Let $(f_n)_{n=1}^\infty$ be a sequence in $L^p ([0,1]), 1\leq p<\infty$. Suppose that $f_n\rightarrow f$ in measure and that $\sup\limits_{n\in\mathbb{N}} \|f_n\|<\infty$. Show that $f_n\rightarrow f$ in the weak topology of $L^p([0,1])$.

As a reminder: a net $\{g_\alpha\}_{\alpha\in I}$ converges in the weak topology of $X$ if and only if $\phi(g_\alpha)\rightarrow \phi(g)$ for all $\phi\in X^*$.

We know convergence in measure and boundness in the $L^p$ norm gives us convergence in $L^p$. I think I could use the Dominated Convergence Theorem but I'm not sure how to proceed. I also found a proof of this Theorem but with the hypothesis $f_n\rightarrow f$ a.e., and they use Egorov's theorem, but I rather not use it.

I thank any suggestion you have.


1 Answer 1


1) For $1< p< \infty$. Let $M = \sup \|f_n\|_{L^p}$. Since $f_n \to f$ in measure, hence there is a subsequence $f_{n_k}$ which converges a.e. to $f$. By Fatou's lemma, we have $f \in L^p$ and $\|f\|_{L^p} \leq M$. We need to check $$\lim_{n\to \infty} \int_0^1 f_n g dx = \int_0^1 f g dx,\qquad\qquad (*)$$ for any $g \in L^{p'}$ with $p' = p/(p-1)$. Indeed, for any $\epsilon >0$, we have $\lim_{n\to \infty} m(\{|f_n -f| > \epsilon\}) = 0$ and $$|\int_0^1 (f_n -f)g dx| = |\int_{\{|f_n -f| >\epsilon\}} (f_n -f) g dx + \int_{\{|f_n -f| \leq \epsilon\}} (f_n -f) g dx,$$ and hence $$|\int_0^1 (f_n -f)g dx| \leq \int_{\{|f_n -f| >\epsilon\}} |f_n -f| |g| dx + \epsilon \int_{\{|f_n -f| \leq \epsilon\}} |g| dx.$$ By Holder inequality, we have $$\int_{\{|f_n -f| \leq \epsilon\}} |g| dx \leq \int_0^1 |g| dx \leq \|g\|_{L^{p'}},$$ and $$\int_{\{|f_n -f| >\epsilon\}} |f_n -f| |g| dx \leq (\int_{\{|f_n -f| >\epsilon\}} |f_n -f|^p dx)^{\frac1p}(\int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx)^{\frac1{p'}} \leq 2M (\int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx)^{\frac1{p'}}.$$ Therefore, it holds $$|\int_0^1 (f_n -f)g dx| \leq 2M (\int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx)^{\frac1{p'}} + \epsilon \|g\|_{L^{p'}}.\qquad\qquad (1)$$ We claim that $$\lim_{n\to \infty} \int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx = 0. \qquad\qquad (**)$$ Indeed, for any $\delta >0$, there exists $A >0$ such that $\int_{\{|g| > A\}} |g|^{p'} dx < \delta$. Hence $$\int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx \leq \int_{\{|f_n -f| >\epsilon\}\cap\{|g| > A\}} |g|^{p'} dx + \int_{\{|f_n -f| >\epsilon\}\cap\{|g| \leq A\}} |g|^{p'} dx \leq \delta + A^{p'} m(\{|f_n -f| > \epsilon\}).$$ Letting $n \to \infty$ and using $m(\{|f_n -f| > \epsilon\}) \to 0$, we get $$\limsup_{n\to \infty} \int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx \leq \delta,$$ for any $\delta >0$. Consequently, we get $$\limsup_{n\to \infty} \int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx \leq 0.$$ Evidently, we have $\liminf_{n\to \infty} \int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx \geq 0$. Hence, the claim $(**)$ holds. Letting $n \to \infty$ and using the claim $(**)$, we obtain $$\limsup_{n\to \infty} |\int_0^1 (f_n -f)g dx| \leq \epsilon \|g\|_{L^{p'}},$$ for any $\epsilon >0$. Therefore, it holds $$\limsup_{n\to \infty} |\int_0^1 (f_n -f)g dx| \leq 0.$$ Evidently, we have $\liminf_{n\to \infty} |\int_0^1 (f_n -f)g dx| \geq 0$. Hence, $$\lim_{n\to \infty} \int_0^1 (f_n -f) gdx =0,$$ which proves $(*)$.

2) For $p =1$, the conclusion does not holds. For example, consider $f_n = n \chi_{(0,1/n)}$, we have $\|f_n\|_{L^1} =1$ for any $n$ and $f_n \to 0$ in measure. However, we have $1 \in L^\infty = (L^1)^*$ and $$\int_0^1 f_n \, 1 dx = 1 \not\to 0.$$


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