Does negative distributive property of convolution over cross correlation holds?

Let $\star$ denote convolution binary operation and $\otimes$ denote cross correlation binary operation between two functions.

Let $f,g,h$ be functions. Does this negative distribution property holds? $$f\star ( g\otimes h) = - (f \star g) \otimes h$$

Edit : It is given that $f$ is an odd function.

Definitions :

$$f\star g = \int \limits_{-\infty}^\infty f(x)g(\tau-x)dx$$ and $$f\otimes g = \int \limits_{-\infty}^\infty f(x)g(x+\tau)dx$$

• wouldn't that be called a negative associative property?
– robjohn
Oct 16, 2013 at 20:59
• Yes. Certainly not distributive. Oct 16, 2013 at 22:45

Let $$\bar{h}(x)=h(-x)\tag{1}$$ then \begin{align} g\otimes h(x) &=\int g(t)h(t+x)\,\mathrm{d}t\\ &=\int g(-t)h(-t+x)\,\mathrm{d}t\\ &=\int\bar{g}(t)h(x-t)\,\mathrm{d}t\\ &=\bar{g}\star h(x)\tag{2} \end{align} Furthermore \begin{align} g\otimes h(x) &=\int g(t)h(t+x)\,\mathrm{d}t\\ &=\int g(t-x)h(t)\,\mathrm{d}t\\ &=h\otimes g(-x)\\ &=\overline{h\otimes g}(x)\tag{3} \end{align} and \begin{align} \overline{f\star g}(x) &=f\star g(-x)\\ &=\int f(t)g(-x-t)\,\mathrm{d}t\\ &=\int\bar{f}(-t)\bar{g}(x+t)\,\mathrm{d}t\\ &=\int\bar{f}(t)\bar{g}(x-t)\,\mathrm{d}t\\ &=\bar{f}\star\bar{g}(x)\tag{4} \end{align} Therefore, \begin{align} (f\star g)\otimes h &=\overline{f\star g}\star h\\ &=\bar{f}\star\bar{g}\star h\\ &=\bar{f}\star(g\otimes h)\tag{5} \end{align} Since $f$ is an odd function, $(5)$ becomes $$(f\star g)\otimes h=-f\star(g\otimes h)\tag{6}$$
• I forgot to mention that $f$ is an odd function. I guess it holds now. Actually $f(t )=1/t$, a kernel of Hilbert transform, I was expecting this result to prove my answer on the metric question. Please correct me if I am wrong.Thanks for the answer. Oct 16, 2013 at 22:40
• @RajeshD: It would be nice to add that $f$ is odd to the question.