This is from page 123 of Analysis I by Amann and Escher. I am trying to verify the details of the excerpt below.
Excerpt:
My attempt:
According to my book, in order to show that End($V$) is an algebra, the first thing I have to show is that $\big( \text{End}(V), +, \circ \big)$ is a ring. Looking at the definition of a ring in my text, I have to show that (i) $\big( \text{End}(V), + \big)$ is an Abelian group, (ii) multiplication is associative, and (iii) the distributive law holds.
The first two are automatic since End($V$) is a $K$-vector space, which requires the additive Abelian group, and function composition is associative. That just leaves distributivity. For linear $S, T, U \in \text{End}(V)$ and $v \in V$, we have
\begin{align*} S \circ (T + U)(v) &= S \big( T(v) + U(v) \big)\\ &= S \big( T(v) \big) + S \big( U(v) \big)\\ &= S \circ T(v) + S \circ U(v)\\ \\ (T + U) \circ S (v) &= (T + U) \big( S(v) \big)\\ &= T \big( S(v) \big) + U \big( S(v) \big)\\ &= T \circ S (v) + U \circ S (v). \end{align*}
That proves distributivity and therefore the definition of a ring is satisfied.
The second step in proving that End($V$) is an algebra is proving the algebra distributive law. For $\lambda, \mu \in K$, we have
\begin{align*} S \circ \big( (\lambda T) + (\mu U) \big) (v) &= S \circ (\lambda T)(v) + S \circ (\mu U)(v)\\ &= S \circ T (\lambda v) + S \circ U(\mu v)\\ &= \lambda (S \circ T)(v) + \mu(S \circ U)(v)\\ \\ \big( (\lambda T) + (\mu U) \big) \circ S (v) &= (\lambda T) \circ S(v) + (\mu U) \circ S(v)\\ &= (\lambda T)\big( S(v) \big) + (\mu U)\big( S(v) \big)\\ &= T \big( \lambda S(v) \big) + U \big( \mu S(v) \big)\\ &= T \big( S (\lambda v) \big) + U \big( S( \mu v) \big)\\ &= \lambda T \big( S ( v) \big) + \mu U \big( S( v) \big)\\ &= \lambda (T \circ S)(v) + \mu (U \circ S)(v). \end{align*}
That proves (algebra) distributivity, and therefore End($V$) is an algebra. (I think the line $ABx = A(Bx)$ in the excerpt has nothing to do with proving that End($V$) is an algebra; it's just a demonstration of the fact that multiplication in End($V$) is composition of linear maps.)
The last thing we have to do is verify that the function $K [X] \to \text{End}(V)$, $p \mapsto p(A)$ is an algebra homomorphism. (I will omit the work I did trying to show that $K[X]$ is an algebra for brevity. Long story short, it's a ring, and can easily be made into a vector space, and the ring and scalar multiplications play nice with each other.)
If we denote the function in question by $f$, then we are first trying to establish that $f(\lambda p) = \lambda f(p)$ for $\lambda \in K$, and $p \in K[X]$. If $v \in V$, then we have
\begin{align*} \big( f(\lambda p) \big) (v) &= \big( (\lambda p_0 A^0) + \dots + (\lambda p_k A^k) \big) (v)\\ &= (\lambda p_0 A^0)(v) + \dots + (\lambda p_k A^k)(v)\\ &= \lambda p_0 A^0(v) + \dots + \lambda p_k A^k(v)\\ %&= p_0 A^0(\lambda v) + \dots + p_k A^k (\lambda v)\\ &= \lambda \big( p_0 A^0(v) + \dots + p_k A^k (v) \big)\\ &= \lambda (p_0 A^0 + \dots + p_k A^k) (v)\\ &= \big( \lambda (p_0 A^0 + \dots + p_k A^k) \big) (v)\\ &= \big(\lambda f(p) \big)(v). \end{align*}
The next thing to establish is that $f(p + q) = f(p) + f(q)$, where $p, q \in K[X]$.
\begin{align*} \big( f(p + q) \big) (v) &= \Big( \big( (p_0 + q_0)A^0 \big) + \dots + \big( (p_k + q_k)A^k \big) \Big) (v)\\ &= \big( (p_0 + q_0)A^0 \big)(v) + \dots + \big( (p_k + q_k)A^k \big)(v)\\ &= (p_0 + q_0)A^0 (v) + \dots + (p_k + q_k)A^k (v)\\ &= A^0 \big( (p_0 + q_0)v \big) + \dots + A^k \big( (p_k + q_k)v \big)\\ &= A^0 (p_0v + q_0v ) + \dots + A^k (p_kv + q_kv )\\ &= A^0 (p_0v) + A^0 (q_0v) + \dots + A^k (p_kv) + A^k(q_kv)\\ &= A^0 (p_0v) + \dots + A^k (p_kv) + A^0 (q_0v) + \dots + A^k (q_k v)\\ &= p_0 A^0 (v) + \dots + p_k A^k (v) + q_0 A^0 (v) + \dots + q_k A^k (v)\\ &= ( p_0 A^0 + \dots + p_k A^0 + q_0 A^0 + \dots + q_k A^k ) (v)\\ &= \big( f(p) + f(q) \big)(v). \end{align*}
The last thing to establish is that $f(pq) = f(p)f(q)$.
\begin{align*} \big( f(p) f(q) \big) (v) &= f(p) \circ f(q) (v)\\ &= (p_0 A^0 + \dots + p_k A^k) \circ (q_0 A^0 + \dots + q_l A^l) (v)\\ &= (p_0 A^0 + \dots + p_k A^k) \big( (q_0 A^0) (v) + \dots + (q_l A^l)(v) \big) \\ &= (p_0 A^0 + \dots + p_k A^k) \big( q_0 A^0 (v) + \dots + q_l A^l(v) \big) \\ &= (p_0 A^0) \big( q_0 A^0 (v) + \dots + q_l A^l(v) \big) + \dots + (p_k A^k) \big( q_0 A^0 (v) + \dots + q_l A^l(v) \big)\\ &= (p_0 A^0) \big( q_0 A^0 (v) \big) + \dots (p_0 A^0) \big( q_l A^l(v) \big) + \dots + (p_k A^k) \big( q_0 A^0 (v) \big) + \dots + (p_k A^k) \big( q_l A^l(v) \big)\\ &= p_0 q_0 A^0 (v) + \dots p_0 q_l A^l (v) + \dots + p_k q_0 A^k (v) + \dots p_k q_l A^{k + l} (v)\\ &= \big( (p_0 A^0 + \dots p_k A^k)(q_0 A^0 + \dots + q_l A^l) \big) (v). \end{align*}
I think the last expression is equal to $\big( f(pq) \big)(v)$ but I'm not 100% sure. I'm not sure how to package it as a discrete convolution in the form $\Big( \sum_n \big( \sum_{j = 0}^n p_j q_{n - j} \big) A^n \Big) (v)$ as my textbook suggests.
Questions:
Is this work (at least mostly) correct? How can I rewrite the last bit of my work in the suggested form, as a discrete convolution?
Is there an easier way to see that End($V$) is an algebra, and that the given function is an algebra homomorphism? I ask because this verification seems like an awful lot of work, but maybe that's because I'm still unfamiliar with all the terminology.
I appreciate any help.