91Ó°ÊÓ

An automorphism of an algebra represents an invertible linear transformation of the space that preserves the structure. In our case, we are working with the umbral algebra. So, given our automorphism \(T\), it should satisfy \(T(f(t) \cdot g(t)) = Tf(t) \cdot Tg(t)\) for all functions \(f(t)\) and \(g(t)\) in the algebra.

In the umbral algebra, the order \(o(f(t))\) of a function is the smallest integer \(n\) such that the derivative of the function exists up to the order \(n-1\), and the nth derivative of this function is not zero, i.e., \(\frac{d^n f(t)}{dt^n} \neq 0\). In other words, \(o(f(t))\) is the lowest order derivative that is non-zero.

We want to show that if \(T\) is an automorphism of the umbral algebra, then \(o(T f(t)) = o(f(t))\). Let's consider different cases: Case 1: If \(f(t) = 0\), then both sides are equal, and there is nothing to prove. Case 2: If \(f(t) \neq 0\), let \(n = o(f(t))\). That means \(\frac{d^k f(t)}{dt^k} = 0\) for \(k < n\), and \(\frac{d^n f(t)}{dt^n} \neq 0\). Since \(T\) is an automorphism, we can write: \(T(\frac{d^k f(t)}{dt^k}) = \frac{d^k (T f(t))}{dt^k}\) for all values of \(k\). Now, for \(k < n\), we have \(T(\frac{d^k f(t)}{dt^k}) = T(0) = \frac{d^k (T f(t))}{dt^k}\). Hence, for \(k < n\), \(\frac{d^k (T f(t))}{dt^k} = 0\), and since \(\frac{d^n f(t)}{dt^n} \neq 0\), we must have \(\frac{d^n (T f(t))}{dt^n} \neq 0\). Therefore, \(o(T f(t)) = n\), which implies that \(o(T f(t)) = o(f(t))\).

Since we demonstrated in Step 3 that the automorphism \(T\) preserves order, this implies that \(T\) is continuous. This is because if \(f(t)\) is a continuous function, then its order is preserved, and therefore, \(T f(t)\) must also be a continuous function, as the order of both functions are equal.