I have been stuck at question 10B.5 from Axler's Measure, Integration and Real Analysis (MIRA) for a long time. Could anyone please give me some hints or post a solution? The exercise reads
Give an example of a bounded operator $T$ on a normed vector space such that for every $\alpha \in F$, the operator $T− \alpha I$ is not invertible.
Here, $F$ is either $\mathbb{R}$ or $\mathbb{C}$.
If the vector space is a Banach space, then $T − \alpha I$ is invertible for $\alpha > \|T\|$ (this is result 10.35b in the book), so I have been considering non-complete spaces. But I haven't had any luck so far.
Thank you in advance!
Not worrying about boundedness for the moment, take $V = F[x]$ and let $T : V \to V$ be the linear map given by multiplication by $x$. Then $T - \alpha I$ is the operator which multiples by $x - \alpha$, which is never invertible because the result is always a polynomial of positive degree so in particular can never be $1$. ($T$ and its close relative the derivative $D = \frac{d}{dx}$ are a great source of counterexamples in infinite-dimensional linear algebra. $T$ is injective but not surjective, $D$ is surjective but not injective, $T$ has no eigenvalues, the only eigenvalue of $D$ is $0$ but $D$ isn't nilpotent...)
It remains to figure out a norm to put on $F[x]$ with respect to which $T$ is bounded. But this is straightforward: we can make $F[x]$ an inner product space (not a Hilbert space) in which the monomials $x^i$ are orthonormal, so that the norm of a polynomial $\sum f_i x^i$ is $\sum | f_i |^2$. Then $T$ is an isometry, so $\| T \| = 1$.