Prove that the convolution product $\star$ on $L^1 ([-\pi,\pi])$ ($2\pi$ periodic Lebesgue integrable functions on $\mathbb{R}$ with respect to the $L^1$ norm satisfies $f\star g(\theta) = \frac{1}{2\pi} \int_{-\pi}^\pi f(\theta – t) g(t)dt,$ by only assuming $f\star g (\theta)$ is defined as such for $2\pi$ periodic continuous functions and that for any sequences of continuous functions $(f_n), (g_n), f_n\to f, g_n\to g, f\star g=\lim\limits_{n\to\infty} f_n\star g_n,$ where the limit is respect to the $L^1$ norm.
Let $(f_n)$ and $(g_n)$ be continuous $2\pi$ periodic functions converging to $f$ and $g$ under the $L^1$ norm.
I thought of using Fubini's theorem for measures to show that $\frac{1}{2\pi} \int_{-\pi}^\pi f_n(\theta – t)g_n(t)dt$ converges to $\frac{1}{2\pi} \int_{-\pi}^\pi f(\theta – t)g(t)dt.$ Basically, $\frac{1}{2\pi} \int_{-\pi}^\pi \frac{1}{2\pi} \int_{-\pi}^\pi |f_n(\theta – t)g_n(t) – f(\theta – t)g(t)|d \theta dt$ can be made arbitrarily close to $0$ (using the fact that $f_n, g_n\to f,g$), which would then show that $\lim\limits_{n\to\infty} f_n(\theta – t)g_n(t) = f(\theta-t)g(t)$ almost everywhere and would imply the result.
Am I missing something?
Best Answer
By $$ \bigg|\frac{1}{2\pi}\int_{-\pi}^\pi (f*g)(\theta)d\theta\bigg|\le\|f\|\|g\| $$ one has \begin{eqnarray} &&\|f_n*g_n-f*g\|\\ &=&\frac{1}{2\pi}\int_{-\pi}^{\pi}|f_n*g_n(\theta)-f*g(\theta)|dt\\ &=&\frac{1}{(2\pi)^2} \int_{-\pi}^{\pi}\bigg|\int_{-\pi}^\pi f_n(\theta - t) g_n(t)dt-\int_{-\pi}^\pi f(\theta - t) g(t)dt\bigg|d\theta\\ &=&\frac{1}{(2\pi)^2} \int_{-\pi}^\pi \bigg|\int_{-\pi}^\pi \bigg[f_n(\theta - t)-f(\theta-t)\bigg]g_n(t) dt+ \int_{-\pi}^\pi f(\theta - t)\bigg[g_n(t)-g(t)\bigg] dt\bigg|d\theta\\ &\le&\frac{1}{(2\pi)^2}\bigg[\int_{-\pi}^\pi \int_{-\pi}^\pi|f_n(\theta - t)-f(\theta-t)||g_n(t)-g(t)| dtd\theta\\ &&+\int_{-\pi}^\pi \int_{-\pi}^\pi|f_n(\theta - t)-f(\theta-t)||g(t)| dtd\theta +\\ &&\int_{-\pi}^\pi\int_{-\pi}^\pi |f(\theta - t)||g_n(t)-g(t)| dt d\theta\bigg]\\ &=&\frac{1}{2\pi}\bigg[\int_{-\pi}^\pi(|f_n-f|*|g_n-g|)(\theta)d\theta+\int_{-\pi}^\pi(|f_n-f|*|g|)(\theta)d\theta+\int_{-\pi}^\pi(|f|*|g_n-g|)(\theta)d\theta\bigg]\\ &\le&\|f_n-f\|\|g_n-g\|+\|f_n-f\|\|g|+\|f\|\|g_n-g\|\\ &\to&0 \end{eqnarray} since $f_n, g_n\to f,g$ in $L^1 ([-\pi,\pi])$ .