If $f_n$ is a sequence weakly convergent to $f$ in $L^1[0,1]$, is the convolution $f_n*g$ weakly convergent to $f*g$?

Question

The question is as in the title.

Let $f_n$ be weakly convergent to $f$ in $L^1[0,1]$. That is, for any $\phi \in L^\infty[0,1]$, we have \begin{equation} \int_{[0,1]} f_n\phi \to \int_{[0,1]}f\phi \end{equation} as $n \to \infty$.

Next, for any $g \in L^1[0,1]$, the convolution $f_n *g$ is in $L^1[0,1]$ by Young's inequality. However, I cannot prove / disprove if $f_n *g$ converges to $f*g$ weakly in $L^1[0,1]$ as $n \to \infty$.

Could anyone please help me?

Answer 1

Let $\phi \in L^{\infty}$. First note that $$ \int |(f_n*g)(x)\phi(x)|\ \mathrm{d}x \le \Vert \phi\Vert_{L^{\infty}}\Vert f_n * g\Vert_{L^1} < \infty $$ by Young's inequality. Hence by Fubini-Tonelli, $$ \begin{align*} \int (f_n*g)(x)\phi(x)\ \mathrm{d}x &= \int\int f_n(y)g(x-y) \phi(x)\ \mathrm{d}y \mathrm{d}x \\ &= \int f_n(y)\int g(x-y) \phi(x)\ \mathrm{d}x \mathrm{d}y \\ \end{align*} $$ Furthermore by Young's inequality, $$ \int g(x-y) \phi(x)\ \mathrm{d}x \in L^{\infty}, $$ being the convolution of $g(-\cdot) \in L^1$ with $\phi \in L^\infty$. Thus $$ \int f_n(y)\int g(x-y) \phi(x)\ \mathrm{d}x \mathrm{d}y \to \int f(y)\int g(x-y) \phi(x)\ \mathrm{d}x \mathrm{d}y $$ by weak convergence of $f_n$.

Answer 2

Observe for $\phi \in L^\infty([0,1])$, $$ \begin{aligned} \int_{[0,1]} (f_n * g) \phi &= \int_{[0,1]} \int_{[0,1]} f_n(x - y) g(y) \phi(x) dy dx \\ &= \int_{[0,1]} g(y) \int_{[0,1]} f_n(x - y) \phi(x) dx dy \\ \end{aligned} $$ where $f_n$ and $f$ are taken to be $0$ outside their domain of definition. Now let $h_n \in L^1([0,1])$ be defined so $h_n(y) := g(y) \int_{[0,1]} f_n(x - y) \phi(x) dx$.

Claim: $h_n \to h$, where $h(y) := g(y) \int_{[0,1]} f(x - y) \phi(x) dx$.

Proof: Let $\psi_y := 1_{[0, 1-y]} \phi(\cdot + y)$. Note $\psi_y \in L^\infty$ and $$ \begin{aligned} \int_{[0,1]} f_n(x - y) \phi(x) dx &= \int_{[0,1-y]} f_n(z) \phi(z+y) dz \\ &= \int_{[0,1]} f_n(z) \psi_y(z) dz \\ & \to \int_{[0,1]} f(z) \psi_y(z) dz \\ &= \int_{[0,1-y]} f(z) \phi(z + y) dz \\ &= \int_{[0,1]} f(x - y) \phi(x) dx \end{aligned}. $$ As $y \in [0,1]$ was arbitrary, $h_n \to h$. $\square$

Next note that because weakly convergent sequences are bounded, we have some $M > 0$ so $M \geq \|f_n\|_1$ for all $n \in \mathbb{N}$. Then

$$\begin{aligned} |h_n| &\leq |g| \int_{[0,1]} |f_n(x - \cdot)| |\phi(x)| dx \\ &\leq |g| \|f_n\|_1 \|\phi\|_\infty \\ &\leq |g| M \|\phi\|_\infty \in L^1([0,1]). \end{aligned} $$

Then, by Dominated Convergence Theorem, we have $$\begin{aligned} \int_{[0,1]} (f_n * g) \phi &= \int_{[0,1]} h_n \\ &\to \int_{[0,1]} h \\ &= \int_{[0,1]} (f * g) \phi. \end{aligned} $$

As $\phi \in L^\infty$ was arbitrary, we are done. $\blacksquare$

Answer 3

Let $\phi \in L^\infty(0, 1)$. We want to prove that $$ \int_0^1 (f_n * g) \phi \xrightarrow{n \to \infty} \int_0^1 (f * g) \phi. $$

By definition of weak topology, it suffices to prove that the linear map $$ T:L^1(0, 1) \to \mathbb R, h \mapsto \int_0^1 (h * g) \phi $$ is continuous. This is indeed true because $$ \begin{align*} |T(h)| &\le \|\phi\|_{L^\infty} \| h * g\|_{L^1} \text{ by Hölder's inequality}\\ & \le \|\phi\|_{L^\infty} \| h\|_{L^1}\|g\|_{L^1} \text{ by Young's inequality}. \end{align*} $$