Can a Non-Analytic Function Have Directional Derivatives in Every Direction?

[g1990]

Homework Statement

f(z) is a complex function (not necessarily analytic) on a domain D in C. The directional derivative is Dwf(z0)=lim(t->0) (f(z0+tw)-f(z0))/t, where w is a unit directional vector in C. There are three parts to the question:
a. Give an example of a function that is not differentiable at any point but that has directional derivatives is very direction w at every point z0. Very that your example satisfies the required properties
b. If f is differentiable at z0, show that there is a constant c such that Dwf(z0)=cw for every w
c.Assume that the real and imaginary parts of f has continuous partials. Show that if there exists a constant c such that Dwf(z0) exists and equals cw for every w, then f is differentiable at zo.

Homework Equations

Cauchy Riemann: du/dx=dv/dy and du/dy=-dv/dx

The Attempt at a Solution

for a, I know that my function can't satisfy the CR eqns, b/c it can't be differentiable, but I don't know how to plug a specific function into my directional derivative definition

 

[Dick]

Ok, here's a practice function. Suppose f(x+iy)=x. What's the directional derivative of f at z0=0 in the direction w=1+0i? How about in the direction w=0+1i? Just plug z0 and w into your definition of directional derivative.

 

[g1990]

the first one would be 1 and the second one zero.

So in general, for this function, Dwf(z0)=f(z0+t(u+iv))-f(z0)/t=(Re(z)+ut-Re(z))/t=u
where u+iv is the unit directional vector. So that would be an example of a function that is not differentiable anywhere but which has directional derivatives everywhere, right?

 

[Dick]

the first one would be 1 and the second one zero.

So in general, for this function, Dwf(z0)=f(z0+t(u+iv))-f(z0)/t=(Re(z)+ut-Re(z))/t=u
where u+iv is the unit directional vector. So that would be an example of a function that is not differentiable anywhere but which has directional derivatives everywhere, right?

Sure. f(x+iy)=x doesn't satisfy Cauchy-Riemann, even though it has directional derivatives everywhere.

 

[g1990]

wow thanks! for part b, what I was thinking is that if f is differentiable at some z0, any two given directional derivatives must be equal. I'm not sure how to spin that into there exists a c such that every directional derivative equals cw though. One thing is that c should be the gradient at z0, but we haven't officially learned that yet, so I shouldn't have to use it.

 

[Dick]

wow thanks! for part b, what I was thinking is that if f is differentiable at some z0, any two given directional derivatives must be equal. I'm not sure how to spin that into there exists a c such that every directional derivative equals cw though. One thing is that c should be the gradient at z0, but we haven't officially learned that yet, so I shouldn't have to use it.

Well, no, any two directional derivatives won't be equal. If Dwf=cw then it depends on w. This isn't the complex derivative. Try expressing the directional derivative in terms of the partial derivatives of f=u+iv. That way you can use Cauchy-Riemann.

 

[g1990]

Okay, so if f(x,y)=u(x,y)+iv(x,y) and w=w1+iw2 is the unit directional number, then
the directional derivative is
lim(t->0)[u(x+tw1,y+tw2)-u(x,y)]/t+[v(x+tw1,y+tw2)-v(x,y)]/t
I'm not sure how to spit that up even more to extract the partials, or to get w out of the equation.

 

[Dick]

du(x,y)/dt=du(x,y)/dx*dx/dt+du(x,y)/dy*dy/dt. That's basically the gradient formula. Some of those 'd's should be partial. And don't forget the 'i' in front of v(x,y).

 

[g1990]

okay so, let me get this straight. I have that the directional derivative equals
du(x,y)/dt+idv(x,y)/dt=du(x,y)/dx*dx/dt+du(x,y)/dy*dy/dt+i(dv(x,y)/dx*dx/dt+dv(x,y)/dy*dy/dt). Now, I know that du(x,y)/dx is just du/dx, one of the terms of the CR eqns, and likewise for the other three. I suspect that dx/dt=w1 and so on, but I'm not sure why. Also, should I be putting another i in front of all of the dy/dt's?

 

[g1990]

Can someone please help me with the last bit of this proof?

 

[Dick]

okay so, let me get this straight. I have that the directional derivative equals
du(x,y)/dt+idv(x,y)/dt=du(x,y)/dx*dx/dt+du(x,y)/dy*dy/dt+i(dv(x,y)/dx*dx/dt+dv(x,y)/dy*dy/dt). Now, I know that du(x,y)/dx is just du/dx, one of the terms of the CR eqns, and likewise for the other three. I suspect that dx/dt=w1 and so on, but I'm not sure why. Also, should I be putting another i in front of all of the dy/dt's?

You are giving up too quickly. You are almost there. Sure dx/dt=w1, x(t)=x0+tw1, dx(t)/dt=w1. No, you don't need any more i's. Why would you? Now just use Cauchy-Riemann. Put c1=du/dx=dv/dy and c2=du/dy=(-dv/dx). Can you factor (w1+iw2) out?

 

[g1990]

okay so- I have that dx/dt=w1 and dy/dt=w2.
Then,
du(x,y)/dt+dv(x,y)/dt=du(x,y)/dx*dx/dt+du(x,y)/dy*dy/dt+i(dv(x,y)/dx*dx/dt+dv(x,y)/dy*dy/dt)=du/dx*w1+du/dy*w2+idv/dx*w1+idv/dy*w2=(du/dx+idv/dx)w1+(-idu/dy+dv/dy)iw2=(c1+ic2)w1+(c1+ic2)iw2=(c1+ic2)(w1+iw2)

so my constant is just c1+ic2=du/dx+idv/dx. That seems right to me- is that what you meant?

 

[Dick]

okay so- I have that dx/dt=w1 and dy/dt=w2.
Then,
du(x,y)/dt+dv(x,y)/dt=du(x,y)/dx*dx/dt+du(x,y)/dy*dy/dt+i(dv(x,y)/dx*dx/dt+dv(x,y)/dy*dy/dt)=du/dx*w1+du/dy*w2+idv/dx*w1+idv/dy*w2=(du/dx+idv/dx)w1+(-idu/dy+dv/dy)iw2=(c1+ic2)w1+(c1+ic2)iw2=(c1+ic2)(w1+iw2)

so my constant is just c1+ic2=du/dx+idv/dx. That seems right to me- is that what you meant?

Yes, the constant c1+ic2 is the complex derivative.