7.9 Given that $ f \in L^1(\mathbb{R}) $ and
$ \int_{\mathbb{R}}\int_{\mathbb{R}} f(4x)f(x+y)dxdy =1 $,
calculate $ \int_{\mathbb{R}} f(x) dx $.
Since $ f \in L^1(\mathbb{R}) $ we can use Fubini:
$ \begin{align} 1 &= \int_{\mathbb{R}}\int_{\mathbb{R}} f(4x)f(x+y)dxdy\\ &= \int_{\mathbb{R}}f(4x)\left(\int_{\mathbb{R}} f(y+x)dy\right)dx\\ &= \int_{\mathbb{R}}f(4x)\left(\int_{\mathbb{R}} f(y)dy\right)dx\\ &= \left(\int_{\mathbb{R}} f(y)dy\right)\left(\int_{\mathbb{R}}f(4x)dx\right)\\ &= \left(\int_{\mathbb{R}} f(y)dy\right)\left(\int_{\mathbb{R}}f(y)\frac{1}{4}dy\right)\\ &= \frac{1}{4}\left(\int_{\mathbb{R}} f(y)dy\right)^2\\ &= \left(\frac{1}{2}\int_{\mathbb{R}} f(y)dy\right)^2\\ \end{align} $
Thus we see that $ \int_{\mathbb{R}} f(x) dx = \pm 2 $.
