How to prove that convolution is associative and distributive with “plus”












1














The convolution is defined as:
$$int_{-infty}^{+infty}f(x-t)g(t)dt=f*g(x)$$
I want to prove the associativity and distributivity of it:$$f*(g*h)=(f*g)*h$$
$$f*(g+h)=f*g+f*h$$



I expand them but I find that it seems difficult to process especially how to change the order of integration.



Thank you a lot!






calculus integration







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    1














    The convolution is defined as:
    $$int_{-infty}^{+infty}f(x-t)g(t)dt=f*g(x)$$
    I want to prove the associativity and distributivity of it:$$f*(g*h)=(f*g)*h$$
    $$f*(g+h)=f*g+f*h$$



    I expand them but I find that it seems difficult to process especially how to change the order of integration.



    Thank you a lot!






    calculus integration







    share|cite|improve this question

























      1












      1








      1







      The convolution is defined as:
      $$int_{-infty}^{+infty}f(x-t)g(t)dt=f*g(x)$$
      I want to prove the associativity and distributivity of it:$$f*(g*h)=(f*g)*h$$
      $$f*(g+h)=f*g+f*h$$



      I expand them but I find that it seems difficult to process especially how to change the order of integration.



      Thank you a lot!






      calculus integration







      share|cite|improve this question













      The convolution is defined as:
      $$int_{-infty}^{+infty}f(x-t)g(t)dt=f*g(x)$$
      I want to prove the associativity and distributivity of it:$$f*(g*h)=(f*g)*h$$
      $$f*(g+h)=f*g+f*h$$



      I expand them but I find that it seems difficult to process especially how to change the order of integration.



      Thank you a lot!






      calculus integration




      calculus integration






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jul 1 '14 at 7:30









      zgzhen

      1013




      1013






















          1 Answer
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          Hint:

          For the first try writing out the integral then using a simple substitution like $u=x-t$



          The second one follows from the fact that the integral is a linear operator. So,
          begin{align*}
          f*(g+h) &= int_{-infty}^infty f(x-t)(g(t)+h(t)),dt\
          &=int_{-infty}^infty f(x-t) g(t) , dt +int_{-infty}^infty f(x-t)h(t) ,dt
          end{align*}






          share|cite|improve this answer























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            1 Answer
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            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            Hint:

            For the first try writing out the integral then using a simple substitution like $u=x-t$



            The second one follows from the fact that the integral is a linear operator. So,
            begin{align*}
            f*(g+h) &= int_{-infty}^infty f(x-t)(g(t)+h(t)),dt\
            &=int_{-infty}^infty f(x-t) g(t) , dt +int_{-infty}^infty f(x-t)h(t) ,dt
            end{align*}






            share|cite|improve this answer




























              4














              Hint:

              For the first try writing out the integral then using a simple substitution like $u=x-t$



              The second one follows from the fact that the integral is a linear operator. So,
              begin{align*}
              f*(g+h) &= int_{-infty}^infty f(x-t)(g(t)+h(t)),dt\
              &=int_{-infty}^infty f(x-t) g(t) , dt +int_{-infty}^infty f(x-t)h(t) ,dt
              end{align*}






              share|cite|improve this answer


























                4












                4








                4






                Hint:

                For the first try writing out the integral then using a simple substitution like $u=x-t$



                The second one follows from the fact that the integral is a linear operator. So,
                begin{align*}
                f*(g+h) &= int_{-infty}^infty f(x-t)(g(t)+h(t)),dt\
                &=int_{-infty}^infty f(x-t) g(t) , dt +int_{-infty}^infty f(x-t)h(t) ,dt
                end{align*}






                share|cite|improve this answer














                Hint:

                For the first try writing out the integral then using a simple substitution like $u=x-t$



                The second one follows from the fact that the integral is a linear operator. So,
                begin{align*}
                f*(g+h) &= int_{-infty}^infty f(x-t)(g(t)+h(t)),dt\
                &=int_{-infty}^infty f(x-t) g(t) , dt +int_{-infty}^infty f(x-t)h(t) ,dt
                end{align*}







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 20 '18 at 7:40









                Kei

                347




                347










                answered Jul 1 '14 at 7:39









                Millardo Peacecraft

                9101619




                9101619






























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