# Inequality for normal operators

Let $(T_1,..., T_n)\in \mathcal{L}(E)^n$ be a tuple of commuting normal operators (i.e. each $T_k$ is normal and $T_iT_j=T_jT_i$ for all $i,j$), where $E$ is a complex Hilbert space.

I want to show that $$\displaystyle\sup_{\|x\|=1}\bigg(\displaystyle\sum_{i=1}^n|\langle T_ix,x\rangle|^2\bigg)\geq\displaystyle\sup_{\|x\|=1}\bigg(\displaystyle\sum_{i=1}^n\|T_ix\|^2\bigg)\;.$$

• Do you also assume $T_i^*T_j=T_jT_i^*$ for all $i,j$ or not? (you need it to get the representation with multiplication operators) – Pietro Majer Feb 17 '18 at 20:17
• @PietroMajer thank you but we don't need it. – Student Feb 17 '18 at 20:41
• I'm just saying that for the representation with multiplication operators to hold, you need all the $2n$ operators $A_1,\dots, A_n,A_1^*,\dots, A_n^*$ to commute, that is, that they generate a commutative normal algebra. However, this is not required for your inequality to hold. – Pietro Majer Feb 18 '18 at 13:35
• @PietroMajer For the representation with multiplication operators to hold, we need only that the operators $A_k$ are normal and commuting. – Student Feb 18 '18 at 14:40
• Yes, right: if they are normal and commuting, then $A_i^*A_j=A_jA_i^*$ follows as a consequence. – Pietro Majer Feb 18 '18 at 15:44

Consider the case of finite dimensional Hilber space. Then may suppose that operators $T_1,\dots,T_d$ are diagonal in the same orthonormal basis $(e_1,e_2,\dots)$, denote the diagonal of $T_i$ by $(p_{i1},p_{i2},\dots)$. The square of RHS is nothing but $\sup_j |p_{1j}|^2+|p_{2j}|^2+\dots+|p_{dj}|^2$. For fixed $j$, taking $x=e_j$ we get the same thing $|p_{1j}|^2+|p_{2j}|^2+\dots+|p_{dj}|^2$ for LHS.

Now do the same in general situation. Note that $$\sup_{\|f\|_{L^2(\mu)}=1}\bigg[\displaystyle\sum_{i=1}^n\left(\int_Y|\varphi_i|^2|f|^2d\mu \right)\bigg]={\rm essup}\,\sum_i |\varphi_i|^2=:A.$$ Note that (by separability of a complex plane and by the definition of essential supremum) we may find discs $D_i$ such that $\mu(\Omega)>0$, where $\Omega=\{x:\varphi_i(x)\in D_i,i=1,\dots,n\}>0$ and $\sum_i |\theta_i|^2>A-\varepsilon$ whenever $\theta_i\in D_i$. Choose any function $f$ with $\|f\|=1$ concentrated in $\Omega$. Then $\int \varphi_i |f|^2d\mu\in D_i$ (indeed, $D_i$ are convex, thus a weighted mean of elements of $D_i$ belongs to $D_i$). It implies that LHS is not less than $A-\varepsilon$.

• This is only the case $\dim E<\infty$ (or $T_j$'s with pure point spectrum), though a version of this should work in general (using the spectral theorem for commuting operators). – Christian Remling Oct 30 '17 at 16:46
• Saying "we may suppose" I mean a certain limit argument involved. – Fedor Petrov Oct 30 '17 at 16:59
• Where you have used that the operators are commuting? Thank you – Student Nov 3 '17 at 17:11
• When I reduce the problem to the case of simultaneously diagonal operators. – Fedor Petrov Nov 3 '17 at 18:22
• @Student yes, exactly. What is not rigorous in this answer? – Fedor Petrov Feb 16 '18 at 18:12

Remark: Note that this is an answer to the original version of the question, for general (normal) $T_j$'s; the OP has now added the assumption that the $T_j$ commute.

This is false, unless $d=1$, when it has an easy proof using the spectral theorem. For a counterexample, consider $$T_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad T_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} .$$ The RHS of your inequality equals $\sqrt{2}$ (take $x=e_1$); on the LHS, $|\langle x, T_j x\rangle |\le \|T_j\|= 1$ for each summand individually for any $x$, so the only chance would be to make them both $=1$, but that's clearly not working since $\langle e_1, T_2 e_1\rangle = 0$.

• they do not commute – Fedor Petrov Oct 24 '17 at 12:44