$\sup_{n}\|f_{n}\|_{2}<\infty$ and $f_{n}\rightarrow f$ a.e. implies weak convergence.

Question

Suppose that each $f_{n}$ and $f$ are Lebesgue measurable functions from $\mathbb{R}$ to $\mathbb{R}$.

Prove that if $\sup_{n}\|f_{n}\|_{2}<\infty$ and $f_{n}\rightarrow f$ a.e., then $$ \int f_{n}g\ dx\rightarrow\int fg\ dx $$ for every $g\in L^{2}\left(\mathbb{R}\right)$.

First, I use Egoroff's theorem. Except small measure, the convergence is uniform. But measure of $\mathbb{R}$ except small measure is $\infty$. So I stuck here. Can anyone help me?

(1st edition)

Let $g$ is in $C_{c}\left(\mathbb{R}\right)$.

Then $m(\text{supp}(g))<\infty$.

So above weak convergence becomes $$ \int_{\text{supp}(g)}f_{n}g\ dx\rightarrow\int_{\text{supp}(g)}fg\ dx. $$ I write this form in the norm notation : $$ \|(f_{n}-f)g\|_{L^{1}\left(\text{supp}(g)\right)}\longrightarrow0. $$ To apply Erogoff on the left hand side : $\forall\epsilon>0$, there exists measurable set $E$ such that $m(E)<\epsilon$ and $f_{n}\rightarrow f$ uniformly on $E^{c}$. Then $$ \|\left(f_{n}-f\right)g\|_{L^{1}\left(\text{supp}(g)\right)}\leq\|f_{n}-f\|_{L^{1}\left(\text{supp}(g)\right)}\|g\|_{L^{1}\left(\text{supp}(g)\right)}\leq\|g\|\left(\int_{E}(f_{n}-f)\ dx+\int_{E^{c}}\left(f_{n}-f\right)\ dx\right). $$

By uniform convergence, $(f_{n}-f)<\epsilon$ on $E^{c}$ by taking $n$ large. And $\int_{E}\left(f_{n}-f\right)\ dx\leq2\sup_{n}\|f_{n}\|m(E)<2\sup_{n}\|f_{n}\|\epsilon$(This is What Sangchul Lee mentioned below)(More detail, I used Holder inequality).

So $$ \|\left(f_{n}-f\right)g\|_{L^{1}\left(\text{supp}(g)\right)}<M\epsilon $$ where $M$ is constant. If anyone has spare time, please check my proof here.

Answer 1

The idea is sound, but there are a few glitches in the execution.

Let's define $S := \sup_n \lVert f_n\rVert_{2}$. First we note that $f\in L^2(\mathbb{R})$ and $\lVert f\rVert_2 \leqslant S$ by Fatou's lemma:

$$\int \lvert f\rvert^2\,dx = \int \liminf_{n\to\infty} \lvert f_n\rvert^2\,dx \leqslant \liminf_{n\to\infty} \int \lvert f_n\rvert^2\,dx \leqslant S^2.$$

Next we note that due to the boundedness of $(f_n)$, it suffices to prove the assertion

$$\lim_{n\to \infty} \int f_n g\,dx = \int fg\,dx\tag{1}$$

for a family $\mathscr{G}\subset L^2(\mathbb{R})$ that spans a dense subspace of $L^2(\mathbb{R})$.

Let $\mathscr{H} = \{ g \in L^2(\mathbb{R}) : (1)\text{ holds}\}$. Trivially $0 \in \mathscr{H}$, and by the linearity of the integral, every linear combination of elements of $\mathscr{H}$ belongs to $\mathscr{H}$, so $\mathscr{H}$ is a linear subspace of $L^2(\mathbb{R})$. Since $(f_n)$ is bounded, $\mathscr{H}$ is closed: Let $h \in \overline{\mathscr{H}}$. For every $\varepsilon > 0$ there is a $g\in \mathscr{H}$ with $\lVert g-h\rVert_2 \leqslant \varepsilon$, and hence

\begin{align} \limsup_{n\to\infty}\: \Biggl\lvert \int (f_n - f)h\,dx\Biggr\rvert &= \limsup_{n\to\infty}\: \Biggl\lvert \int (f_n - f)g\,dx + \int (f_n-f)(h-g)\,dx\Biggr\rvert \\ &\leqslant \limsup_{n\to \infty}\: \Biggl\lvert \int (f_n - f)g\,dx\Biggr\rvert + \limsup_{n\to \infty}\: \Biggl\lvert \int (f_n - f)(h-g)\,dx\Biggr\rvert \\ &= \limsup_{n\to \infty}\: \Biggl\lvert \int (f_n - f)(h-g)\,dx\Biggr\rvert \tag{$g\in\mathscr{H}$} \\ &\leqslant \limsup_{n\to\infty} \:\lVert f_n -f \rVert_2\lVert h-g\rVert_2 \\ &\leqslant 2S\varepsilon. \end{align}

Since that holds for all $\varepsilon > 0$, it follows that $h \in \mathscr{H}$.

So if $(1)$ holds for all $g \in \mathscr{G}$, it holds for all $h \in \overline{\operatorname{span} \mathscr{G}}$. For a proof using Egorov's theorem, one chooses $\mathscr{G}$ as a family of bounded functions that vanish outside some set of finite measure. Besides the family $C_c(\mathbb{R})$, another good choice would be the family of characteristic functions of sets of finite measure. The latter has the advantage that we can use that family on arbitrary measure spaces. The proof works the same for both families.

