Talk:Resolvent set

there's a separate article resolvent formalism. these two could be merged. Mct mht 12:13, 19 January 2007 (UTC)

Not stated mathematically

From the article:

$\lambda$ is said to be a regular value if $R(\lambda ,L)$, the inverse operator to $L_{\lambda }$

1. exists;
2. is a bounded linear operator;
3. is defined on a dense subspace of X.

The resolvent set of L is the set of all regular values of L.

This is not really a mathematical definition. You cannot speak of the inverse of a function f:A-->B "existing" if f is not surjective. A non-surjective function can have a right inverse, but not an inverse.

If you insist on using the word "exists", you might get away with requiring that the inverse of f "exist" on its range, that is, the function f:A-->f(A) is invertible. But then it is quicker and more precise simply to say that f is injective.

The above discussion assumes that a straightforward set-theoretic inverse is intended, which is certainly the case when discussing the resolvent of a bounded operator. Such an inverse is, itself, automatically a bounded operator.

But when it comes to unbounded operators, I'm a little unclear on what we would even require of a "inverse". Can it also be unbounded? What is the exact definition? The link in the article to "inverse operator" is worthless — it only gives the most basic definition of inverse. No partial functions or dense subspace there! This concept — the inverse of an unbounded operator — needs a definition in the current article.

178.38.67.19 (talk) 00:19, 14 May 2015 (UTC)

I fixed it. The inverse is when the operator is seen as a bijection onto its range. --nBarto (talk) 19:57, 17 August 2018 (UTC)

No, an operator is not "when" anything.

Are a couple of words missing?

The section Definitions contains this passage:

"Let X be a Banach space and let $L\colon D(L)\rightarrow X$ be a linear operator with domain $D(L)\subseteq X$. Let id denote the identity operator on X. For any $\lambda \in \mathbb {C}$, let

$L_{\lambda }=L-\lambda \,\mathrm {id} .$

"A complex number $\lambda$ is said to be a regular value if the following three statements are true:"

[This is followed =by three statements depending heavily on both $\lambda$ and L.

Therefore complex number $\lambda$ satisfying the three conditions is not merely a "regular value".

It must be technically defined as a regular value vis-à-vis L.

Sure, we can just call it a "regular value" as long as we're talking about a fixed L.

But for the sake of an encyclopedia article, we need to tell readers the correct technical term for such a $\lambda$ vis-à-vis L. 2601:200:C000:1A0:8459:386D:315C:95BC (talk) 03:36, 3 September 2022 (UTC)

Is R(𝜆, L) the resolvent?

The operator R(𝜆, L) discussed in this article uses the identical notation to that used in the article about resolvents of operators.

Because of this, I hope that someone knowledgeable about this subject can clarify if R(𝜆, L) in fact is the resolvent (or is the resolvent under certain conditions).