I am trying to express the convolution of two function in abstract braket notation. Here by abstract I mean that a function $|f>$ is abstract and is not yet assigned a basis like $<x|$. So, I can write a convolution as, $$<x|f*g>=[f*g](x)=<\hat{T}(x) \hat{R}g| f> $$ where $\hat{T}(x)$ is translation by $x$ operator and $\hat{R}$ is reversal operator. I would like to know if this can be written as $|f*g> = \hat{C_g}|f>$, and if $\hat{C}_g$ is some kind of projection operator. Here is my attempt at this:
Let $|g'> = \hat{R} |g>$, then I can create a continuous family of vectors as, $|g'_x>=\hat{T}(x) |g'>$. Then I may consider the projection operator: $$\hat{P}_g = \int dx |g'_x><g'_x| $$ and then we have, $$\hat{P}_g |f>= \int dx |g'_x><g'_x| f>$$ Then I can put this in the $\tilde{x}$ basis (using tilde to make it different from the other $x$), we have then $$<\tilde{x}|\hat{P}_g |f>= \int dx<\tilde{x}|g'_x><g'_x| f>$$ I don't know where to go from here or if this is correct...
Another approach which I think is the right direction is to consider the operator, $$ \hat{Q}_g = \int dx |x><g'_x|$$ then, $$ <\tilde{x}|\hat{Q}_g|f> = \int dx <\tilde{x}|x><g'_x|f>$$ $$<\tilde{x}|\hat{Q}_g|f> = \int dx \ \delta(x-\tilde{x})<g'_x|f> $$ $$<\tilde{x}|\hat{Q}_g|f> = <g'_\tilde{x}|f> = <\hat{T}(\tilde{x}) \hat{R} g|f> $$ So is this then he correct answer?