If a function $f$ is analytic and $f(x,y)=f(x,-y)=f(x+y,y)$ then the function depends on $y$ alone

Question

Let $f:\mathbb{R^2}\rightarrow\mathbb{R}$ a function such that $f(x,y)=f(x,-y)=f(x+y,y)$.

I need to show that if $f$ is an analytic function, then $f$ depends only on $y$

What I did was this:

Given that $f$ is analytic, the Taylor series at the point $(x+h, y+k)$ is equal to $f(x+h,y+k)$, so:

$f(x+h,y+k)=f(x,y)+hf_x(x,y)+kf_y(x,y)+\frac12(h^2f_{xx}+2hkf_{xy}+k^2f_{yy})+...$

for $k=0$ and $h=y$ we'll get:

$f(x+y,y)=f(x,y)+yf_x(x,y)+\frac12(y^2f_{xx}(x+y,y)+...$

I'll define $g:\mathbb{R^2}\rightarrow\mathbb{R}$ by $g(x,y)=f(x+y,y)-f(x,y)$. Based on the initial data $g(x,y)\equiv0$, and so:

$0\equiv f(x+y,y)-f(x,y)=yf_x(x,y)+\frac12y^2f_{xx}(x+y,y)+...$

It follows then that $f_x \equiv f_{xx} \equiv f_{xxx} \equiv ... \equiv0$

What I want now is to show that every mixed derivative, i.e $f_{xy},f_{xyy}$, etc is also equivalent to $0$, but I don't how to do this, or really why this would be true.

Edit: Because $f_{xy}=f_{yx}$, all the mixed derivatives would be taking the derivative of a constant function (where the constant is just $0$), And so they'll all be equivalent to the $0$ function.

And then I'll get that $f(x+h,y+k)=f(x,y)+kf_y+\frac12k^2f_{yy}+...$ is an expression that doesn't depend on $h$ and then $f$ is supposedly a function that only depends on $y$. But I don't really understand why this means that $f$ is only dependent on $y$. I can still have $x$'s in the partial derivatives of $f$ according to $y$, no?

Presumably because $f(x,y)=f(x,-y)$ it'll then mean that $f$ only depends on $|y|$

Answer 1

This is probably not the easiest approach, but it should work. $$0=f(x+y,y)-f(x,y)=\sum_{n=1}^\infty \frac{\partial_x^n f(x,y)}{n!} \, y^n = \sum_{n=1}^\infty \frac{y^n}{n!} \sum_{m=0}^\infty \frac{\partial_y^m\partial_x^n f(x,y)\big|_{y=0}}{m!} \, y^m \\ =\sum_{k=1}^\infty y^k \sum_{\substack{n=1 \\ m=0\\n+m=k}}^\infty \frac{\partial_y^m\partial_x^n f(x,y)\big|_{y=0}}{n!m!} = \sum_{k=1}^\infty y^k \sum_{n=1}^{k} \frac{\partial_y^{k-n}\partial_x^n f(x,y)\big|_{y=0}}{n!(k-n)!}$$ This implies $$\sum_{n=1}^{k} \frac{\partial_y^{k-n}\partial_x^n f(x,y)\big|_{y=0}}{n!(k-n)!}=0 \, . \tag{1}$$

For $k=1$ we have $$\partial_x f(x,0)=0 \quad \Rightarrow \quad \partial_x^n f(x,0)=0$$ for all $n\in\mathbb{N}$. With this the $n=k$ term always vanishes. This then gives for $k=2$ $$\partial_y\partial_x f(x,y)\big|_{y=0} = 0 \quad \Rightarrow \quad \partial_y\partial_x^n f(x,y)\big|_{y=0}=0 \quad \forall n\in\mathbb{N}.$$ It then follows that the $n=k-1$ term always vanishes. Now suppose $$\partial_y^m\partial_x^n f(x,y)\big|_{y=0}=0 \quad \forall n\in\mathbb{N} \, \text{ and } \, 0\leq m\leq k-2$$ by which (1) reduces to $$\partial_y^{k-1}\partial_x f(x,y)\big|_{y=0}=0 \quad \Rightarrow \quad \partial_y^{k-1}\partial_x^n f(x,y)\big|_{y=0}=0 \quad \forall n \in \mathbb{N} \, .$$ As a result $$\partial_y^{m}\partial_x^n f(x,y)\big|_{y=0}=0 \quad \forall n\in\mathbb{N} \, \text{ and } \, \forall m \in \mathbb{Z}_{\geq 0} \, . \tag{2}$$ Now using (2) $$f(x,y)=\sum_{m=0}^\infty \frac{\partial_y^m f(x,y)\big|_{y=0}}{m!} \, y^m \\ \Rightarrow \quad \partial_x f(x,y)=\sum_{m=0}^\infty \frac{\partial_y^m \partial_x f(x,y)\big|_{y=0}}{m!} \, y^m = 0$$ for all $x$ and $y$, which is the required result.

Just wondering: The evenness in $y$ wasn't used here. Why did you think it was necessary?

Answer 2

Remark
Is this result true for general functions $f : \mathbb R^2 \to \mathbb R$, not necessarily analytic? no.
For continuous functions $f$? no.
For differentiable functions? no.
For $C^n$ functions? no.
For $C^\infty$ functions? no.

Example.
$$ f(x,y) = \begin{cases} e^{-1/y^2}\cos\frac{2\pi x}{y},\qquad && y \ne 0 , \\ 0,\qquad && y=0 .\end{cases} $$ Now $f$ is $C^\infty$ and satisfies: $f(x,y)=f(x,-y)=f(x+y,y)$.
But $f(x,y)$ does not depend on $y$ alone.