2

According to [wikipedia][1]

Let $T$ be a bounded linear operator acting on a Banach space $X$ over the complex scalar field $\mathbb{C}$ and $I$ be the identity operator on $X$. The spectrum of $T$ is the set of all $\lambda \in \mathbb{C}$ for which the operator $T-\lambda I$ does not have an inverse that is a bounded linear operator

This definition seems a like unprecise to me because of the following. Because $X$ is Banach, if $T$ has an inverse, [this inverse must be bounded][2]. But (in my opinion) the definition on wikipedia might be misleading because one could think that it could happen that $T-\lambda I$ is invertible but not bounded, in which case $\lambda$ seems also to be an element of the spectrum of $T$ according to the above definition. I think a better definition of the spectrum, in this case, would be the set of all complex numbers such as $T-\lambda I$ is not invertible.

Question: If $X$ is assumed to be normed instead of Banach, what is the best definition of spectrum? Does one demand $T-\lambda I$ not to be invertible or not to be invertible and bounded?

[1]: https://en.wikipedia.org/wiki/Spectrum_(functional_analysis)#:~:text=%2C%20for%20all%20).-,Basic%20properties,subset%20of%20the%20complex%20plane.&text=would%20be%20defined%20everywhere%20on%20the%20complex%20plane%20and%20bounded.&text=The%20boundedness%20of%20the%20spectrum,bounded%20by%20%7C%7CT%7C%7C. [2]: The inverse of bounded operator?

Idontgetit
  • 1,891
  • 2
    I don't know what presentation you are following but indeed by the open mapping theorem, given that $T-\lambda I$ is continuous from a Banach space to itself, if it is bijective then its inverse must be continuous. I think the formulation of the definition had in mind the more general case where $T$ is unbounded (partially defined) operator. In that case adding the adjective continuous for the inverse is not superfluous. – Abdelmalek Abdesselam Aug 19 '20 at 19:14

2 Answers2

1

If $T-\lambda I$ is injective, then $T-\lambda I$ will have an inverse on $\mathcal{R}(T-\lambda I)$, but that does not guarantee that $(T-\lambda I)^{-1} : \mathcal{R}(T-\lambda I)\subset X\rightarrow X$ is bounded. For example, consider $T : L^2[0,1]\rightarrow L^2[0,1]$ defined by $$ Tf = \int_0^x f(t)dt. $$ $T$ is bounded. Even though the inverse $T^{-1}g = g'$ is closed, it is defined only on functions $g \in L^2[0,1]$ that are

$\;\;\;$(i) absolutely continuous,
$\;\;\;$(ii) vanish at $0$, and
$\;\;\;$(iii) have a square-integrable derivative on $[0,1]$.

Furthermore $T^{-1}$ is not bounded on its domain; so it is not possible to extended $T^{-1}$ in such a way that it will be continuous. If the range of $T$ were all of $X$, so that the inverse of $T$ were defined everywhere on $L^2[0,1]$, then your argument would apply because $T$ would be defined on a Banach space and would have a closed graph. But that doesn't have to happen, even if $T^{-1}$ exists, as it does not happen in this case.

Disintegrating By Parts
  • 87,459
  • 5
  • 65
  • 149
  • Thanks for the answer! But I'm still confused with the proper definition of spectrum when $X$ is normed. As you beatufilly explained, $T-\lambda I$ can be invertible but not bounded and, also, it might be not even invertible for some $\lambda$. In this case, does $\lambda \in \mbox{sp}(T)$ if any of the above occur? – Idontgetit Aug 21 '20 at 00:30
  • 1
    Suppose the operator $T$ is norm bounded. One definition of inverse: $T-\lambda I$ has an inverse that is defined on a dense subset of $X$ and that inverse is norm bounded on $X$. I believe that sort of definition would allow you to recast everything in the completion. – Disintegrating By Parts Aug 21 '20 at 00:40
0

The issue is an injective bounded operator $T\in\mathcal{B}(X)$ might not be surjective, and therefore its set-theoretic inverse $T^{-1}: T(X)\to X$ might not be defined everywhere on $X$.

You are right in that, if $T$ is bijective and bounded, then, by the open mapping theorem, $T^{-1}$ is well-defined everywhere on $X$ and is also bounded. But the definition of the spectrum leaves open all possible scenarios. It could be $T$ is bounded but not injective, or injective but its image is smaller than the whole space (in which case, there are the possibilities of the image being closed or dense or contained within a proper closed subspace).

user760
  • 1,169