I'm familiar with this definition of convolution :

$$f*g(x)=\int f(x-y)g(y) dy$$

But anyone help me see the link between this one and the two definitions hereafter? :

Definition 1 : $$f*g(A)=\int_Xf(A-x)dg(x)$$

where $A-x=\{a-x|a\in A\}$

Definition 2: $$f*g(A)=\int\int 1_A (x+y)dg(x)df(y)$$

thank you


Your definition 1 makes sense for measures: $$\mu*\nu(E) = \int\mu(E-y)d\nu(y).$$ For each $y$, $E-y$ is a set, $\mu(E-y)$ is a number and $y\mapsto\mu(E-y)$ is a function that can be integrated.

For your definition 2, think in the previous formula in the case $$\mu(A) = \int_A f(x)\,dx,\qquad\nu(A) = \int_A g(x)\,dx.$$

  • $\begingroup$ thank you. but what does $E-y$ mean when $E$ is a set and $y$ a number? how do you get the definition 2? i mean the proof? $\endgroup$ – Dave ddd Mar 3 '16 at 14:50
  • $\begingroup$ $E-y$ is a translate of the set $E$ (remember that the context is $\Bbb R^n$). For understanding the definition 2, try with a simple example, like $A =$ an interval in $\Bbb R$. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 3 '16 at 15:01

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