# weak convergence and pointwise implies $L_p$ convergence

Suppose $$f_i \to f$$ weakly in $$L^p(X, M, \mu)$$, $$1 < p < \infty$$, and that $$f_i \to f$$ pointwise $$\mu$$-a.e. Prove that $$f_i^+ \to f^+$$ and $$f_i^- \to f^-$$ weakly in $$L^p$$.

My proof:

Since $$f^\pm = |f| \pm f$$ then for a $$g \in L^q(X,\mu)$$ we have

\begin{align} \int (f_i^+ -f^+) g d\mu \leq \frac{1}{2} \int |f_i-f| gd\mu + \frac{1}{2} \int (f_i -f)g d\mu \end{align}

So I need to show that $$f_i \to f$$ in $$L^p$$.

Since $$L^p$$ is reflexive and $$f_i\to f$$ weakly, $$f_i$$ is bounded in $$L^p$$, that is , $$\exists M>0$$ such that $$|f_i| \leq M, \forall i$$, so $$|f|\leq M$$ by Fatou's lemma.

Since $$f^p \in L^1$$, the tail is finite, i.e., there is $$A \subset X$$ such that $$\int_{A^c} |f|^p \leq \epsilon^p$$, and $$\mu(A) <\infty$$. Now, we use Egorofs theorem to obtain a uniform convergence on $$F\subset A$$ and $$\mu(A-F) <\epsilon$$, now we have;

\begin{align*} \int |f_i-f|^p &\leq \int_F |f_i-f|^p + \int_{A-F} |f_i -f|^p + \int_{A^c} |f_i -f|\\ & \epsilon \mu(F) + 2M \mu(A-F) + 2M \mu(A^c)\\ & \lesssim \epsilon \end{align*}

Is boundedness the only reason the weak convergence holds?

• Do we know anything about $X, \mu$? Commented Jul 17 at 15:48
• @HyperbolicPDEfriend No, it is just a measure space. Commented Jul 17 at 15:49
• This seems wrong. You estimated the integrand in $\int_{A-F} \lvert f_i - f \rvert^p$ pointwise, although you only have an $L^p$ bound. So $\mu(A-F)$ does not remain Commented Jul 17 at 16:05
• But then why does $\mu(A-F)$ remain. The bound from Banach-Alaoglu is for the norm, not pointwise. Commented Jul 17 at 16:17
• @Mr.Proof: your idea is correct but the execution is not quite right. You want to show that $\lim_n\int_X g_nh=\int_Xgh$ for all $h\in L_q$ and $g_n=f^{\pm}_n$, $g=g^{\pm}$. I wrote a solution with as many details as possible. Commented Jul 17 at 20:41

In general, the statement in the title of the OP is false. In $$((0,\infty),\mathscr{B}(\mathbb{R}),m)$$, $$m$$ is Lebesgue measure, define $$f_n(x)=\frac1{x-n}\mathbb{1}_{(1,\infty)}(x-n)$$. $$f_n\xrightarrow{n\rightarrow\infty}0$$ pointwise and for any $$g\in L_2$$, \begin{align} \int^\infty_0f_n(x)g(x)\,dx&=\int^\infty_{n+1}\frac{g(x)}{x-n}\,dx=\int^\infty_1\frac{g(x+n)}{x}\,dx\\ &\leq\Big(\int^\infty_0|g(x+n)|^2\,dx\big)^{1/2}=\Big(\int^\infty_{n}|g(x)|^2\,dx\big)^{1/2}\xrightarrow{n\rightarrow\infty}0 \end{align} Yet, as $$\|f_n\|_2=1$$., $$f_n$$ does not converge in $$L_2$$.

The problem described in the body of the OP however, is of a different nature, and the answer to that is in the affirmative.

The assumption $$f_n\xrightarrow{n\rightarrow\infty}f$$ $$\mu$$-a.s. implies that $$f^{\pm}_n\xrightarrow{n\rightarrow\infty}f^{\pm}$$ $$\mu$$-a.s.

As pointed out by the OP, the $$L_p$$-weak convergence of $$f_n$$ implies that $$f_n$$ is $$L_p$$-bounded. Since $$f^{\pm}_n\leq |f_n|$$, it follows that $$(f^{\pm}_n:n\in\mathbb{N})$$ is $$L_p$$-bounded. The conclusion follows from the following result:

Theorem: If $$1, $$(g_n:n\in\mathbb{N})$$ is $$L_p$$-bounded and $$g_n\xrightarrow{n\rightarrow\infty}g$$ $$\mu$$-a.s., then $$g_n\xrightarrow{n\rightarrow\infty}g$$ weakly in $$L_p$$.

This result is discussed in Hewitt., E, and Stromberg., K., Real and Abstract Analysis, GTM, Springer Verlag, Berlin Heidelberg, 1965, pp. 207.

Here is a proof using similar arguments as those provided in the OP.

Let $$M=\sup_n\|g_n\|_p$$. Let $$q$$ be the conjugate of $$p$$, i.e., $$\frac1p+\frac1q=1$$. Fix $$h\in L_q$$ We show that $$\lim_n\int_X g_nh\,d\mu=\int_X gh\,d\mu$$. Given $$\varepsilon>0$$, there is $$\delta>0$$ such that $$\|h\mathbb{1}_E\|_q<\frac{\varepsilon}{3M}\quad\text{whenever}\quad \mu(E)<\delta$$ Since $$\{|h|=0\}=\bigcap_n\{|h|\leq\frac1n\}$$ and $$\lim_n\int_{\{|h|\leq\frac1n\}}|h|^q\,d\mu=0$$ There is $$n_\varepsilon\in\mathbb{N}$$ such that $$\|h\mathbb{1}_{\{|h|\leq\frac1{n_\varepsilon}\}}\|_q<\frac{\varepsilon}{6M}$$ The set $$A_{\varepsilon}=\{|h|>\frac1{n_\varepsilon}\}$$ has finite measure: $$\mu(A_\varepsilon)\leq n_\varepsilon\|h\|^q_q<\infty$$. Bu Egorov's theorem, there exists $$B\subset A_\varepsilon$$ such that $$\mu(A_\varepsilon\setminus B)<\delta$$ so that on $$B$$, $$f_n$$ converges to $$f$$ uniformly. Thus, there is $$N\in\mathbb{N}$$ such that for $$n\geq N$$ $$\|(g_n-g)\mathbb{1}_{B}\|_u<\frac{\varepsilon}{3\|h\|_q(\mu(B))^{1/p}}$$ Putting things together, we have that for all $$n\geq N$$ \begin{align} \Big|\int (g-g_n)h\,d\mu\Big|&\leq \int_{A_\varepsilon}|g-g_n||h|\,d\mu + \int_{X\setminus A_\varepsilon}|g-g_n||h|\,d\mu\\ &\leq \int_{B}|g-g_n||h|\,d\mu+ \int_{A_\varepsilon\setminus B}|g-g_n||h|\,d\mu +\|g-g_n\|_p\|h\mathbb{1}_{X\setminus A_\varepsilon}\|_q\\ &\leq \|(g-g_n)\mathbb{1}_B\|_p\|h\|_q+\|g-g_n\|_p\|h\mathbb{1}_{A_\varepsilon\setminus B}\|_q+\frac{\varepsilon}{3}\\ &<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3} \end{align} Hence $$g_n$$ converges to $$g$$ weakly in $$L_p$$.