This got me thinking about what sort of functions f would satisfy U(f)(x) = f(x + k). Clearly it worked for polynomials, but certainly not so for all smooth functions. So what about analytic functions? It doesn't., and the proof of this is below.
A quick correction to the above proof: About a week after I made this post and had moved on from the problem, I mentioned the result and the proof to a colleague of mine. When it came to the series representation I had for U_k, he pointed out potential convergence issues with chosen k. Now, this was something that
did occur to me while working on the problem, but I erroneously thought I had it handled with the fact that U_{k+l} = U_k o U_l as in "Statement and Initial Results". However, looking at U_k more carefully: I realized the error, and its subtlety I believe is something worth mentioning.
Take the expression for U_k given in "Statement and Initial Results". See that if one applies: k = ( (x + k) - x), the series becomes precisely the Taylor series of f centered at x evaluated at x+k. This illuminates a few things:
- It is obvious then that for entire functions whose power series representation applies to the entire real line one has that U_k(f)(x) = f(x + k) for any x and k - as seen in polynomials and the exponential function for example.
- The failure of U_k(f)(x) to evaluate correctly with k > 0 at x=0 for f(x) = exp(-1/x^2) makes sense as what U_k(f)(x) is literally doing is becoming the Taylor series of f(x) at x=0 which is just 0.
- Furthermore, for functions such as f(x) = log(x + 1), taking them on their domain (-1,infty), with a center x = 0 one then clearly has k for which U_k(f)(x) = f(x+k) fails, for example k = 3. Now, while f in this case is not analytic R ---> R, the same technique applies to show U_k(f)(x) = f(x+k) failure for analytic-but-not-entire functions as your k can exceed a local radius of convergence.
This is all seen with a simple substitution. Now with this perspective, the potential error in U_{k+l} = U_k o U_l becomes clear: Take U_k o U_l(f)(x). Suppose the power series of f at x is defined at x+l and the power series of f at x+l is defined at x+k+l. Then one has U_l(f)(x) = f(x + l) and thus U_k o U_l(f)(x) = U_k(f)(x+l) = f(x + k + l). However, one might have k,l such that the power series of f at x is
not defined at x + k + l. Therefore, it
cannot be that U_{k+l}(f)(x) = U_k o U_l(f)(x). The subtlety of the error is that though U_{k+l} and U_k o U_l are
algebraically the same power series,
analytically they are different - because U_{k+l} fixes its center at x while U_k o U_l has a
dynamic center that moves in the stages of composition from x to x+l. Leveraging this is why the rhs can produce the desired shift by convergent increments while the lhs fails to do it all at once.
Now, this error doesn't change anything in the proof, it just changes the result. I have
not shown that for analytic f and any x and k one has U_k(f)(x) = f(x + k). By the above that cannot be true. However, what I
have shown is that under a
family of such operators {U_k_i} any shift is possible. So pick any x and k, one has U_k_1,...,U_k_n for which U_k_n o ... o U_k_1(f)(x) = f(x + k). And once you understand that U_k(f)(x) is the power series of f centered at x evaluated at x+k, this makes perfect sense once one has the "Second Lemma". You start at a point and move forward within the interval of convergence. You stop and take a new power series representation with its own interval convergence and repeat until you get where you want to go. What the "Second Lemma" confirms is that you may always move forward by a certain amount, and thus, inevitably, get there.
Looking at U_k(f)(x) as a Taylor series does present a clean set of sufficient conditions for U_k(f)(x) = f(x + k). For any smooth function f if: (1) f has a power series representation center at x and (2) x+k is in the interval of convergence - then one has the desired equality. I'm not saying "necessary" since I'm not sure of that, but I can see that for any smooth f if one must have a neighborhood of k about 0 for which U_k(f)(x) = f(x + k) then f must be analytic.
