Let $\Omega \subset \mathbb{R}^n$ be an open set and let $1 < p < \infty$. Consider a sequence $(f_n)_n \subset L^p(\Omega)$ and $f \in L^p(\Omega)$ such that $f_n$ converges to $f$ weakly in $L^p(\Omega)$. Let $g_n: \Omega \rightarrow \mathbb{R}$ and $g: \Omega \rightarrow \mathbb{R}$ be measurable functions such that $g_n \rightarrow g$ in measure and $||g_n||_{L^\infty} \le C$, $\forall n$ and for some constant $C > 0$. Prove that $g \in L^\infty (\Omega)$ and $f_n g_n \rightharpoonup fg$ in $L^p(\Omega)$.

$\textbf{Note}$: $g_n \rightarrow g$ in measure if $\forall \epsilon > 0$, $\mu\{x \in \Omega : |g_n(x) - g(x)| > \epsilon\} \rightarrow 0$, where $\mu$ in this case denotes Lebesgue measure.

Here is my attempt.

Since $g_n \rightarrow g$ in measure, there exists a subsequence $(g_{n_k})_k$ such that $g_{n_k} \rightarrow g$ a.e. in $\Omega$. It follows that $|g_{n_k}(x)| \rightarrow |g(x)|$ for a.e. $x \in \Omega$. By hypothesis, $|g_{n_k}(x)|\le ||g_{n_k}||_{L^{\infty}} \le C$ for all $k \in \mathbb{N}$ $\Rightarrow$ $|g(x)| \le C$ for a.e. $x \in \Omega$. This implies that $g \in L^{\infty}(\Omega)$.

We aim to prove that $f_n g_n \rightharpoonup fg$ in $L^p(\Omega)$ $\Longleftrightarrow$ $\forall \phi \in L^q(\Omega)$ (where $\frac{1}{p} + \frac{1}{q} = 1$) , $\int_{\Omega} f_n g_n \phi dx \rightarrow \int_{\Omega} fg \phi dx$. First of all, let us check $f_n g_n \in L^p(\Omega)$.

$||f_n g_n||_{L^p}^p = \int_{\Omega} |f_n g_n|^p dx \le ||g_n||_{L^{\infty}}^p ||f_n||_{L^p}^p < \infty$ since, by hypothesis, $f_n \in L^p(\Omega), g_n \in L^\infty(\Omega), \, \forall n$. (The same argument holds true also for $fg$).

Note that $f_n g_n - fg = (f_n -f)g_n + f (g_n - g)$. Then

$|\int_{\Omega} f_n g_n \phi dx - \int_{\Omega} fg \phi dx| = |\int_{\Omega} (f_n g_n -fg) \phi dx| \le |\int_{\Omega} (f_n-f) g_n \phi dx| + |\int_{\Omega} f(g_n -g) \phi dx|$.

Let us focus on the first integral on the right hand-side of the expression above. $|\int_{\Omega} (f_n-f) g_n \phi dx| \le ||g_n||_{L^{\infty}} |\int_{\Omega} (f_n-f) \phi dx| \le C |\int_{\Omega} (f_n-f) \phi dx| \rightarrow 0$, since $f_n \rightharpoonup f $.

Let $\epsilon > 0$ and define $A_{\epsilon} := \{x \in \Omega: |g_n(x) - g(x)| < \epsilon\}$. Convergence in measure implies that $|\int_{\Omega} f(g_n -g) \phi dx| \le |\int_{A_{\epsilon}} f(g_n - g) \phi dx| + |\int_{\Omega \setminus A_{\epsilon}} f(g_n -g) \phi dx| < \epsilon \int_{A_{\epsilon}} |f| \phi dx + \int_{\Omega \setminus A_{\epsilon}} |f(g_n -g) \phi| dx$ $< \epsilon \int_{\Omega} |f| \phi dx + \int_{\Omega \setminus A_{\epsilon}} |f(g_n -g) \phi| dx$. The second integral tends to zero for $n \rightarrow \infty$ because $g_n \rightarrow g$ in measure (which means that $\mu(\Omega \setminus A_{\epsilon}) \rightarrow 0$, the thesis is a consequence of the absolute continuity of the integral), while $\int_{\Omega} |f| \phi dx$ is merely a constant.

If anyone could check the above reasoning, it would be greatly appreciated.

