Bound of trace norm

Question

Let $\mathcal{H}$ be a separable Hilbert space with complex scalar, $T_1$ be a self-adjoint positive linear operator with finite trace norm, that is $$\|T_1\|_{tr} := \sum_i\langle T_1 e_i,e_i\rangle <\infty.$$ For another linear operator $T_2$, if for any $e\in \mathcal{H}$ we have $$|\langle T_2 e,e\rangle|\leq \langle T_1 e,e\rangle,$$ will we have $\|T_2\|_{tr}\leq \|T_1\|_{tr}$?

Answer 1

I can at least give a partial answer here. First check that $T_2$ is of trace class: for every orthonormal basis $\{e_n\}_{n\in\mathbb N}$ of $\mathcal H$ one has $$ \sum_{n=1}^\infty |\langle T_2e_n,e_n\rangle|\leq\sum_{n=1}^\infty \langle T_1e_n,e_n\rangle=\|T_1\|_1<\infty\,. $$ This implies that $T_2$ is of trace-class. Now assume that $T_2$ is normal so there exists a complex null sequence $(\lambda_n)_{n\in\mathbb N}$ and an orthonormal system $(f_n)_{n\in\mathbb N}$ in $\mathcal H$ such that $T_2=\sum_{n=1}^\infty \lambda_n\langle \cdot,f_n\rangle f_n$ and thus $$ \|T_2\|_1=\sum_{n=1}^\infty |\lambda_n|=\sum_{n=1}^\infty|\langle T_2f_n,f_n\rangle|\leq \sum_{n=1}^\infty \langle T_1f_n,f_n\rangle\leq \operatorname{tr}(T_1)=\|T_1\|_1 $$ as desired. One then gets the following partial result:

Let $T_2\in\mathcal B(\mathcal H)$ and $T_1\in\mathcal B_+^1(\mathcal H)$ with $|\langle T_2e,e\rangle|\leq\langle T_1e,e\rangle$ for all $e\in\mathcal H$. Then $T_2$ is of trace class. If $T_2$ is normal then $\|T_2\|_1\leq \|T_1\|_1$. For general $T_2$ one has $\|T_2\|_1\leq 2\|T_1\|_1$. The latter upper bound seems to not be sharp (see further below).

Proof. Indeed for all $e\in\mathcal H$ $$ \Big|\Big\langle \frac12(T_2+T_2^*)e,e\Big\rangle\Big|\leq \frac12( |\langle T_2e,e\rangle|+|\langle e,T_2e\rangle| )=|\langle T_2e,e\rangle|\leq\langle T_1e,e\rangle $$ so, because $\frac12(T_2+T_2^*)$ is self-adjoint (hence normal) $\|\frac12(T_2+T_2^*)\|_1\leq\|T_1\|_1$ (analogously for $\frac1{2i}(T_2-T_2^*)$). Then \begin{align} \|T_2\|_1&=\Big\| \frac12(T_2+T_2^*)+i\Big(\frac1{2i}(T_2-T_2^*)\Big) \Big\|_1\\ &\leq \Big\| \frac12(T_2+T_2^*)\Big\|_1+\Big\|\frac1{2i}(T_2-T_2^*)\Big\|_1\leq \|T_1\|_1+\|T_1\|_1=2\|T_1\|_1\,.\tag*{$\square$} \end{align}

However I am not sure whether this constant "2" can be removed in the non-normal case. The classic counter-example popping up in a number of related problems is the matrix $$ T_2=\begin{pmatrix}0&1\\0&0\end{pmatrix} $$ but even for $T_1=\frac1{2}\operatorname{id}_2$ (the "extreme" case where one still has $|\langle T_2e,e\rangle|\leq\langle T_1e,e\rangle$ for all $e\in\mathbb C^2$) one gets $1=\|T_2\|_1\leq\|T_1\|_1=1$. To ensure this is not an artefact of dimension two I looked at $$ T_1=\begin{pmatrix}a^2+d^2&0&0\\0&b^2+f^2&0\\0&0&c^2+g^2\end{pmatrix}\qquad T_2\begin{pmatrix}0&2ab&2cd\\0&0&2fg\\0&0&0\end{pmatrix} $$ which for all $a,b,c,d,f,g\in\mathbb R$ and all $e\in\mathbb C^3$ satisfy $|\langle T_2e,e\rangle|\leq\langle T_1e,e\rangle$ (obvious). However even through numerical simulation I could not find any $a,b,c,d,f,g$ which violate $\|T_2\|_1\leq\|T_1\|_1$. Maybe someone else can complete the puzzle here.

