Talk:Resolvent set

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

Not stated mathematically

From the article:

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

exists;
is a bounded linear operator;
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)[reply]

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)[reply]

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)[reply]

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).