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Convexity of exponential function

If C, D are n \times n symmetric real matrices we write C \geq D if-f the matrix C-D is non negative definite. Examine if

    \[\exp \left ( \frac{A+B}{2} \right ) \leq \frac{\exp A +\exp B}{2}\]

for each pair 2 \times 2 real symmetric matrices A, B such that AB=BA.


Since A, B commute it follows from the properties of the exponential function

(1)   \begin{equation*} \exp{\frac{A+B}{2}}=\exp{\frac{A}{2}} \exp{\frac{B}{2}} \end{equation*}

Noting that

(2)   \begin{equation*} \frac{\exp A +\exp B}{2}-\exp \left ( \frac{A+B}{2} \right )=\frac{1}{2}\left(\exp {\frac{A}{2}}-\exp {\frac{B}{2}} \right)^{2}  \end{equation*}

Setting \displaystyle M= \exp {\frac{A}{2}}-\exp {\frac{B}{2}} we must prove that M^2 \geq 0. Since A, B commute so are \displaystyle \exp \frac{A}{2} \; , \; \exp \frac{B}{2}. Hence M is symmetric. Thus,

(3)   \begin{equation*} x^{\top }M^{2}x=x^{\top }M^{\top }Mx=\left \| Mx \right \|^{2}  \end{equation*}

and that’s all!

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