Si tengo la función $u: \mathbb{R}^n \longrightarrow \mathbb{R}$ suave, ¿siempre se mantiene así: $$\nabla^2(\nabla u)= \nabla(\nabla^2 u)$$
¿esto es cierto en general?
Respuesta
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Esto es válido para un $C^3$ definida en $\mathbb{R}^n$ . Para comprobarlo, observe que $$\tag{1}\Delta (\nabla u)=\Delta\Big(\frac{\partial u}{\partial x_1},\frac{\partial u}{\partial x_2},...,\frac{\partial u}{\partial x_n}\Big) =\Big(\Delta(\frac{\partial u}{\partial x_1}),\Delta(\frac{\partial u}{\partial x_2}),...,\Delta(\frac{\partial u}{\partial x_n})\Big).$$ Por otra parte, tenemos $$\tag{2}\nabla(\Delta u)=\Big(\frac{\partial }{\partial x_1}(\Delta u),\frac{\partial}{\partial x_2}(\Delta u),...,\frac{\partial}{\partial x_n}(\Delta u)\Big).$$
Desde $u$ est $C^3$ tenemos para $1\leq i\leq n$ $$\Delta(\frac{\partial u}{\partial x_i})= \sum_{j=1}^n\frac{\partial^2}{\partial x_j^2}(\frac{\partial u}{\partial x_i})=\sum_{j=1}^n\frac{\partial^3u}{\partial x_j^2\partial x_i}\\ =\sum_{j=1}^n\frac{\partial^3u}{\partial x_i\partial x_j^2} =\frac{\partial}{\partial x_i}\Big(\sum_{j=1}^n\frac{\partial^2u}{\partial x_j^2}\Big) =\frac{\partial}{\partial x_i}(\Delta u), $$ lo que implica que $$\Delta (\nabla u)=\nabla(\Delta u)$$ por $(1)$ y $(2)$ .
