# operator concavity of a function involving trace and logarithm

I want to see if $$f(A)= \operatorname{trace}(C\log(I+\sqrt{A}B\sqrt{A}))$$ is operator concave with respect to a Hermitian positive definite matrices $A$? $\log$ is matrix logarithm, $B$ and $C$ are arbitrary Hermitian positive definite matrices, and $I$ is unit matrix.

I checked it numerically with many randomly generated positive definite matrices, using the following condition from Bhatia's matrix analysis: $$f\left({\frac{X+Y}{2}}\right)>\frac{f(X)}{2}+\frac{f(Y)}{2}$$ and it seems that the condition is satisfied.

I want to prove it using the same condition. I am trying to show: $$\operatorname{trace}\left(C\log\left(I+\sqrt{\frac{X+Y}{2}}B\sqrt{\frac{X+Y}{2}}\right)\right)>\frac{\operatorname{trace}\left(C\left(\log(I+\sqrt{X}B\sqrt{X})+\log(I+\sqrt{Y}B\sqrt{Y})\right)\right)}{2}, (1)$$

Update:

First, I solve it for special case $C=I$ (i.e. $C$ is identity matrix). Using $\operatorname{trace}(\log x)=\log \det(x)$, left side of (1) is, $$\operatorname{trace}\left(\log\left(I+\sqrt{\frac{X+Y}{2}}B\sqrt{\frac{X+Y}{2}}\right)\right) \\=\log\left(\det\left(I+\sqrt{\frac{X+Y}{2}}B\sqrt{\frac{X+Y}{2}}\right)\right) \\=\log\left(\det\left(I+{\frac{X+Y}{2}}B\right)\right), (2)$$ Third line is from Sylvester's determinant identity(https://en.wikipedia.org/wiki/Sylvester%27s_determinant_identity). Right side of (1) becomes $$\frac{1}{2}\operatorname{trace}\left(\left(\log(I+\sqrt{X}B\sqrt{X})+\log(I+\sqrt{Y}B\sqrt{Y})\right)\right)\\ =\frac{1}{2}\left(\log\left(\det(I+\sqrt{X}B\sqrt{X})\right)+\log\left(\det(I+\sqrt{Y}B\sqrt{Y})\right)\right)\\ =\frac{1}{2}\left(\log\left(\det(I+XB)\right)+\log\left(\det(I+YB)\right)\right), (3)$$

From concavity of log-determinant (Log-Determinant Concavity Proof), I conclude (2) is greater than(3) and thus the condition (1) is satisfied for $C=I$. I have two questions:

1) Is my conclusion correct?(I am not 100% sure)

2)How can I extend this result (if it is correct) to a general positive definite $C$?

Update 2: I also tried the approach suggested in the answer to this question: Is the trace of inverse matrix convex? ...but the second derivative was too complicated.

Any help is very appreciated.

If you are aware of any other way to prove concavity of $f(A)= \operatorname{trace}(C\log(I+\sqrt{A}B\sqrt{A}))$ please let me know.

• Don't you think you get more answers if you explain the notions you use? – amsmath Jul 17 '18 at 21:59
• I updated my question and tried to explain things clearly. – Mah Jul 19 '18 at 12:42

I also tried this numerically and quickly stumbled upon counterexamples

>> A1
A1 =
0.485971910201617   0.392980465973661
0.392980465973661   0.323621544804127
>> A2
A2 =
0.864345280133754   0.504105821991426
0.504105821991426   0.596439049805159
>> B
B =
1.659740432896511   0.467165733847868
0.467165733847868   0.487112266549647
>> C
C =
0.132518904109893   0.298931955806039
0.298931955806039   1.161589772124985
>> A3 = (A1+A2)/2;
>> f1 = trace(C*logm(I+sqrtm(A1)*B*sqrtm(A1)))
f1 = 0.6891
>> f2 = trace(C*logm(I+sqrtm(A2)*B*sqrtm(A2)))
f2 = 0.8090
>> f3 = trace(C*logm(I+sqrtm(A3)*B*sqrtm(A3)))
f3 = 0.7355