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I am stumped in the following question from Brian Hall's Quantum Theory for Mathematicians.

Assume that $\dim(\mathcal{H})=\infty$. Show that if $A$ has finite rank, then $\|A+cI\|\ge|c|$ for any $c\in\mathbb{C}$. (With $c=-1$, this shows that $I$ is not an operator-norm limit of finite-rank operators.)

Showing the above operator norm inequality amounts to showing that $$\|A\psi+c\psi\|\ge|c|\|\psi\|$$ for $\psi\in\mathcal{H}$. So I began by expanding the norm as an inner product ($\mathcal{H}$ is a hilbert space): $$\|A\psi+c\psi\|^2=\|A\psi\|^2+|c|^2\|\psi\|^2+\langle c\psi,A\psi\rangle+\langle A\psi,c\psi\rangle$$ $$=\|A\psi\|^2+|c|^2\|\psi\|^2+2\Re(\langle c\psi,A\psi\rangle)$$ where $\Re(x)$ denotes the real part of $x$. But from here I wasn't sure if I could show that $\|A\psi\|^2+2\Re(\langle c\psi,A\psi\rangle)\ge0,$ which would complete the proof. My next attempt was using the reverse triangle inequality: $$\|A\psi+c\psi\|=\|A\psi-(-c\psi)\|\ge\bigg|\|A\psi\|-\|-c\psi\|\bigg|$$ $$=\bigg|\|A\psi\|-\|c\psi\|\bigg|.$$ If $\|A\psi\|\ge\|c\psi\|,$ then $\bigg|\|A\psi\|-\|c\psi\|\bigg|\ge\|c\psi\|$ and we are done. However, I don't see how to finish the proof when $\|A\psi\|\le\|c\psi\|$.

Note that I have not used the fact that $A$ has finite rank in either of these attempts, so if either of them is the correct approach, then the solution should involve this.

zbrads2
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  • You only need to show that $|A\psi+c\psi|\ge|c||\psi|$ for some $\psi\in\mathcal{H}$. Note that equality holds if $A \psi = 0$. – Martin R Oct 11 '20 at 20:36
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    @MartinR Good point. So if I can show that the kernel of $A$ is nontrivial, then I am done. Since the hilbert space is infinite dimensional, and the range of $A$ is finite, then this must be true by the Rank-Nullity theorem, right? – zbrads2 Oct 11 '20 at 21:36

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Your attempted proof does not work out because you are trying to prove to much: $\Vert A+cI \Vert \ge |c|$ does not amount to show that $\Vert Ax+cx \Vert \ge |c| \Vert x \Vert$ for all $x \in \cal H$, but for some (non-zero) $x \in \cal H$.

$\operatorname{rank} A < \infty = \dim \cal H$ implies that $\ker A$ is non-trivial. One can use the rank-nullity theorem for the infinite-dimensional case (see for example Does the rank-nullity theorem hold for infinite dimensional $V$?), or simply argue that otherwise $A$ would be a vector space isomorphism from $\cal H$ to a finite dimensional subspace.

So there is a non-zero $y \in \cal H$ with $Ay = 0$, and therefore $$ \Vert A +cI\Vert = \sup_{x \ne 0} \frac{\Vert (A+cI)x \Vert}{\Vert x \Vert} \ge \frac{\Vert (A+cI)y \Vert}{\Vert y \Vert} = |c| \, . $$

Martin R
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