Usually norm is not differentiable function. But norm has directional derivative. What is the form dirctional derivative of a norm in a direction d?
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Usually a given norm is not differentiable. What kind of norms are you thinking of? – Siminore Nov 10 '16 at 10:24
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The norm$$f(x):=\|x\|_2=\sqrt{x_1^2+x_2^2+\ldots+x_n^2}$$ is differentiable at all points $x\ne0$. One has $${\partial f\over\partial x_i}={2x_i\over 2\sqrt{x_1^2+x_2^2+\ldots+x_n^2}}={x_i\over\|x\|_2}\qquad(1\leq i\leq n)\ ,$$ and therefore $$\nabla f(x)={x\over\|x\|_2}\qquad(x\ne0)\ .$$ If $u$ is a given unit vector representing a certain direction then $$D_uf(x):=\lim_{t\to0+}{f(x+t u)-f(x)\over t}=\nabla f(x)\cdot u={x\cdot u\over\|x\|_2}\ .$$ The directional derivatives as defined here exist even at $0$: One has $$D_uf(0)=1$$ for all unit vectors $u$.
Christian Blatter
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Your result is even true if the norm comes from an inner product $\langle.,.\rangle$, i.e., $|x|=\sqrt{\langle x,x\rangle}$. Using the chain rule one finds that $$D_u|x|=\frac{\langle x,u\rangle}{|x|}.$$ And that allows a nice interpretation: it is the component of $u$ in direction $x$. – Michael Hoppe Nov 10 '16 at 13:08