# Convergence of product of (weakly) converging sequences in $L^{p}$

Preparing for an exam, I was wondering about general statements about the convergence of products.

1) Let $p, q \in ]1, \infty[$ such that $\frac{1}{p}+\frac{1}{q} = 1$ and

• $a_n \rightharpoonup a$ in $L^p(\mathbb{R}^n)$,
• $b_n \rightharpoonup b$ in $L^q(\mathbb{R}^n)$.

What can we say about the convergence of $a_nb_n$?

2) What if $p$ and $q$ don't satisfy $\frac{1}{p}+\frac{1}{q}=1?$ Is there anything you can say about the convergence?

3) What if one (or both) of the sequences converge strongly?

The attempts:

1) I think $a_nb_n$ converge weakly in $L^1$.

To show: $\int{a_nb_nc} \rightarrow \int{abc}$ for all $c \in L^\infty$.

Using the trick $|\int{a_nb_nc} - \int{abc}| \leq |\int{a_nb_nc} - \int{a_nbc}| + |\int{a_nbc} - \int{abc}| = |\int{(b_n-b)a_nc}| + |\int{(a_n-a)bc}|$

you see that both summands converge to 0 because $a_nc \in L^p$ and $bc \in L^q$ respectively (a function in $L^\infty$ multiplied by a function in $L^p$ is in $L^p$).

Is that correct? What about strong convergence?

In all examples I have studied so far, one of the two sequences converges strongly. So I wondered what I was missing in my calculations above.

Hint: In the following example the sequence $a_n$ converges strongly in $L^2$ and $b_n$ weakly in $L^2$. The accepted poster suggests that $a_nb_n$ converge strongly: weak convergence of product of weakly and strongly convergent $L^{2}$ sequences in $L^{2}$

2) I don't think you can say anything about convergence in general? Only if $p$ or $q$ are $\infty$, as seen here: Product of weak/strong converging sequences

If I understand right, you need the strong convergence of the $L^\infty$ sequence because $2$ isn't dual to $\infty$.

3) Clearly, having strong convergence for both sequences does not harm, but does it change anything? I think it doesn't.

(I already asked above if we require one of them to converge strongly.)

Thanks a lot for your input!

Saying $b_n\to b$ weakly in $L^q$ says that $\int(b_n-b)\phi\to0$ if $\phi$ is an element of $L^p$. But your $a_nc$ is not an element of $L^p$, it's a sequence of element$s$ of $L^p$.

Example on $\Bbb R$: Say $a_n=b_n=\chi_{[n,n+1]}$. Then $a_n\to0$ weakly in $L^q$ and $b_n\to$ weakly in $L^p$ (if $1<p<\infty$) but $\int a_nb_n$ does not converge to $0$.

• So you are saying, I require $b_n$ to converge strongly, so I can split up the first integral (after the last equation) using Hölder? – andreas Jan 28 '16 at 0:30
• @andreas I wasn't saying that, but yes it seems like that would be sufficient. – David C. Ullrich Jan 28 '16 at 0:37
• I know, you were indirectly implying it. ;-) Thanks for clearing my mistake up. – andreas Jan 28 '16 at 0:44

So here is my attempt: Let $f, f_k \in L^p(\mathbb R^n)$, $1 \leq p < \infty$, $g, g_k \in L^q(\mathbb R^n)$, such that $\frac 1 p + \frac 1 q = 1$. Let's first assume that $f_k \to f$ in $L^p(\mathbb R^n)$ and $g_k \to g$ in $L^q(\mathbb R^n)$. We claim that $f_k g_k \to fg$ in $L^1(\mathbb R^n)$. By Hölder we know that $fg, f_k g_k \in L^1(\mathbb R^n)$, and we calculate \begin{align*} \int_{\mathbb R^n} \vert f_k g_k - fg \vert \, \mathrm dx &\leq \int_{\mathbb R^n} \vert f_k g_k - f_k g \vert \, \mathrm dx + \int_{\mathbb R^n} \vert f_k g - fg \vert \, \mathrm dx \\ &= \int_{\mathbb R^n} \vert f_k \vert \vert g_k - g \vert \, \mathrm dx + \int_{\mathbb R^n} \vert f_k - f \vert \vert g \vert \, \mathrm dx \\ & \leq \Vert f_n \Vert_{L^p} \Vert g_k - g \Vert_{L^q} + \Vert f_k - f \Vert_{L^p} \Vert g \Vert_{L^q} \\ & \leq \underbrace{ \sup_{k \in \mathbb N} \Vert f_k \Vert_{L^p} }_{< \infty} \underbrace{\Vert g_k - g \Vert_{L^q}}_{\to 0} + \underbrace{\Vert f_k - f \Vert_{L^p}}_{\to 0} \underbrace{ \Vert g \Vert_{L^q} }_{< \infty} \; , \end{align*} so we have $\Vert f_k g_k - fg \Vert_{L^1} \to 0$, i.e. $f_k g_k \to fg$ in $L^1(\mathbb R^n)$.

Now let's change the assumption $g_k \to g$ in $L^q(\mathbb R^n)$ to $g_k \rightharpoonup g$ in $L^q(\mathbb R^n)$ for $1 \leq q < \infty$. We claim that $f_k g_k \rightharpoonup fg$ in $L^1(\mathbb R^n)$. So we have $\Vert f_n - f \Vert_{L^p} \to 0$ and $$\int_{\mathbb R^n} g_n \phi \, \mathrm dx \to \int_{\mathbb R^n} g\phi \, \mathrm dx \quad \text{for each } \phi \in L^p(\mathbb R^n) \; .$$ We have to show, that $$\int_{\mathbb R^n} f_k g_k \psi \, \mathrm dx \to \int_{\mathbb R^n} fg \psi \, \mathrm dx \quad \text{for each } \psi \in L^\infty(\mathbb R^n) \; .$$ Again, we calculate \begin{align*} \bigg\vert \int_{\mathbb R^n} f_k g_k \psi \, \mathrm dx - \int_{\mathbb R^n} fg \psi \, \mathrm dx \bigg\vert &= \bigg \vert \int_{\mathbb R^n} f_k g_k \psi - fg \psi \; \mathrm dx \bigg\vert \\ &\leq \bigg\vert \int_{\mathbb R^n} (f_k g_k - f g_k) \psi \, \mathrm cx \bigg\vert + \bigg\vert \int_{\mathbb R^n} f g_k \psi - fg\psi \, \mathrm dx \bigg\vert \\ &\leq \Vert \psi \Vert_{L^\infty} \underbrace{\int_{\mathbb R^n} \vert g_k \vert \vert f_k - f \vert \, \mathrm dx }_{\leq \Vert f_k - f \Vert_{L^p} \Vert g_k \Vert_{L^q}}+ \underbrace{\bigg\vert \int_{\mathbb R^n} (g_k - g) \underbrace{(f \psi)}_{\in L^p(\mathbb R^n)} \, \mathrm dx \bigg\vert}_{\to 0} \end{align*}

Now, there is a theorem, which states, that if $g_k \rightharpoonup g$, then $\Vert g_k \Vert_{L^q}$ is bounded (maybe someone can elaborate this?), so it follows $f_k g_k \rightharpoonup fg$

I think $a_nb_n$ weakly converges to $ab$. \begin{aligned} \left|\int a_n b_n c-abc\right|&\leq \left|\int a_n(b_n-b)c\right|+\left|\int(a_n-a)bc\right|\\ &\leq\|a_n\|_\infty\left|\int (b_n-b)c\right|+\left|(a_n-a)bc\right|\\ \end{aligned} As long as $\|a_n\|_\infty$ is bounded, it is obvoius.