# Convolution of $L_1$ functions is $L_1$

Let $f,g\in L_1(\mathbb R, m)$ where m is the Lebesgue measure.

Prove that:

a) $f(x-t)g(t) \in L_1(\mathbb R, m)$ as a fuction of $t$ almost all $x$

b) $h\in L_1(\mathbb R, m)$ where $h=(f \ast g)(t)=\int_{\mathbb R}f(x-t)g(t)dm(t)$

c)$||h||_{L_1}\le ||f||_{L_1}||g||_{L_1}$

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## 1 Answer

Hint: Use Fubini on $\mathbb{R}^2$ for the function $G(x,t)=|f(x-t)g(t)|$.

If you need more help, see p.16 here: http://people.math.gatech.edu/~heil/6338/summer08/section4c_convolve.pdf

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