# Young-type inequality for bounded operator

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 $$p=q=2$$.

• 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? Jul 2 at 6:15
• Can you develop ? I don't see how it simplifies.
– Chr
Jul 2 at 10:54
• This might be interesting to you : "Matrix Young Inequalities" by T. Ando , link.springer.com/chapter/10.1007/978-3-0348-9076-2_5 . Jul 2 at 13:31
• 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.
– Chr
Jul 3 at 7:17

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.