Proving a convolution inequality

Question

If $\int_{\mathbb{R}}|f(x)|dx$ and $\int_{\mathbb{R}}|g(x)|dx$ are bounded, show that

$\int_{\mathbb{R}}|(f\star g)(x)|dx \leq \int_{\mathbb{R}}|f(x)|dx \cdot \int_{\mathbb{R}}|g(x)|dx $.

From the definition we have $(f\star g)(x)=\int_{\mathbb{R}}f(x-\tau)g(\tau)\,d\tau$. I guess I need to do work on this right side, but I don't see how to simplify to get the desired inequality

Answer 1

Applying the integral to the convolution:

$$\Bigr | \int_\mathbb{R} (f\star g) (x) dx \Bigr | = \Bigr |\int_\mathbb{R} \int_\mathbb{R} f(x-\tau)g(\tau)d\tau dx \Bigr | = \Bigr |\int_\mathbb{R} \int_\mathbb{R} f(x-\tau)g(\tau)dx d\tau \Bigr |$$ $$ \leq \int_\mathbb{R} \int_\mathbb{R} |f(x-\tau)||g(\tau)| dx d\tau = \int_\mathbb{R} |f(x)|dx \cdot \int_\mathbb{R} |g(x)|dx $$