This should be very easy, but I can't get my head around it: given $1\leq p < \infty$, and a point x with $\|x\|_p = 1$, how do I get a (or the unit, or any) vector which is perpendicular to the $\ell_p$-circle of all $\{s : \|s\|_p = 1$} at that point $x$? I'm more than happy with even just 2D - I tried parameterizing the circle of unit $\ell_p$ norm, then parameterizing whatever is perpendicular to it by setting the scalar product to zero, and then filling in a certain point, but somehow I never end up with a meaningful equation :/
Clearly, for $p=1$ it is always on the diagonal. For $p=2$, it always is collinear with the point $x$. But what about all other cases of $p$ - what relation holds between $x$ and the perpendicular vector at that point? What's the best approach even - parameterization and simple analysis, or tangent and cotangent spaces, differential analysis, or how would one even start?
UPDATE:
Thanks a lot to anon and Leonid. The gradient of the norm was simply what i was looking for. Since I don't even care about the length and am in the positive orthant, I can simply take $y_i = x_i^{p-1}$. Also, I quickly verified in Python/Reinteract2, quick code is below, also image for $p=3.82$. It seems to do exactly what I needed, and is correct for $p=1$ and $p=2$ by intuition. Thanks again.
from refigure2 import * from numpy import * import math ps = linspace(1.2,6,12) t = linspace(0,1, 21) x = zeros(len(t)) y = zeros(len(t)) dx = zeros(len(t)) dy = zeros(len(t)) for i in xrange(len(ps)): p = ps[i] for i in xrange(len(t)): x[i] = t[i] y[i] = 1.0-t[i] pnorm = (x[i]**p+y[i]**p)**(1.0/p) x[i] = x[i]/pnorm y[i] = y[i]/pnorm dx[i] = x[i]**(p-1) dy[i] = y[i]**(p-1) with figure() as f: ax = f.add_subplot(111, aspect='equal') ax.plot(x,y) for i in xrange(len(t)): ax.plot([x[i],x[i]+dx[i]],[y[i],y[i]+dy[i]],c='r') ax.axis([0,1.1,0,1.1]) f
And here a Resulting plot sample.
UPDATE2: Ah, i see - I forgot to mention that i'm only interested in the positive orthant. Thanks for clarifying, leonid.