Appendix A:

In the deconvolution problem, for a Gaussian additive noise and in the case of real functions we must compute the gradient of the function

$$J(x)=\frac12 \int^{+\infty}_{-\infty}(y-h\otimes x)^2{\rm d}u$$ (A1)

y, h, x depends on u.

Considering for the variable x, an increase "$\rho$'' in the general direction "s'', we have:

$$J(x+\rho s)$$

$$=\frac 12\int^{+\infty}_{-\infty}\left[y^2-2y(h\otimes (x+\rho s))+(h\otimes (x+\rho s))^2\right].$$ (A2)

Expanding this function and neglecting the second order terms, we have:

$$J(x+\rho s)=J(x)+\rho\int^{+\infty}_{-\infty}(h\otimes s)[(h\otimes x)-y]{\rm d}u$$ (A3)

this expression must be identical to: $J(x+\rho s)=J(x)+\rho J'(x,s)$.

With: $H(u)=[(h\otimes x)-y](u)$, expanding the convolution product between h and s, we have:

$$J'(x,s)=\left[\int^{u=+\infty}_{u=-\infty}H(u)\int^{t=+\infty}_{t=-\infty} s(t)h(u-t){\rm d}t {\rm d}u\right]$$ (A4)

or, inverting the integrals

$$J'(x,s)=\left[\int^{t=+\infty}_{t=-\infty}s(t)\int^{u=+\infty}_{u=-\infty} H(u)h(u-t){\rm d}u {\rm d}t\right]$$ (A5)

letting: z=u-t, we have:

$$J'(x,s)=\left[\int^{t=+\infty}_{t=-\infty}s(t) \int^{z=+\infty}_{z=-\infty}H(z+t)h(z){\rm d}z {\rm d}t\right]$$ (A6)

which can be written in the form:

$$J'(x,s)=\left[\int^{t=+\infty}_{t=-\infty}s(t)\{h(-t)\otimes H(t)\}{\rm d}t\right].$$ (A7)

Using the expression of the derivative of J(x) at x in the direction s we obtain:

$$\nabla J(x)=h(-u)\otimes \{[h(u)\otimes x(u)]-y(u)\}\cdot$$ (A8)