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Suppose that $T$ is a bounded operator on a Hilbert space $\mathsf{H}$. Fix $\varepsilon > 0$. Assume that I have two unit vectors $\xi$ and $\eta$ such that $\xi$ belongs to the spectral subspace $\mathbb{1}_{|T| \leqslant \varepsilon}$ (spectral projection of $|T|:=\sqrt{T^{\ast}T}$ corresponding to the interval $[0,\varepsilon]$) and $\eta$ belongs to the spectral space $\mathbb{1}_{|T - Id| \leqslant \varepsilon}$. We then get $\|T\xi\|=\| |T|\xi\| \leqslant \varepsilon$ and $\|T\eta - \eta\| = \| |T-Id| \eta\| \leqslant \varepsilon$. It therefore follows from the triangle inequality that $1-2\varepsilon < \|T(\xi - \eta)\|$, so $\|\xi - \eta\| \geqslant \frac{1-2\varepsilon}{\|T\|}$. My question is -- can one achieve a bound that does not depend on the norm of $\|T\|$?

Let me explain, why this feels true. Inequality $\||T| \eta\| \leqslant 1+\varepsilon$ shows that $\eta$ is almost supported on the spectral subspace $\mathbb{1}_{|T|\leqslant M}$ if we take a big enough $M$. Then, up to a small error, we should be able to replace $\eta$ by a vector supported in this spectral subspace and therefore get rid of the dependence on $\|T\|$ in the estimate. I tried to play with this idea for a couple of hours but each time I thought I got it, something went wrong at some point. If this is not true, that would also be interesting.

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I'm afraid there is no such estimate. Let $\delta>0$ and let $\xi$ and $\eta$ be any linearly independent norm-one vectors in ${\mathbb C}^2$ with $\parallel \xi-\eta\parallel<\delta$. Let $S$ be the matrix with columns $\xi$ and $\eta$ and let $T=S\left(\begin{array}{cc}0&0\\ 0&1\end{array}\right)S^{-1}$. Then $T\xi=0$ and $T\eta=\eta$.

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  • $\begingroup$ Oh, silly me. Thanks a lot for the answer. $\endgroup$ – Mateusz Wasilewski Feb 3 '17 at 16:18

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