\,)^{1/q} = [\mu (A_n)]^{1-1/q} (\mu(A_n))^{1/q}=\mu(A_n)$. $\endgroup$ – David Mitra Mar 17 '14 at 8:40
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  • If $g_n$ is defined as $\mu(A_n)^{1-1/q}\chi_{A_n}$, then we only need Hölder's inequality, since $\lVert g_n\rVert_q=\mu(A_n)$.

  • It's more subtle if $g_n:=\mu(A_n)^{-1/q}\chi_{A_n}$. In this case, $\lVert g_n\rVert_q=1$. Using an approximation argument, we are reduced to show the result when $f=\chi_B$ for some measurable set $B$ of finite measure. Then $$0\leqslant\int fg\mathrm d\mu=\mu(B\cap A_n)\mu(A_n)^{-1/q}\leqslant \mu(B\cap A_n)^{1-1/q}\leqslant \mu(A_n)^{1/p}.$$

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  • $\begingroup$ I made a mistake, it was $\mu(A_n)^{-\frac{1}{q}}$. But exactly that's the point where I'm stuck, because I can't show that the statement is true just because it is true for simple functions. $\endgroup$ – Brandon Mar 17 '14 at 14:32
  • $\begingroup$ If $f\in L^p$ and $\delta$ is fixed, take $f'$ simple such that $||f-f'||<\delta$. Then $|\int fg_n|\lt \delta +|\int f'g_n|$ and taking the limit, $\limsup_n|\int fg_n|\leq \delta$. $\endgroup$ – Davide Giraudo Mar 17 '14 at 14:34
  • $\begingroup$ Oh, I got it. I forgot that I could use simple functions in that way. Thank you $\endgroup$ – Brandon Mar 17 '14 at 14:51
  • $\begingroup$ You are welcome. $\endgroup$ – Davide Giraudo Mar 17 '14 at 15:01

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