“…Qpyqq, which implies that ∆ hm¨¨¨∆h2 ∆ h1 P pxq and ∆´c´1 m hm¨¨¨∆´c´1 2 h2 ∆´c´1 1 h1 Qpyq are both constant functions. In particular, taking h m`1 P R d and applying the operator ∆ phm`1,0q to both sides of the equation, we get ∆ hm`1 ∆ hm¨¨¨∆h2 ∆ h1 P pxq¨1pyq " 0 and, analogously, if we apply ∆ p0,hm`1q to both sides of the equation, we get 1pxq¨∆ hm`1 ∆´c´1 m hm¨¨¨∆´c´1 2 h2 ∆´c´1 1 h1 Qpyq " 0 Thus ∆ hm`1 ∆ hm¨¨¨∆h2 ∆ h1 P pxq " ∆ hm`1 ∆´c´1 m hm¨¨¨∆´c´1 2 h2 ∆´c´1 1 h1 Qpyq " 0 for all h 1 ,¨¨¨, h m`1 in R d (with equality in the sense of DpR d q 1 ), and the result follows from the corresponding Fréchet's type theorem for distributions, which is known (see again [1], [2], [5] or [7]). Note that, if we assume Q " 0 in the hypotheses of the theorem, then equation (16) takes the form 0 " p∆ hm¨¨¨∆h2 ∆ h1 P pxqq¨1pyq, which leads to ∆ hm¨¨¨∆h2 ∆ h1 P pxq " 0 and henceforth, in that case, P is a polynomial of degree ď m´1.…”