# Thread: differentiable but not analytic function

1. ## differentiable but not analytic function

Questiion: show that $x^3 + 3xy^2 -3x + i(y^3+3x^2y-3y)$ is differentiable on the coordinate axis but nowhere analytic.

Attempt: $u_{x} =3x^2 + 3y^2-3$ and $v_{y}= 3y^2+3x^2-3$

so for $u_{x} = v_{y}$ , $3y^2+3x^2-3 = 3x^2+3y^2-3$, which is true.

Is this enough to show that it is diffentiable? Also what does on the co-ordinate axis mean?

Then to show it wasn't analytic i did, $u_{y} = 6xy$ , $-v_{x}=-6xy$

for $u_{y}=-v_{x}$, $x=y=0$ is the only way. Is this enough?

2. ## Re: differentiable but not analytic function

Hi, linalg123.

First, nice work so far. Looking to the Cauchy-Riemann Equations is a good idea.

so for

$u_{x} = v_{y}$ ,

$3y^2+3x^2-3 = 3x^2+3y^2-3$, which is true.

Is this enough to show that it is diffentiable?
this is not enough. Satisfying only one of the Cauchy-Riemann Equations is NOT enough to prove differentiability. Satisfying BOTH equations at a point is part of what is needed to deduce differentiability. We will further discuss what is needed to conclude differentiability at a point below.

You used the term "analytic" in your previous post, have you seen the definition for holomorphic before?

A function $f(z)$ is said to be holomorphic at $z_{0}$ if it is differentiable at every point in neighborhood of $z_{0}$.

The key here is that the function is differentiable not just at $z_{0}$, but at EVERY point in some neighborhood around $z_{0}$.

Now, it turns out that a function is holomorphic at a point if and only if it is analytic at that point.

So, if we can show that your function is not holomorphic anywhere, then (by the above if and only if) it is not analytic anywhere. This is what we will outline below.

The Cauchy-Riemann Equations are a necessary condition for complex differentiability; meaning that if $f(z)$ is differentiable at $z_{0}$, then $f$ satisfies the Cauchy-Riemann Equations at $z_{0}$.

However, there is a partial converse to the previous statement. It says that:

Theorem

If $u$ and $v$ are differentiable at $z_{0}=x_{0}+iy_{0}$ and they satisfy the Cauchy-Riemann Equations at $z_{0}$, then $f(z)$ is differentiable at $z_{0}$.

First, we observe that $u$ and $v$ are polynomials and so are differentiable everywhere.

Now you noted that $u_{x}=v_{y}$ for all $(x,y)$. However, You also observed that $u_{y}=-v_{x}$ only along the coordinate axes (i.e. when $x=0$ or when $y=0$). So the Cauchy-Riemann Equations are only satisfied along the coordinate axes. By our above Theorem, the function $f$ is differentiable at points along the coordinate axis.

Now here's the key: We only have that $f$ is differentiable at points along the axes, we do NOT have that $f$ is differentiable on neighborhoods/little disks of such points. Since we can't find entire disks on which the function is differentiable that means the function is not holomorphic (by definition of holomorphic - see above) and so is not analytic.

For example, the point $z_{0}=i$ sits on the imaginary axis, because $x=0$ there. Since $z_{0}=i$ sits on the coordinate axis, $f$ is differentiable at this point (by our above argument). However, if we draw a little disk around $z_{0}=i$ (NO MATTER HOW SMALL) we will include points that do not sit on the coordinate axis. Since the points that don't sit on the coordinate axis fail to satisfy the Cauchy-Riemann Equations (specifically such points WON'T satisfy $u_{y}=-v_{x}$), $f$ is not differentiable at them.

To sum up, the important feature that makes the given function NOT analytic/holomorphic is the fact that we can't say $f$ is differentiable on ENTIRE neighborhoods surrounding points, all we can say is that $f$ is differentiable at specific points.

Does this help clear things up?

Good luck!

3. ## Re: differentiable but not analytic function

Very succinctly, thanks so much.

I wish our teachers were more like this.

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# give example of function which is differentiable but not analytic

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