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?
Best Answer
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.