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Edit (2020/02/13): Today I thought about using Wielandt's trick${}^1$ to extend the normal case to the general case: for arbitrary $T_2\in\mathbb C^{n\times n}$ the larger matrix $$ \tilde T_2:=\begin{pmatrix} 0&T_2\\T_2^*&0\end{pmatrix} $$ is obviously hermitian and the (non-zero) eigenvalues of $\tilde T_2$ are the (non-zero) singular values of $T_2$ and their negatives. Now if it were true that $$ |\langle \tilde T_2z, z\rangle|\leq \langle \tilde T_1z,z\rangle\qquad\text{ for all }z\in\mathbb C^{2n}\tag{1} $$ (where $\tilde T_1:=T_1\oplus T_1=\operatorname{diag}(T_1,T_1)\geq 0$) then the result for hermitian matrices (resp. normal operators) would imply that $\|T_2\|_1=\frac12\|\tilde T_2\|_1\leq\frac12\|\tilde T_1\|_1=\|T_1\|_1$ as desired.

Unfortunately, however, $|\langle T_2x,x\rangle|\leq\langle T_1x,x\rangle$ for all $x\in\mathbb C^n$ in general does not imply (1). For this take $T_1,T_2\in\mathbb C^{2\times 2}$ as above (so $|\langle T_2x,x\rangle|\leq\langle T_1x,x\rangle$ holds) but for $z=(1,0,0,1)^T$ $$ \Big|\Big\langle \begin{pmatrix}0&T_2\\T_2^*&0\end{pmatrix}z,z\Big\rangle\Big|=2\not\leq 1=\Big\langle\begin{pmatrix}T_1&0\\0&T_1\end{pmatrix}z,z\Big\rangle $$ as is readily verified. One can trace back the failure of this trick to the fact that $|\langle T_2x,x\rangle|\leq\langle T_1x,x\rangle$ for general $T_2$ does not imply that $\|T_2\|\leq\|T_1\|$ where $\|\cdot\|$ is the usual operator norm on the Hilbert space $\mathbb C^n$. ${}^2$ This is consistent with the inequality $\frac12\|A\|\leq r(A)\leq\|A\|$ (with $r(A)=\sup_{x\in\mathcal H,\|x\|=1}|\langle Ax,x\rangle|$ being the numerical radius) and by assumption we only know that $$ \frac12\|T_2\|\leq r(T_2)\leq r(T_1)\overset{T_1\text{ normal}}=\|T_1\|\,. $$ Note that this factor $\frac12$ emerges for the same reason the factor $2$ emerges in the above statement, so getting rid of it (assuming that is possible) will require a different idea altogether.

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_{Footnote 1: This idea of relating $T_2$ to the hermitian "extension" $\tilde T_2$ in terms of eigenvalues / singular values is due to Wielandt (refer to footnote 4 in "Some metric inequalities in the space of matrices" (Ky Fan, A.J. Hoffman, 1955)), hence the name.}

_{Footnote 2: If $\|T_2\|\leq\|T_1\|$ then the matrix $\begin{pmatrix}T_1&\lambda T_2\\\lambda T_2^*&T_1\end{pmatrix}$ is positive semi-definite for all $\lambda\in [-1,1]$ (follows, e.g., from Proposition 1.3.2 in Bhatia's 2007 book "Positive Definite Matrices" because $T_1\geq 0$). Evaluating this at $\lambda=\pm1$ readily implies $2|\operatorname{Re}\langle T_2x,y\rangle|\leq\langle T_1x,x\rangle+\langle T_1y,y\rangle$ for all $x,y\in\mathbb C^n$ which is equivalent to (1). Therefore, in this special case, Wielandt's trick would lead to success.}