Derivatives and exponential derivatives quotient operators on two variables

Question

I consider for example the following function of two variables given by $$f(x,y)=\sum_{n=0}^{+\infty}\frac{\delta_{x}^{-m} \exp(\delta_{x})}{\delta_{y}^{-m}\exp(\delta_{y})}\left(\frac{x}{y}\right)^{n}$$ Where $\delta_{x}=\frac{d}{dx}$ and $\delta_{y}=\frac{d}{dy}$ and $m$ is a positive integer. My question is: when $x < y$ can we write f as $$f(x,y)=\frac{\delta_{x}^{-m} \exp(\delta_{x})}{\delta_{y}^{-m}\exp(\delta_{y})}\sum_{n=0}^{+\infty} \left(\frac{x}{y}\right)^{n}=\frac{\delta_{x}^{-m} \exp(\delta_{x})}{\delta_{y}^{-m}\exp(\delta_{y})}g(x,y)?$$ such that $g(x,y)=\frac{1}{1-\frac{x}{y}}=\frac{y}{y-x}$. If the formula above is correct can we prove it more explicitly. Precisely, constructing a corollary for which $\frac{\delta_{x}^{-m} \exp(\delta_{x})}{\delta_{y}^{-m}\exp(\delta_{y})}$ is a well-defined operator denoted by $A_{x,y}^{m}$. I need a response, if someone have an idea.

Best regards and thank you.

Answer 1

To make sense of this, you might Fourier transform $$A_m=\frac{\delta_{x}^{-m} \exp(\delta_{x})}{\delta_{y}^{-m}\exp(\delta_{y})}=\delta_{x}^{-m} \delta_{y}^{m}e^{\delta_x-\delta_{y}},$$ which would give the operator $$\hat{A}_m=(k_y/k_x)^m e^{i(k_x-k_y)}.$$ Then $A_m$ acting on a function $f(x,y)$ could be obtained by inverse Fourier transformation, $$A_m f(x,y)=\int_{-\infty}^\infty \frac{dk_x}{2\pi}\int_{-\infty}^\infty \frac{dk_y}{2\pi}(k_y/k_x)^m \hat{f}(k_x,k_y)e^{i(x+1)k_x+i(y-1)k_y}.$$ The question whether $A_m$ is a well-defined operator would then boil down to whether this integration is well-defined. The function $\hat{f}(k_x,k_y)$ would need to vanish when $k_x\rightarrow 0$ to avoid the pole.