# Proving that $\varphi$ is operator convex.

Let $$\mathbb P_n$$ be the space of all $$n \times n$$ self-adjoint positive definite matrices. Consider the function $$\varphi: \mathbb P_n \longrightarrow \mathbb R$$ defined by $$\varphi (A) = \text {tr}\ (A \log A).$$ Show that $$\varphi$$ is operator convex.

I need to show that for any $$A,B \in \mathbb P_n$$ $$(1-t) \varphi (A) + t \varphi (B) \geq \varphi ((1-t) A + t B)$$ for all $$t \in [0,1].$$

I can able to show the result when $$A$$ and $$B$$ are both diagonal by using the fact that the function $$x \mapsto x \log x$$ is convex on $$(0,\infty).$$ Now for any arbitrary $$A$$ and $$B$$ there exist unitary matrices $$U$$ and $$V$$ such that $$U^* A U = D$$ and $$V^* B V = D',$$ where $$D$$ and $$D'$$ are both diagonal matrices. Now it is easy to see that $$\varphi (A) = \varphi (D)$$ and $$\varphi (B) = \varphi (D').$$ Hence we have \begin{align*} (1-t) \varphi (A) + t \varphi (B) & = (1-t) \varphi (D) + t \varphi (D') \\ & \geq \varphi ((1-t) D + t D') \end{align*} If we can somehow show that $$\varphi ((1-t) D + t D') = \varphi ((1-t) A + t B)$$ then we are through. If $$A$$ and $$B$$ commute then it is true since then $$A$$ and $$B$$ are simultaneously diagonalizable. But how can it shown if $$A$$ and $$B$$ don't commute? Could anyone suggest something needful?

Thanks a bunch!