# How to prove that the composition $T_{b_2} \circ T_{b_1}$ of the convolution operators $T_{b_2}$ and $T_{b_1}$ is the convolution operator $T_b$?

Suppose $b_1, b_2 \in l^2(Z_N)$. How to prove that the composition $T_{b_2} \circ T_{b_1}$ of the convolution operators $T_{b_2}$ and $T_{b_1}$ is the convolution operator $T_b$ with $b=b_2 \cdot b_1$? Hint use previous question where I asked: How to prove that convolution is associative? and also $b_2 \cdot b_1(m) = \sum_{n=0}^{N-1} b_2(m-n)b_1(n)$.

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tagged it as homework... –  Djaian Nov 22 '10 at 12:23
Just compute it. –  Rasmus Nov 22 '10 at 18:55
I want to thanka you for encouraging words which were helpful in solving this problem. –  laovultai Nov 23 '10 at 9:12

$T_{b_2} \circ T_{b_1} = T_{b_2}(T_{b_1}(z))= T_{b_2}(b_1*z)=b_2*(b_1*z)=(b_2*b_1)*z = b*z$.