4
$\begingroup$

Suppose $T$ is a compact self-adjoint operator and $f$ is a unit vector such that $\| (T -3)f \| \leq 1/2$. Denote by $p$ the orthogonal projection onto the direct sum of eigenspaces of $T$ with eigenvalue $2 \leq \lambda \leq 4$. I want to show that $$\| p f \| \geq \frac{\sqrt{3}}{2}.$$

I'm unsure of how to consider the norm of $p$ and how to relate it to the eigenvalues. Perhaps $$\| Tf \| = \lambda \| p \circ T f \|?$$

1 Answers 1

2

hint: $\Vert(T-3)f\Vert\geq\Vert(T-3)(1-p)f\Vert\geq\Vert(1-p)f\Vert$

  • 0
    I'm still confused. If we have $\| (1-p)f \| \leq \| (T-3)(1-p)f \| \leq \| (T-3)f \| \leq 1/2$, do we use the $\| (1-p)f \| \leq \frac{1}{2}$ and maybe $\| (1-p)f \| \leq 1 + \| pf \|$? – 2017-02-23
  • 1
    then we just use that $f=(1-p)f +pf$ and that $(1-p)f$ and $pf$ are orthogonal to each other – 2017-02-24
  • 0
    What? I don't see how that gets the result... – 2017-02-25
  • 0
    $\Vert f\Vert^2=\Vert(1-p)f\Vert^2+\Vert pf\Vert^2$ – 2017-02-25