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If $\mu(X)<\infty$, $f_n\rightarrow0$ in measure, and $\limsup\limits_{n \rightarrow \infty} \int (f_n)^2 < 1$, then $f_n \rightarrow f$ in $L^r$ for all $r \in [1,2)$.

Attempt: We want to show that $\int |f_n|^r \rightarrow 0$. Now, note that $\int |f_n|^r = \int_{|f_n| > 1} |f_n|^r + \int_{\epsilon < |f_n| \leq 1} |f_n|^r + \int_{|f_n| \leq \epsilon} |f_n|^r$. Let's bound each term:

(i) Here, we note that since $\limsup_{n \rightarrow \infty} \int |f_n|^2 < \infty $, then for all $n$ sufficiently large, $\sup_{n \geq N} \int |f_n|^2 < \infty$, so that $(f_n)^{1+\delta}_{n \geq N}$ are uniformly integrable for all $\delta \in [0,1)$, hence uniformly absolutely continuous (by this theorem). But then, since convergence in measure implies that $\mu(|f_n| > 1) \rightarrow 0$, we have that $\int_{|f_n| > 1} f^r \rightarrow 0$ by the uniformly absolutely continuous condition, whenever $r \in [1,2)$.

(ii) $\int_{\epsilon < |f_n| \leq 1} |f_n|^r \leq \mu(|f_n| > \epsilon) \rightarrow 0$ as $n \rightarrow \infty$.

(iii) $\int_{|f_n| \leq \epsilon} |f_n|^r \leq \mu(X) \epsilon^r$.

And since $\epsilon$ is arbitrary, we are done.

Doubts: Is the above argument valid? Is there an alternate method which doesn't involve uniform integrability? In particular, I'm unsure because the argument above works for $\limsup_n \int f_n^2 < \infty$.

Mittens
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algebroo
  • 1,406

2 Answers2

2

Here is a perhaps more elementary argument using Hölder's inequality. Let $r\in{[1,2)}$. We take $p=\frac2{2-r}>1$ and $q=\frac2r>1$ (so indeed $\frac1p+\frac1q=1$). Then, for all $\varepsilon>0$ and $n\ge1$, \begin{align*} \int|f_n-f|^r\,\mathrm d\mu&=\int\mathbf1_{\{|f_n-f|\le\varepsilon\}}|f_n-f|^r\,\mathrm d\mu+\int\mathbf1_{\{|f_n-f|>\varepsilon\}}|f_n-f|^r\,\mathrm d\mu\\[.4em] &\le\varepsilon^r\mu(X)+\mu(|f_n-f|>\varepsilon)^{\frac1p}{\left(\int|f_n-f|^{qr}\,\mathrm d\mu\right)}^{\!\frac1q}\\[.4em] &=\varepsilon^r\mu(X)+\mu(|f_n-f|>\varepsilon)^{\frac1p}{\left(\int|f_n-f|^2\,\mathrm d\mu\right)}^{\!\frac r2}, \end{align*} where $\int|f_n-f|^2\,\mathrm d\mu\le2(\int|f|^2\,\mathrm d\mu+\int|f_n|^2\,\mathrm d\mu)\le2(\int|f|^2\,\mathrm d\mu+2)$ eventually (for $n\ge1$ sufficiently large), because $\limsup_{n\to\infty}\int|f_n|^2\,\mathrm d\mu<1$. Since $f_n\to f$ in measure, it follows that $$\limsup_{n\to\infty}\int|f_n-f|^r\,\mathrm d\mu\le\varepsilon^r\mu(X),$$ and it suffices to let $\varepsilon\to0$ to conclude.

nejimban
  • 4,107
2

The statement in the OP holds for in fact $0<r<2$.

Showing that $\|f_n\|_r\xrightarrow{n\rightarrow\infty}0$ for all $0<r<2$ can be shown by splitting the integral in a similar fashion as the OP's and then application of Hölder's inequality along with the assumptions in the OP. To be more precise,

  • The assumption that $\limsup_n\|f_n\|_2<1$ implies that $M=\sup_n\|f_n\|_2<\infty$.

  • If $0<r<2$, $f\in L_r$ implies that $L^r\in L_{2/r}$. Then $$\int_{\{|f_n|>\varepsilon\}} |f_n|^r\,d\mu\leq\|f_n\|^r_2\big(\mu(|f_n|>\varepsilon\big)^{\tfrac{2-r}{2}}\leq M^r\big(\mu(|f_n|>\varepsilon\big)^{\tfrac{2-r}{2}}\xrightarrow{n\rightarrow\infty}0$$

  • On the other hand, $$\int_{\{|f_n|\leq\varepsilon\}}|f_n|^r\,d\mu\leq \varepsilon^r\mu(X)$$

  • Putting things together, we obtain that $\lim_n\|f_n\|_r=0$.

Mittens
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