Is there a generalization of Araki-Lieb-Thirring inequality for four matrices?

Question

It is known that $$ Tr[(AB)^n] \leq Tr (A^n B^n)$$ for $A$, $B$ positive semi-definite matrices. I am looking of a generalization of it for a product of $4$ positive semi-definite matrices of the the form, $$Tr [(ABCD)^n] \leq Tr [(AB)^{2n} ] Tr [(DC)^{2n}]$$ Is there any such inequality? Can someone suggest an approach if the above thing can be proved or disproved?

Answer 1

• The inequality you propose seems not very homogeneous ... in your case, just take $A=B=C=D$ and it becomes $$\mathrm{Tr}(A^{4n}) ≤ \mathrm{Tr}(A^{4n})^2,$$ which is obviously false for general matrices.

• The homogeneous version could be $$\mathrm{Tr}((ABCD)^{n}) ≤ \mathrm{Tr}((AB)^{2n})^\frac{1}{2}\, \mathrm{Tr}((CD)^{2n})^\frac{1}{2}.$$ But it seems to also be false in the general case. To prove that, take for example $CD = BA$ and you will obtain $$\mathrm{Tr}(|BA|^{2n}) ≤ \mathrm{Tr}((AB)^{2n})^\frac{1}{2}\, \mathrm{Tr}((BA)^{2n})^\frac{1}{2}$$ where I used the notation for the absolute value value of a matrix defined by $|A|^2 = A^* A$, which gives $|BA|^2 = AB^2A$. By using the cyclicity of the trace, it is easy to see that $\mathrm{Tr}((BA)^{2n}) = \mathrm{Tr}((AB)^{2n})$, so your inequality becomes $$\mathrm{Tr}(|BA|^{2n}) ≤ \mathrm{Tr}((BA)^{2n}),$$ which is false in general.

• An inequality in the idea of what you are looking for could be $$\mathrm{Tr}((ABCD)^{n}) ≤ \mathrm{Tr}(|AB|^{2n})^\frac{1}{2}\, \mathrm{Tr}(|CD|^{2n})^\frac{1}{2},$$ which follows from Hölder's inequality for matrices. It can equivalently be written $\mathrm{Tr}((ABCD)^{n}) ≤ \mathrm{Tr}((A^2B^2)^{n})^\frac{1}{2}\, \mathrm{Tr}((C^2D^2)^{n})^\frac{1}{2}$ using the definition of the absolute value and the cyclicity of the trace. The Araki-Lieb-Thirring inequality then tells you that you can also obtain $$\mathrm{Tr}((ABCD)^{n}) ≤ \mathrm{Tr}(A^{2n}B^{2n})^\frac{1}{2}\, \mathrm{Tr}(C^{2n}D^{2n})^\frac{1}{2}.$$