this function has to be continuous at the origin,
have finite directional derivatives there,but they are not bounded.
(meaning for some vectors v with |v|=1 the directional derivatives at 0 can be as large as we want)
I first thought about (x2+y2)1/3 but here the directional derivatives are infinite.
Any ideas would be welcomed,
thanks~
Answer
Here's a geometric answer that should lead to a formula after some thought. Imagine the graph of a function f(x) which changes from 0 to x0 basically linearly as x varies from 0 to x20, and then exponentially decays back down to 0 as x→∞. Clearly this can be made into a continuous family of functions as x0 varies, with the limit of the zero function as x0 approaches zero. Also the derivatives become arbitrarily large at 0 as x0 approaches 0. So just define your two-argument function this way, with the family of functions extended radially from the origin, with sinθ being the value of x0 for a given point at angle θ from the origin. The directional derivatives will be unbounded at 0 as sinθ→0 but be zero (the derivative of the zero function) for sinθ=0.
No comments:
Post a Comment