Now we come to your argument. You write

$$\|\left(f_{n}-f\right)g\|_{L^{1}\left(\text{supp}(g)\right)}\leq\|f_{n}-f\|_{L^{1}\left(\text{supp}(g)\right)}\|g\|_{L^{1}\left(\text{supp}(g)\right)}\leq\|g\|\left(\int_{E}(f_{n}-f)\ dx+\int_{E^{c}}\left(f_{n}-f\right)\ dx\right),$$

and say you used Hölder's inequality. This may be a typo, but as written, it is incorrect. Hölder's inequality uses a pair of conjugate exponents, it doesn't bound the $L^1$-norm of a product by the product of $L^1$-norms. Since we're dealing with $L^2$-functions, a natural choice would be to take the $L^2$-norm of both factors. But that would lead to something including the integral

$$\int_E \lvert f_n - f\rvert^2\,dx,$$

and I don't know how to show that that is uniformly (in $n$) small for small $\varepsilon > 0$ without an assumption of uniform integrability of the $f_n$. It works better to use the $L^1$-norm for $f_n - f$ and the $L^{\infty}$-norm for $g$. Also, in the last integrals, you need to take the absolute value, $\lvert f_n - f\rvert$.

With the pair $(1,\infty)$ of conjugate exponents, we get the inequality

$$\lVert (f_n - f)g\rVert_{L^1} \leqslant \lVert g\rVert_{L^{\infty}}\Biggl(\int_E \lvert f_n - f\rvert\,dx + \int_{E^c} \lvert f_n - f\rvert\,dx\Biggr). \tag{2}$$

On the right hand side of $(2)$, we have

$$\int_{E^c}\lvert f_n - f\rvert\,dx \leqslant m(E^c)\sup \:\{ \lvert f_n(x) - f(x)\rvert : x \in E^c\} \leqslant m(\operatorname{supp} g)\sup \:\{ \lvert f_n(x) - f(x)\rvert : x \in E^c\},$$

and that tends to $0$ because of the uniform convergence on $E^c$. For the other integral, you state the bound

$$\int_{E}\left(f_{n}-f\right)\ dx\leq2\sup_{n}\|f_{n}\|m(E)<2\sup_{n}\|f_{n}\|\epsilon,$$

but that isn't correct. If, say, $f_n(x) = x^{-\alpha}\cdot \chi_{[1/n,1]}(x)$, and $E = [0,\varepsilon]$, then we have

$$\int_E \lvert f_n - f\rvert\,dx = \int_0^{\varepsilon} x^{-\alpha}\,dx = \frac{\varepsilon^{1-\alpha}}{1-\alpha}$$

for $n \leqslant 1/\varepsilon$, and for every fixed $n$, that tends to $0$ slower than a constant multiple of $\varepsilon$. However, the Cauchy-Schwarz inequality yields

$$\int_E \lvert f_n - f\rvert\,dx \leqslant \lVert f_n -f \rVert_2 \sqrt{m(E)} \leqslant 2S\sqrt{\varepsilon},$$

so we get the same type of inequality, just with a smaller exponent on $\varepsilon$. This then yields

$$\limsup_{n\to\infty} \:\lVert (f_n - f)g\rVert_{L^1} \leqslant M\sqrt{\varepsilon}$$

for every $\varepsilon > 0$, i.e. $\lVert (f_n - f)g\rVert_{L^1} \to 0$.

Answer 2

This is a proof that the hypothesis imply $f \in {L}^{2} \left(\mathbb{R}\right)$. Let us consider the set ${A}_{k} = \left\{x \in \left[{-k} , k\right] , \left|f \left(x\right)\right| \leqslant k\right\}$. Let ${\epsilon} > 0$ and, according to Egoroff's theorem, let ${E}_{{\epsilon}} \subset {A}_{k}$ be a set having measure $ \leqslant {\epsilon}$ such that ${f}_{n} {{1}}_{{A}_{k}}$ converges uniformly to $f {{1}}_{{A}_{k}}$ on ${A}_{k} \setminus {E}_{{\epsilon}}$. Let ${d}_{n} = \sup \left\{\left|{f}_{n} \left(x\right)-f \left(x\right)\right| , x \in {A}_{k} \setminus {E}_{{\epsilon}}\right\}$ One has

$$\left|f {{1}}_{{A}_{k} \setminus {E}_{{\epsilon}}}\right| \leqslant \left|{f}_{n} {{1}}_{{A}_{k} \setminus {E}_{{\epsilon}}}\right|+{d}_{n} {{1}}_{{A}_{k}} \in {L}^{2} \left({A}_{k}\right)$$

and

$${\left\|f {{1}}_{{A}_{k} \setminus {E}_{{\epsilon}}}\right\|}_{2} \leqslant {\left\|{f}_{n} {{1}}_{{A}_{k} \setminus {E}_{{\epsilon}}}\right\|}_{2}+{d}_{n} \sqrt{2 k} \leqslant M+{d}_{n} \sqrt{2 k}$$

By letting $n$ tend to infinity, one deduces ${\left\|f {{1}}_{{A}_{k} \setminus {E}_{{\epsilon}}}\right\|}_{2} \leqslant M$ Now

$$\int_{{E}_{{\epsilon}}}^{}{\left|f\right|}^{2} d {\mu} \leqslant \int_{{E}_{{\epsilon}}}^{}{k}^{2} d {\mu} \leqslant {k}^{2} {\epsilon}$$

hence ${\left\|f {{1}}_{{A}_{k}}\right\|}_{2} \leqslant M+k \sqrt{{\epsilon}}$ By letting ${\epsilon}$ tend to 0, one deduces that ${\left\|f {{1}}_{{A}_{k}}\right\|}_{2} \leqslant M$. Since ${\left|f\right|}^{2} {{1}}_{{A}_{k}}$ converges monotonically to ${\left|f\right|}^{2}$ when $k \rightarrow \infty $, it follows that $f \in {L}^{2} \left(\mathbb{R}\right)$ and that ${\left\|f\right\|}_{2} \leqslant M$.