[proofplan] We first check that the pointwise operations make $I(P;R)$ an $R$-module and that convolution is well-defined because every interval $[x,y]$ is finite. Bilinearity follows by expanding the finite convolution sum and using distributivity in $R$. Associativity is proved by expanding both parenthesizations and identifying both finite sums with the same sum over pairs $(u,v)$ satisfying $x\leq u\leq v\leq y$. Finally, substituting the delta function into the convolution formula proves the left and right identity laws. [/proofplan]

[step:Check the pointwise module structure and the finiteness of convolution]Let \begin{align*} D=\{(x,y)\in P\times P:x\leq y\} \end{align*} denote the domain of the functions in $I(P;R)$. Since $R$ is a commutative unital ring, the set of all functions $D\to R$ is an $R$-module under pointwise addition and pointwise scalar multiplication. Explicitly, for $f,g\in I(P;R)$, $a\in R$, and $(x,y)\in D$, define \begin{align*} (f+g)(x,y)=f(x,y)+g(x,y) \end{align*} and \begin{align*} (af)(x,y)=a f(x,y). \end{align*} The $R$-module axioms follow pointwise from the corresponding ring axioms in $R$. Now let $f,g\in I(P;R)$ and let $x,y\in P$ satisfy $x\leq y$. The indexing set in the convolution formula is the interval \begin{align*} [x,y]=\{z\in P:x\leq z\leq y\}. \end{align*} By local finiteness this set is finite, so \begin{align*} (f*g)(x,y)=\sum_{z\in [x,y]} f(x,z)g(z,y) \end{align*} is a finite sum in $R$. Thus $f*g$ is a well-defined function $D\to R$, namely an element of $I(P;R)$.[/step]

[guided]The first point is that there is no hidden convergence issue in the definition of convolution. Define \begin{align*} D=\{(x,y)\in P\times P:x\leq y\}. \end{align*} Then $I(P;R)$ is exactly the set of all functions from $D$ to $R$. Since $R$ is a commutative unital ring, the usual pointwise formulas \begin{align*} (f+g)(x,y)=f(x,y)+g(x,y) \end{align*} and \begin{align*} (af)(x,y)=a f(x,y) \end{align*} make this function set an $R$-module: associativity and commutativity of addition, the zero element, additive inverses, and the scalar multiplication laws are all checked by evaluating at an arbitrary pair $(x,y)\in D$ and using the same law in $R$. For convolution, fix $f,g\in I(P;R)$ and fix $x,y\in P$ with $x\leq y$. The possible middle elements $z$ are exactly the elements of the interval \begin{align*} [x,y]=\{z\in P:x\leq z\leq y\}. \end{align*} The hypothesis is local finiteness, so this interval is finite even if the whole poset $P$ is infinite. Therefore \begin{align*} (f*g)(x,y)=\sum_{z\in [x,y]} f(x,z)g(z,y) \end{align*} is a finite sum in the ring $R$. This proves that the displayed formula defines an element $f*g\in I(P;R)$.[/guided]

[step:Prove bilinearity of convolution over $R$] Let $f_1,f_2,g,h\in I(P;R)$, let $a\in R$, and let $x,y\in P$ satisfy $x\leq y$. Using the finite interval $[x,y]$ and distributivity in $R$, we have \begin{align*} ((f_1+f_2)*g)(x,y)=\sum_{x\leq z\leq y}(f_1(x,z)+f_2(x,z))g(z,y) \end{align*} and hence \begin{align*} ((f_1+f_2)*g)(x,y)=(f_1*g)(x,y)+(f_2*g)(x,y). \end{align*} Similarly, \begin{align*} (f*(g+h))(x,y)=\sum_{x\leq z\leq y}f(x,z)(g(z,y)+h(z,y)) \end{align*} so \begin{align*} (f*(g+h))(x,y)=(f*g)(x,y)+(f*h)(x,y). \end{align*} For scalar multiplication, \begin{align*} ((af)*g)(x,y)=\sum_{x\leq z\leq y}a f(x,z)g(z,y)=a(f*g)(x,y). \end{align*} Also, using commutativity of $R$ to move the scalar through the product, \begin{align*} (f*(ag))(x,y)=\sum_{x\leq z\leq y}f(x,z)a g(z,y)=a(f*g)(x,y). \end{align*} Since these identities hold for every $(x,y)\in D$, convolution is $R$-bilinear. [/step]

[step:Identify both associative products with the same finite double sum] Let $f,g,h\in I(P;R)$ and let $x,y\in P$ satisfy $x\leq y$. Expanding the left parenthesization gives \begin{align*} ((f*g)*h)(x,y)=\sum_{x\leq v\leq y}(f*g)(x,v)h(v,y). \end{align*} For each $v\in [x,y]$, the interval $[x,v]$ is finite, and so \begin{align*} (f*g)(x,v)=\sum_{x\leq u\leq v} f(x,u)g(u,v). \end{align*} Substituting this finite sum into the previous display and distributing in $R$ gives \begin{align*} ((f*g)*h)(x,y)=\sum_{\substack{x\leq u\leq v\leq y}} f(x,u)g(u,v)h(v,y). \end{align*} The indexing set $\{(u,v)\in P\times P:x\leq u\leq v\leq y\}$ is finite because it is a subset of $[x,y]\times [x,y]$. Expanding the right parenthesization gives \begin{align*} (f*(g*h))(x,y)=\sum_{x\leq u\leq y}f(x,u)(g*h)(u,y). \end{align*} For each $u\in [x,y]$, the interval $[u,y]$ is finite, and so \begin{align*} (g*h)(u,y)=\sum_{u\leq v\leq y} g(u,v)h(v,y). \end{align*} Substituting and distributing gives \begin{align*} (f*(g*h))(x,y)=\sum_{\substack{x\leq u\leq v\leq y}} f(x,u)g(u,v)h(v,y). \end{align*} Both parenthesizations have the same value for every $x\leq y$, so convolution is associative. [/step]

[step:Verify that the delta function is a two-sided identity] Define $\delta\in I(P;R)$ by $\delta(x,x)=1_R$ for every $x\in P$ and $\delta(x,y)=0_R$ whenever $x,y\in P$ satisfy $x<y$. Let $f\in I(P;R)$ and let $x,y\in P$ satisfy $x\leq y$. Then \begin{align*} (\delta*f)(x,y)=\sum_{x\leq z\leq y}\delta(x,z)f(z,y). \end{align*} If $z\in [x,y]$ and $z\neq x$, then $x<z$, so $\delta(x,z)=0_R$. The only nonzero term occurs at $z=x$, and therefore \begin{align*} (\delta*f)(x,y)=\delta(x,x)f(x,y)=1_R f(x,y)=f(x,y). \end{align*} Similarly, \begin{align*} (f*\delta)(x,y)=\sum_{x\leq z\leq y}f(x,z)\delta(z,y). \end{align*} If $z\in [x,y]$ and $z\neq y$, then $z<y$, so $\delta(z,y)=0_R$. The only nonzero term occurs at $z=y$, and therefore \begin{align*} (f*\delta)(x,y)=f(x,y)\delta(y,y)=f(x,y)1_R=f(x,y). \end{align*} Thus $\delta*f=f$ and $f*\delta=f$ for every $f\in I(P;R)$. [/step]

[step:Conclude the algebra structure] The pointwise operations make $I(P;R)$ an $R$-module, convolution is a well-defined $R$-bilinear multiplication, convolution is associative, and $\delta$ is a two-sided multiplicative identity. Hence $I(P;R)$ is an associative unital $R$-algebra with multiplicative identity $\delta$. [/step]