I am trying to understand Theorem 2 ($F(A,K) := \text{tr}(A^{-r}K^{\ast} A^{-p} K)$ is jointly convex for $p,r \geq 0$ and $p+r \leq 1$) of the paper Convex trace functions and the Wigner-Yanase-Dyson conjecture. I follow the steps and the argument (I think).
In my understanding, in order to prove that the function $F(A,K) := \text{tr}(A^{-r}K^{\ast} A^{-p} K)$ is convex, we differentiate the function $N(A,B,K,L)/D(A,B,K,L)$ where $$N(A,B,K,L):= F(\lambda A + (1-\lambda)B, \lambda K + (1-\lambda)L)$$ and $$D(A,B,K,L) := \lambda F(A,K) + (1-\lambda)F(B,L)$$
and show that if the maxima exists then $F$ must be convex. I do not understand why a maxima must exist.
Update (3 Apr, 2023): Here is a possible argument: In order to maximise $N/D$ both $K$ and $L$ can be chosen such that their norms (say $2$-norm) are at most $1$. To see this, suppose $\Vert K\Vert_2 \geq \Vert L\Vert_2$, then $$\frac{N(A,B,K,L)}{D(A,B,K,L)} = \frac{N(A,B,K/\Vert K\Vert_2,L/\Vert K\Vert_2)}{D(A,B,K/\Vert K\Vert_2,L/\Vert K\Vert_2)}$$ and the operators $K/\Vert K\Vert_2,L/\Vert K\Vert_2$ both have norm at most $1$. In finite dimensions (which is assumed to be the case) there exists a maximum in the arguments $K,L$ for $N/D$ over the unit ball (call it $K_m, L_m$). Moreover, as a result of the above equation every value obtained by $N/D$ is obtained in the unit ball, which means a global maxima should also exist and must coincide with $K_m, L_m$. Hence, one can assume that a maxima exists for the function and in particular the derivatives can be set to $0$.
Is this argument correct?
