Dear virtual lecturers,

while reading through Lecture 11, I was rather puzzeled by your notion of a resolvent in Definition 10.4, that just has to fulfill the resolvent identity. When I hear “resolvent”, I expect something that is at least in some sense invertible, which is in no way guaranteed by the resolvent equation alone. Indeed, your definition perfectly allows to call the constant zero function a resolvent.

A usual way to clarify this, is to call a map that satisfies the resolvent identity a “pseudoresolvent”. Then one shows, that a pseudoresolvent is actually a resolvent, i.e. it is the inverse of a shifted operator, if and only if, it is injective for one argument, which, by the resolvent identity is again equivalent to injectivity for all arguments.

Best

Robert

Hello :),

regarding the introduction to lecture 11, in which we discuss the solvability of the heat equation on a graph $(b,c)$ over $(X,m)$:

Here we must also assume that the graph is connected to guarantee the existence and uniqueness of a solution right?

Greetings Ragon

Dear virtual lecturers,

thank you very much for this interesting lecture.

I have some questions regarding my understanding of the proof of Theorem 10.7.

• uniqueness: As I understand the proof, the key step is that $C_C\left(X\right)$ is dense in ${\ell }^p(X,m)$ for every $p\in [1,\infty )$. Since $S^{\left(p\right)}:C_C\left(X\right)\to {\ell }^p(X,m)$ is a bounded linear operator there exists a unique extension for $S^{\left(p\right)}$ which is defined on the whole space ${\ell }^p(X,m)$. Right?

• existence: Am I right that the existence of $p_t$ follows from the Riesz representation theorem? My thought were as follows. For $t\ge 0,x\in X$ the map ${\ell }^2(X,m)\ni f\mapsto S\left(t\right)f\left(x\right)\in R$ is linear and since ${\ell }^2(X,m)$ is a Hilbert space the prerequisites for Riesz theorem are satisfied. Thus there exists an $g_{t,x}\in {\ell }^2(X,m)$ with $⟨f,g_{t,x}⟩S\left(t\right)f\left(x\right)=\sum _{y\in X}f\left(y\right)g_{t,x}\left(y\right)m\left(y\right)$. Thus we can define $p_t:X\times X\to R$ by $p_t(x,y)=g_{t,x}\left(y\right)$.

Kind regards, Kai