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I am reading a proof of the Convolution Theorem and don't understand this part:

$$\int |f(z)|\int |g(z-x)| \, dx \, dz = \int|f(z)|\|g\|_1$$

Why does $\int |g(z-x)|dx = \|g\|_1$ ?

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up vote 7 down vote accepted

You have $$\int_{-\infty}^\infty|g(z-x)|\,dx.$$ Do a substitution: $u = z-x$ and $du=-dx$.

You get $$ \int_\infty^{-\infty} |g(u)| \, (-du). $$

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