# Sufficient condition for “Weak Convergence”

This is a sufficient condition for something that resembles “Weak convergence”: $\int f_kg\to \int fg$ for all $g\in L^{p'}$
Suppose that $f_k\to f$ a.e.\ and that $f_k, f\in L^p$, $1. If $\|f_k\|_p\leq M<\infty$, we have $\int f_kg\to\int fg$ for all $g\in L^{p'}$, $1/p+1/p'=1$. Note that the result is false if $p=1$.

Proof:
(Case: $|E|<\infty$, where $E$ is the domain of integration).

We may assume $|E|>0$, $M>0$, $\|g\|_{p'}>0$ otherwise the result is trivially true. Also, by Fatou’s Lemma, $\displaystyle \|f\|_p\leq\liminf_{k\to\infty}\|f_k\|_p\leq M.$

Let $\epsilon>0$. Since $g\in L^{p'}$, so $g^{p'}\in L^1$ and there exists $\delta>0$ such that for any measurable subset $A\subseteq E$ with $|A|<\delta$, $\int_A |g^{p'}|<\epsilon^{p'}$.

Since $f_k\to f$ a.e.\ ($f$ is finite a.e.\ since $f\in L^p$), by Egorov’s Theorem there exists closed $F\subseteq E$ such that $|E\setminus F|<\delta$ and $\{f_k\}$ converge uniformly to $f$ on $F$. That is, there exists $N(\epsilon)$ such that for $k\geq N$, $|f_k(x)-f(x)|<\epsilon$ for all $x\in F$.

Then for $k\geq N$,
\begin{aligned} \left|\int_E f_kg-fg\right|&\leq\int_E|f_k-f||g|\\ &=\int_{E\setminus F}|f_k-f||g|+\int_F|f_k-f||g|\\ &\leq\left(\int_{E\setminus F}|f_k-f|^p\right)^\frac{1}{p}\left(\int_{E\setminus F}|g|^{p'}\right)^\frac{1}{p'}+\epsilon\int_F |g|\\ &<\|f_k-f\|_p(\epsilon)+\epsilon\left(\int_F|g|^{p'}\right)^\frac{1}{p'}\left(\int_F |1|^p\right)^\frac{1}{p}\\ &\leq 2M\epsilon+\epsilon\|g\|_{p'}|E|^\frac{1}{p}\\ &=\epsilon(2M+\|g\|_{p'}|E|^\frac{1}{p}). \end{aligned}

Since $\epsilon>0$ is arbitrary, this means $\int_E f_g\to \int_E fg$.

(Case: $|E|=\infty$). Error: See correction below.

Define $E_N=E\cap B_N(0)$, where $B_N(0)$ is the ball with radius $N$ centered at the origin. Then $|E_N|<\infty$, so there exists $N_1>0$ such that for $N\geq N_1$, $\int_{E_N}|f_k-f||g|<\epsilon$.

Since $|g|^{p'}\chi_{E_N}\nearrow|g|^{p'}$ on $E$, by Monotone Convergence Theorem, $\displaystyle \lim_{N\to\infty}\int_{E_N}|g|^{p'}=\int_E |g|^{p'}<\infty.$
Thus there exists $N_2>0$ such that for $N\geq N_2$, $\int_{E\setminus E_N} |g|^{p'}<\epsilon^{p'}$.

Then for $N\geq\max\{N_1, N_2\}$,
\begin{aligned} \int_E |f_kg-fg|&=\int_{E_N}|f_k-f||g|+\int_{E\setminus E_N}|f_k-f||g|\\ &<\epsilon+\left(\int_{E\setminus E_N}|f_k-f|^p\right)^\frac{1}{p}\left(\int_{E\setminus E_N}|g|^{p'}\right)^\frac{1}{p'}\\ &<\epsilon+\|f_k-f\|_p(\epsilon)\\ &\leq\epsilon+2M\epsilon\\ &=\epsilon(1+2M). \end{aligned}
so that $\int_E f_kg\to\int_E fg$.

(Show that the result is false if $p=1$).

Let $f_k:=k\chi_{[0,\frac 1k]}$. Then $f_k\to f$ a.e., where $f\equiv 0$. Note that $\int_\mathbb{R} |f_k|=1$, $\int_\mathbb{R} |f|=0$ so that $f_k, f\in L^1(\mathbb{R})$. Similarly, $\|f_k\|_1\leq M=1$.

However if $g\equiv 1\in L^\infty$, $\int_\mathbb{R} f_kg=1$ for all $k$ but $\int_\mathbb{R} fg=0$.

Correction for the case $|E|=\infty$:

Define $E_N=E\cap B_N(0)$, where $B_N(0)$ is the ball with radius $N$ centered at the origin.

Since $|g|^{p'}\chi_{E_N}\nearrow |g|^{p'}$ on $E$, by Monotone Convergence Theorem, $\displaystyle \lim_{N\to\infty}\int_{E_N}|g|^{p'}=\int_E|g|^{p'}<\infty.$

Thus there exists $N_1>0$ such that $\int_{E\setminus E_{N_1}}|g|^{p'}<\epsilon^{p'}$.

Since $|E_{N_1}|<\infty$, by the finite measure case there exists $N_2$ such that for $k\geq N_2$, $\displaystyle \int_{E_{N_1}}|f_k-f||g|<\epsilon.$

So for $k\geq N_2$,
\begin{aligned} \int_E|f_kg-fg|&=\int_{E_{N_1}}|f_k-f||g|+\int_{E\setminus E_{N_1}}|f_k-f||g|\\ &<\epsilon+\left(\int_{E\setminus E_{N_1}}|f_k-f|^p\right)^{1/p}\left(\int_{E\setminus E_{N_1}}|g|^{p'}\right)^{1/p'}\\ &<\epsilon+\|f_k-f\|_p(\epsilon)\\ &\leq\epsilon+2M\epsilon\\ &=\epsilon(1+2M). \end{aligned}

so that $\int_Ef_kg\to\int_E fg$.

## Author: mathtuition88

https://mathtuition88.com/

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