Apologies if this is a silly question... but I am thinking about the multiplication operator $Tv = e^{ax}$ where $a$ is a positive real number, and I am curious to know whether or not this operator is bounded when applied to a function from the function space $L^2(\mathbb{R})$.
I have been considering various scattering problems and lattices, which takes it to be an unbounded operator, but an unbounded operator can be bounded regarding the terminology.
My reasons to suspect this is bounded is that functions in $L^2(\mathbb{R})$ can be written as sums as Hermite functions where the Gaussian part is stronger than $e^{ax}$. However, playing around with an epsilon-delta proof
Let $\epsilon >0 $, I need to find $|v - 0 |< \delta$ such that \begin{equation} || e^{2ax}(v) - e^{2ax} (0)|| \leq \epsilon \end{equation}...
A little frazzled about this integral!
Thank you!
I am assuming you are talking about the operator $Tv= e^{ax}v$ for some $v \in L^2(\mathbb{R})$.
If you are just talking about $L^2(\mathbb{R})$ with respect to the Lebesgue measure, then this is an unbounded (or rather not bounded) operator. For example $1/(1+x^2) \in L^2(\mathbb{R})$, but $e^{ax}/(1+x^2) \notin L^2(\mathbb{R})$.