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Let $ 0<p\le \theta<1$. Let $X$ be a self-adjoint bounded linear operator on a Hilbert space $H$ and $P$ is a projection on $H$. Why do we have $$\|P|X|^\theta\|_{p/\theta} \le \|P|X|\|_p^{\theta}?$$ It is clear that $\||X|^\theta\|_{p/\theta} = \||X|\|_p^{\theta}$. But I don't know how to use $P$ here. The author says that by using complex interpolation, we obtain this. However, I don't know how to use interpolation theory when $p<1$. Here, $\|X\|_p =Tr(|X|^p)^{1/p}$, where $Tr$ is the standard trace on $B(H)$.

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1 Answer 1

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Reference

Lemma 12 in 1 sates that: let $0<\theta <1$, $0<p_0,p_1<\infty$ (I think one of $p_i$ can be $\infty$). Then $$ (S_{p_0},S_{p_1})_\theta =S^r, ~ \frac1r =\frac{1-\theta}{p_0} +\frac{\theta}{p_1}.$$ Definitions and properties can be found in Page 416 of 1. Let $f(z)=P|X|^z$, $p_1=p$ and $p_0=\infty$. Then, $f$ satisfies the conditions required in Page 416. Then, $P|X|^\theta = f(\theta)$. Hence, $P|X|^\theta \in S_{p/\theta}$ and $\|P|X|^\theta\|_{p/\theta} \le const \max_{z\in \mathbb{R}} \|f(zi)\|^{1-\theta}_\infty \max_{z\in \mathbb{R}}\|f(1+z_i)\|_p^\theta$ (see Page 416 in 1). Since $\|f(zi)\|_\infty \le 1$ and $\|f(1+z_i)\|_p \le \|P |X|\|_p$ (by functional calculus), it follows that $\|P|X|^\theta\|_{p/\theta} \le const \|P |X|\|_p^\theta$. I think the const here should be 1. Things should be done like this. I can not find good references for $p<1$.

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