Are there any glaring problems with this proof?

Question

kainui · Answer

$$\nabla^2 f(x,y) = \frac{\partial^2 f}{\partial x^2} +  \frac{\partial^2 f}{\partial y^2} $$
By the definition of derivative,

$$ \frac{\partial f}{\partial x} =\lim_{r \to 0} \frac{f(x+\frac{r}{2}) -f(x-\frac{r}{2})}{r}$$

Applied twice gives:

$$\frac{\partial^2 f}{\partial x^2} = \lim_{r \to 0} \frac{2}{r^2} \left( \frac{f(x+r) +f(x-r)}{2}-f(x) \right)$$

None of that is anything new, I literally pulled it out of a PDE textbook. Next this is what I've done is combined the limits: (just for the sake of brevity I've contracted $f(x+r,y) = f(x+r)$ since it's clear what variable we're looking at, however it is a two variable function the whole time):

$$\nabla^2 f(x,y)= \lim_{r \to 0} \frac{4}{r^2} \left( \frac{f(x+r) +f(x-r)+f(y+r)+f(y-r)}{4}-f(x,y) \right)$$

Geometrically what we're looking at is the average of f evaluated at 4 nearby points minus f at that point which gives us some interpretation of curvature, flow, acceleration, whatever but I'm not interested in this right now.

At this point I want to derail into the complex plane, since I can rewrite this as a function of a single complex variable,

$$\Delta f(z) = \lim_{r \to 0} \frac{4}{r^2}\left(\frac{\displaystyle{\sum_{n=0}^3} f(z+ri^n)}{4} - f(z)\right)$$

We could gneeralize this further and put an arbitrary phase factor on the $ri^ne^{i \theta}$ since it shouldn't matter where these 4 points are in space. It won't really add anything new though either.

For small values of $\epsilon$ of once differentiable functions the first derivative is this since epsilon is so small as to be nearly 0. $\epsilon^2=0$:

$$\frac{f(x+\epsilon)-f(x)}{\epsilon} = f'(x)$$
We can algebraically rearrange to get the first two terms of the taylor series,

$$f(x+\epsilon) = f(x) + \epsilon f'(x)$$

Now we can use this since r is also small in our limit to 0:

$$f(z+ri^n)= f(z) + ri^n f'(z)$$

If we plug this into our sum we end up with:


$$\Delta f(z) = \lim_{r \to 0} \frac{4}{r^2}\left(\frac{\displaystyle{\sum_{n=0}^3} f(z) + ri^n f'(z)}{4} - f(z)\right)$$
$$\Delta f(z) = \lim_{r \to 0} \frac{4}{r^2}\left(\frac{\displaystyle{4f(z)+rf'(z)\sum_{n=0}^3}i^n}{4} - f(z)\right)$$
After realizing $\sum_{n=0}^3i^n=0$ we then realize everything left simplifies to 0 and end up with

$$\Delta f(z) = 0$$

kainui · Answer

uhh I sorta just typed this up as I went and realize it's probably not the most readable thing in the world so I'm mainly concerned about if my assumption of once differentiability of f(z) ends up being some kind of circular reasoning about leading to the result $\Delta f(z) = 0$.

kainui · Answer

I wouldn't've bothered to put this if the end result wasn't true, so that's what I'm interested in cause I mean to apply this in more general cases.

bobo-i-bo · Answer

I haven't looked though it all, but I believe that "Next this is what I've done is combined the limits" is not a valid thing to do. This is because, if you combine it as you have, you are in fact calculating the directional derivative in the direction (1,1). Also, the fact that the existence of partial derivatives does not imply that f is differentiable at the point also justifies why you are incorrect to do this.

bobo-i-bo · Answer

I can explain my answer above a bit better if you want :P