  • 2
    $\begingroup$ Everything looks good except for one small detail: to conclude that $\int_{\Omega\setminus A_\epsilon} |f(g_n-g)\phi| \to 0$ you need an integrand which does not depend on $n$. You can do so by first using the hypothesis $\lVert g_n \rVert_{L^\infty}\leq C$: $$\int_{\Omega\setminus A_\epsilon} |f(g_n-g)\phi| \leq \lVert g_n - g\rVert_{L^\infty} \int_{\Omega\setminus A_\epsilon} |f\phi| \leq 2C\int_{\Omega\setminus A_\epsilon} |f\phi|,$$ and now you can use continuity of the integral. $\endgroup$
    – Michh
    Jan 15, 2022 at 20:25
  • $\begingroup$ Thank you very much! $\endgroup$ Jan 15, 2022 at 20:50
  • 2
    $\begingroup$ @Hilbert1234 I don't think $|\int_{\Omega} (f_n-f) g_n \phi dx| \le ||g_n||_{L^{\infty}} |\int_{\Omega} (f_n-f) \phi dx|$ is correct .. Usually we don't have $|\int fg| \le |g|_\infty|\int f|$ .. $\endgroup$
    – r9m
    Jan 16, 2022 at 13:35
  • $\begingroup$ I am using the fact that, by definition of $||\cdot||_{L^{\infty}}$, $g_n(x) \le ||g_n||_{L^{\infty}}$ a.e. in $\Omega$. Honestly, I don't see why it is not correct. Thanks. $\endgroup$ Jan 17, 2022 at 11:00
  • $\begingroup$ And by hypothesis, $||g_n||_{L^{\infty}} \le C$, so I could have used straightforwardly $C$ instead of the $L^{\infty}$ norm. $\endgroup$ Jan 17, 2022 at 11:17

1 Answer 1


As pointed out by r9m, the step $$|\int_{\Omega} (f_n-f) g_n \phi dx| \le ||g_n||_{L^{\infty}} |\int_{\Omega} (f_n-f) \phi dx|$$ may not be valid, for example, if $f=0$, $f_n=\mathbf{1}_{[n,n+1)}-\mathbf{1}_{(-n-1,-n]}$ and $g_n=\frac{1}{n}f_n$, the right hand side is zero if $\phi$ is even.

It is actually better to slip in the following way: $$ \int f_ng_n\phi-\int fg\phi=\int f_n\left(g_n-g\right)\phi+\int f_ng\phi-\int fg\phi $$ because one can directly exploit the weak convergence assumption to get the convergence to $0$ of the difference of the last two integrals. It remains to check that $a_n:=\int f_n\left(g_n-g\right)\phi\to 0$. To do so, define for a fixed $\varepsilon$ and a fixed $n$ the set $A_{n,\varepsilon}=\{\lvert g_n-g\rvert>\varepsilon\}$. Then $$ \lvert a_n\rvert\leqslant \left\lvert \int_{A_{n,\varepsilon}}f_n\left(g_n-g\right)\phi \right\rvert+\left\lvert \int_{A_{n,\varepsilon}^c}f_n\left(g_n-g\right)\phi \right\rvert\leqslant 2C\int_{A_{n,\varepsilon}}\lvert f_n\rvert\lvert \phi\rvert +\varepsilon\int_{A_{n,\varepsilon}^c}\lvert f_n\rvert\lvert \phi\rvert $$ and by Hölder's inequality, we get $$\lvert a_n\rvert\leqslant 2C\sup_{N\geqslant 1}\lVert f_N\rVert_p\lVert \phi\mathbf{1}_{A_{n,\varepsilon}}\rVert_q+\varepsilon \sup_{N\geqslant 1}\lVert f_N\rVert_p\lVert \phi \rVert_q.$$ Since $\mu(A_{n,\varepsilon})\to 0$ for each positive $\varepsilon$, it follows that $\lVert \phi\mathbf{1}_{A_{n,\varepsilon}}\rVert_q\to 0$ hence $\limsup_n \lvert a_n\rvert\leqslant \varepsilon \sup_{N\geqslant 1}\lVert f_N\rVert_p\lVert \phi \rVert_q$, giving the result.

  • $\begingroup$ Thank you very much for the detailed explanation! Now, it is all clear. $\endgroup$ Jan 17, 2022 at 19:52

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