3
$\begingroup$

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.

$\endgroup$
0
0
$\begingroup$

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 $$\int_0^1 f_n \, 1 dx = 1 \not\to 0.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.