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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 $(x^2+y^2)^{1/3}$ but here the directional derivatives are infinite.

Any ideas would be welcomed, thanks~

Guy
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Staley89
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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 $x_0$ basically linearly as $x$ varies from $0$ to $x_0^2$, and then exponentially decays back down to 0 as $x \to \infty$. Clearly this can be made into a continuous family of functions as $x_0$ varies, with the limit of the zero function as $x_0$ approaches zero. Also the derivatives become arbitrarily large at $0$ as $x_0$ approaches 0. So just define your two-argument function this way, with the family of functions extended radially from the origin, with $\sin \theta$ being the value of $x_0$ for a given point at angle $\theta$ from the origin. The directional derivatives will be unbounded at 0 as $\sin \theta \to 0$ but be zero (the derivative of the zero function) for $\sin \theta = 0$.

user2566092
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  • Hi,that`s very insightful. I was wondering how did you get the intuition to approach this problem? – Staley89 Mar 06 '14 at 10:56
  • @user80649 Since your function needed to be continuous at the origin, I thought about how to come up a continuous family of functions that are 0 at the origin and that approach the zero function as some parameter approaches 0, but have derivatives at 0 that approach infinity as the parameter approaches 0. Then that just left the simple issue of defining this family of functions radially about the origin. – user2566092 Mar 10 '14 at 20:11