Can the norm of the resolvent function be equal to $0$?

Question

Let $\lambda\in\rho(A)$ and define $R_\lambda=(\lambda-A)^{-1}$. Can we have that $\|R_\lambda\|= 0$?

This may not be the brightest question to ask, but I believe this is false by giving the following proof:

Suppose we had $\|R_\lambda\|= 0$, then we must have that $R_\lambda=0$, but then we have that $(\lambda-A)^{-1}=0$, which of course is impossible, as $0$ is not invertible.

However, I am not too sure about $R_\lambda=0$ being true, for example it could as well be that $R_\lambda=0$ almost everywhere or something unexpected. Is my proof correct?

Answer 1

I guess it could depend on the setting, but if $\| \cdot \|$ is a norm, it is axiomatic that $\|v\| = 0 \implies v = 0$.

Now, in the case of $L^p$ spaces, "$ = 0$" is a little subtle, as the elements of $L^p$ spaces are equivalence classes of functions, not functions themselves. Two functions are "equal" in an $L^p$ space if they are equal almost everywhere. So, being "equal" to $0$ implies they take the value $0$ almost everywhere.

In this case, it looks like you're dealing with the space of operators on some kind of normed linear space, so the previous paragraph doesn't apply. The operator $R_\lambda$ doesn't represent an equivalence class, so $R_\lambda$ would literally have to equal (the very non-invertible) $0$ for its norm to be $0$.

Answer 2

Your proof is correct. The statement "$R_\lambda=0$ almost everywhere" does not make sense unless you have a measure defined. Even then, $\|R_\lambda\|=0$ means $R_\lambda=0$, with no almost everywhere qualification required.