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Let $A$ and $B$ be two (non commuting) self-adjoint bounded operator acting on a Hilbert space and let $p,q>1$ such that $\frac1p+\frac1q=1$

Do we have a Young-type inequality such as $ \frac12|AB+BA| \leq \frac{|A|^p}{p}+ \frac{|B|^q}{q}$ in the sense of quadratic form ? It is of course obvious for $p=q=2$.

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  • $\begingroup$ Does not the $B=|A|^{1/2}C|A|^{1/2}$ change of notations (suppose for a moment that $|A|$ is invertible) reduce it to the obvious $A=I$ case? $\endgroup$ Jul 2 at 6:15
  • $\begingroup$ Can you develop ? I don't see how it simplifies. $\endgroup$
    – Chr
    Jul 2 at 10:54
  • $\begingroup$ This might be interesting to you : "Matrix Young Inequalities" by T. Ando , link.springer.com/chapter/10.1007/978-3-0348-9076-2_5 . $\endgroup$
    – jjcale
    Jul 2 at 13:31
  • $\begingroup$ Sure it is interesting . However, it seems limited to matrices or compact operators. Aldo, my question might be easier since the inequality only involve selfadjoint operators. $\endgroup$
    – Chr
    Jul 3 at 7:17

1 Answer 1

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As a general criterion, these kind of inequalities hold at the level of singular values (which implies a version of the inequality where one of the sides is conjugated by a unitary/partial isometry), but they tend to fail at the plain operator level.

Here is a counterexample for $p=3$, $q=3/2$. Take $$ A=\begin{bmatrix}1&0\\0&3\end{bmatrix},\qquad\qquad B=\begin{bmatrix}1&1\\1&1\end{bmatrix}. $$ Then $$ \frac{A^3}3+\frac{2B^{3/2}}3=\frac13\,\Big(\begin{bmatrix}1&0\\0&27\end{bmatrix} +\begin{bmatrix}2\sqrt2&2\sqrt2\\2\sqrt2&2\sqrt2\end{bmatrix}\Big) =\frac13\,\begin{bmatrix}1+2\sqrt2&2\sqrt2\\2\sqrt2&27+2\sqrt2\end{bmatrix}. $$ We have $$ AB+BA=\begin{bmatrix}2&21\\21&40\end{bmatrix}, $$ so $$ (AB+BA)^2=\begin{bmatrix} 20&32\\32&52\end{bmatrix} $$ and then Wolfram Alpha tells us that $$ |AB+BA|=\begin{bmatrix} 20&32\\32&52\end{bmatrix}^{1/2}=\frac2{\sqrt5}\,\begin{bmatrix}3& 4\\ 4& 7 \end{bmatrix}. $$

Then, using Wolfram Alpha again, we see that $$ \frac{A^3}3+\frac{2B^{3/2}}3-\frac12\,|AB+BA| $$ is not positive, as it has a negative eigenvalue.

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