Derivado de la norma $l_p$

Question

La norma $l_p$del vector $\mathbf{x}$ se define como $$\Vert \mathbf{x} \Vert_p = \left(\sum_i |x_i|^p\right)^{1/p}$$ Quiero calcular la siguiente derivada. Cualquier pista es apreciada. $$\frac{\partial}{\partial \mathbf{x}}\Vert \mathbf{x} \Vert_p $$

gracias.

Answer 1

Para <span class="math-container">$j = 1, 2, \ldots, N$</span>, por regla de cadena, tenemos <span class="math-container">$$\partial_j |\mathbf{x}|_p = \frac{1}{p} \left(\sum_i \vert x_i \vert^p\right)^{\frac{1}{p}-1} \cdot p \vert x_j \vert^{p-1} \operatorname{sgn}(x_j) = \left(\frac{\vert x_j \vert}{|\mathbf{x}|_p}\right)^{p-1} \operatorname{sgn}(x_j)$$</span>

Answer 2

Para todos los $j\in\lbrace1,\,\dots,\,n\rbrace$, \begin{align} \frac{\partial}{\partial xj}{||\mathbf{x}||}{p} &= \frac{\partial}{\partial xj} \left( \sum{i=1}^{n} |xi|^p \right)^{1/p}\ &= \frac{1}{p} \left( \sum{i=1}^{n} |x_i|^p \right)^{\left(1/p\right)-1} \frac{\partial}{\partial xj} \left(\sum{i=1}^{n} |xi|^p\right)\ &= \frac{1}{p} \left( \sum{i=1}^{n} |xi|^p \right)^{\frac{1-p}{p}} \sum{i=1}^{n} p|x_i|^{p-1} \frac{\partial}{\partial x_j} |xi|\ &= {\left[\left( \sum{i=1}^{n} |xi|^p \right)^{\frac{1}{p}}\right]}^{1-p} \sum{i=1}^{n} |xi|^{p-1} \delta{ij}\frac{x_i}{|xi|}\ &= {||\mathbf{x}||}{p}^{1-p} \cdot |x_j|^{p-1} \frac{x_j}{|x_j|}\ &= \frac{x_j |xj|^{p-2}}{{||\mathbf{x}||}{p}^{p-1}} \end{align}