I have an other question for a function different to the example given before in the link below: Exponential derivative operator and continuous functions We define for instance a function as: $$H(y)=\frac{1}{y^{n}(d/dy)^{n}e^{(d/dy)y(d/dy)}T(y)}$$ where $T$ is not an operator but a function and $n$ is an integer. What can we say about the function and the operator? Is it possible to have $$H(y)=e^{(d/dy)y(d/dy)}(d/dy)^{n}y^{n}\frac{1}{T(y)}?$$
1 Answer
You haven't really defined $(d/dy)^{1} $, but let's pick, for instance, $(d/dy)^{1} f(y) = \int_{0}^{y} dy' f(y')$. Then, your two expressions for $H(y)$ in general are not equal. A simple counterexample is the case $T(y)=y$. In this case, \begin{eqnarray} \frac{1}{y^{n} (d/dy)^{n} e^{(d/dy)y(d/dy)} T(y)} &=& \frac{1}{y^{n} (d/dy)^{n} (y1)} \\ &=& \frac{1}{y^{n} (y^{n+1}/(n+1)! y^n /n!)} \\ &=& \frac{(n+1)!}{y(n+1)} \end{eqnarray} but $$ e^{(d/dy)y(d/dy)} (d/dy)^{n} y^n \frac{1}{T(y)} = e^{(d/dy)y(d/dy)} (d/dy)^{n} y^{n1} =0 $$

$\begingroup$ The question seemed to be "Is it possible to have …", which I interpreted to mean whether the equality ever holds, as opposed to whether the equality always holds. But perhaps @AdamHammam can make it clearer which interpretation was meant. $\endgroup$– LSpiceSep 30, 2021 at 2:43

1$\begingroup$ @LSpice  the questions by AdamHammam tend to concentrate on the effect of the operator, I haven't seen any that concentrate on the function being acted on (here, $T$) in terms of being a solution to a kind of generalized differential equation. I think what is meant is "could $H$ possibly be simplified into the form ...". But yes, AdamHammam should clarify. $\endgroup$ Sep 30, 2021 at 3:17