I have an equation
$$y_\mu y_\nu=By^2\delta_{\mu\nu}+Cy^2{p_\mu p_\nu \over p^2}$$.
Can I take convolution of this equation with respect to ${p_\mu p_\nu} \over p^2 $? On doing so, I get
$\Sigma_{\mu \nu }{{p_\mu p_\nu} \over p^2} y_\mu y_\nu=By^2\Sigma_{\mu \nu }{{p_\mu p_\nu} \over p^2}\delta_{\mu\nu}+Cy^2\Sigma_{\mu \nu }{{p_\mu p_\nu} \over p^2}{p_\mu p_\nu \over p^2}$
this gives
${(p.y)^2 \over p^2}=By^2\Sigma_{\mu \nu }{{p_\mu p_\nu} \over p^2}\delta_{\mu\nu}+Cy^2\Sigma_{\mu \nu }{{p_\mu p_\nu} \over p^2}{p_\mu p_\nu \over p^2}$
this gives
$$B+C =cos^2(\theta)$$
I know about convolution but only little know about discrete convolution, So my doubt is Can I take convolution as shown above? If so, My derived relation is true or not?