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Let $H$ be a Hilbert space and $T$ a normal operator on $H$. In the sequel, ${\rm tr}$ denotes the trace for trace class operators.

Do we have $$ \vert\vert T \vert\vert= \sup |{\rm tr} (TP)| $$

where the supremum is taken over the projections $P$ of rank 1?

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There is some evidence that you actually mean to say that $H$ is a Hilbert space. Will you please clarify? What is the source of the problem? Because $T$ is normal, its norm equals its spectral radius, and for all $\lambda\in\mathbb C$, $T-\lambda I$ is invertible if and only if it is bounded below. (If $T-\lambda I$ is injective then so is $T^*-\overline{\lambda} I$, which implies that $T-\lambda I$ has dense range.) –  Jonas Meyer Jul 8 '12 at 4:18
    
Thank you for the mistake. Unfortunately for me, I does not understand why your argument imply the result. –  Zouba Jul 8 '12 at 6:54
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2 Answers

up vote 2 down vote accepted

Let $\lambda$ be an element of the spectrum of $T$. If $\lambda$ is an eigenvalue for $T$ with eigenvector $v$, let $P$ be the projection onto $\mathbb C v$, so that $\mathrm{tr}(TP)=\lambda$. If $T-\lambda I$ is injective, then because $T$ is normal, $(T-\lambda I)^*$ is also injective, which implies that $T-\lambda I$ has dense range. Since $T-\lambda I$ has dense range and is not invertible, it is not bounded below, so there is a sequence $(v_n)$ of unit vectors in $H$ such that $\|Tv_n-\lambda v_n\|\to 0$. If $P_n$ is the projection onto $\mathbb C v_n$, then $\mathrm{tr}(TP_n)\to \lambda$.

This shows that the spectral radius of $T$ is bounded by the supremum in question, and in fact the spectrum of $T$ is contained in the closure of $\{\mathrm{tr}(TP):P=P^*=P^2,\mathrm{rank}(P)=1\}$. Since the spectral radius of a normal operator is equal to its norm, this gives $\|T\|\leq \sup\limits_P |\mathrm{tr}(TP)|$. The other inequality holds without normality of $T$.


Alternatively, if $H$ is separable, then there is a $\sigma$-finite measure $\mu$ with underlying set $X$ and an essentially bounded measurable function $f:X\to\mathbb C$ such that $T$ is unitarily equivalent to the multiplication operator $M_f$ on $L^2(\mu)$. The norm of $M_f$ is the essential supremum of $|f|$, so for every $\varepsilon>0$ there is a set $E$ of positive finite measure such that for almost all $x\in E$, $|f(x)|\geq \|M_f\|-\varepsilon$. If $P$ is the projection onto the span of $ \chi_E$, then $|\mathrm{tr}(M_fP)|\geq \|M_f\|-\varepsilon$. (The property is preserved by unitary equivalence.)

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Yes. Your equality is easily seen to hold for a diagonal operator. And any normal operator is a norm-limit of diagonals ( by the Weyl-von Neumann-Berg theorem).

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