This a question from Functional analysis-By Kreyszig

Let $T:X\to X$ be a compact linear operator on a normed space. If dim $X=\infty$, then show that $0 \in \sigma(T)$.

For a operator $T:X \to X$ , he defines the resolvent $\rho(T)$ of T as follows:

if $\lambda \in \rho(T)$, then

(1) $T-\lambda I$ is one-one

(2)$(T-\lambda I)^{-1}$ from the range of $T-\lambda I$ to $X$ is bounded operator.

(3) Range($T-\lambda I$) is dense in X.

And then he defines the spectrum $\sigma$(T) of T as complement of $\rho(T)$ in the complex plane!!

Now the problem is that $X$ is not given to be Banach. I know that this is true in infinite dimensional Banach space because of the fact that identity operator from on $X$ is not compact. Is the question correct if we donot assume X to be Banach.

Thanks in advance!!

  • 1
    $\begingroup$ If you are not assuming the space to be complete. What is your definition of the spectrum? $\endgroup$ – Severin Schraven Jun 3 '17 at 16:03
  • $\begingroup$ It can in fact be shown that on an infinite dimensional normed linear space, the identity operator is not compact. $\endgroup$ – Sahiba Arora Jun 3 '17 at 16:10
  • $\begingroup$ @SahibaArora How would one prove this? Do you have a reference for this cool result? $\endgroup$ – Severin Schraven Jun 3 '17 at 16:13
  • 1
    $\begingroup$ It follows from that fact the unit ballin a normed linear space is compact iff the space is finite dimensional. You can find a proof of this here $\endgroup$ – Sahiba Arora Jun 3 '17 at 16:21
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    $\begingroup$ See, as we have Banach space, if 0 is not in the spectrum, then 0 is in the resolvent, and then, $T^{-1}$ is a map from X $\to$ X(bounded) , and hence $T \circ T^{-1}=id_{X}$, and then by the ideal property of compact operator, $id_{X}$, is compact - a contradiction!! $\endgroup$ – Riju Jun 3 '17 at 16:39

Were $0 \not\in \sigma(T)$, then $T$ would be invertible. You don't have to work to hard to figure out that the unit ball of $X$ must be compact. This renders $X$ finite-dimensional.


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