Interpolation of normed spaces *vs* geometrical mean of positive matrices Two $n\times n$ positive definite symetric matrices $A,B$ define two normed spaces $E_A=({\mathbb R}^n;\|\cdot\|_A)$ and $E_B=({\mathbb R}^n;\|\cdot\|_B)$, where
$$\|x\|_A=\sqrt{x^TAx},\qquad \|x\|_B=\sqrt{x^TBx}.$$

Is it true that the interpolation space $[E_A,E_B]_{1/2}$ equals $E_{A\sharp B}$, where $A\sharp B$ is the geometrical mean of $A$ and $B$ ?

Let me recall that $A\sharp B$ is the 'middle' of the geodesic from $A$ to $B$, with respect to the Riemannian metric for which ${\bf SPD}_n$ is a symmetric space:
$$D(A,B)=\left({\rm Tr}(\log(BA^{-1}))^2\right)^{\frac12}.$$
A pedestrian way to ask the question is

Let $f$ be a linear form such that $|f(x)|^4\le(x^TAx)(x^TA^{-1}x)$ for every $x\in{\mathbb R}^n$. Is it true that $|f(x)|\le\|x\|$ for the standard euclidian norm ?

 A: Yes. The proof of Theorem 1.1 from John E. McCarthy's "Geometric interpolation between Hilbert spaces," Ark. Mat. 30, 321-330 (1991) works for this case. Let $A_i$, $B_i$, be SPD matrices, $i = 1, 2$, and let $$X_i = A_i^{1/2}(A_i^{-1/2} B_i A_i^{-1/2})^{1/2}A_i^{1/2}$$ be the geometric means of each pair $(A_i,B_i)$. Given any matrix $L$ with compatible dimensions, the norm $\lVert L \rVert_{X_2,X_1} = \sup_{\lVert u \rVert_{X_1} \le 1} \lVert L u \rVert_{X_2}$ has square bounded by
$$ \begin{aligned} \lVert L \rVert_{X_2,X_1}^2
&= \rho(X_1^{-1} L^* X_2 L) \le
\lVert X_1^{-1} L^* X_2 L \rVert_{A_1} \\
&\le \lVert X_1^{-1} L^* X_2 \rVert_{A_1,A_2}
\lVert L \rVert_{A_2, A_1} \\
&= \lVert X_2 L X_1^{-1} \rVert_{A_2^{-1},A_1^{-1}}
\lVert L \rVert_{A_2, A_1} \\
&= \lVert L \rVert_{B_2,B_1}
\lVert L \rVert_{A_2, A_1}
 \end{aligned} $$
so that $(E_{X_1}, E_{X_2})$ is an exact interpolation pair of exponent $1/2$ for the couples $(E_{A_1}, E_{B_1})$ and $(E_{A_2}, E_{B_2})$. Note the last step above used that, e.g., $\lVert X_1 u \rVert_{A_1^{-1}} = \lVert u \rVert_{B_1}$, which follows since $X_1^*A_1^{-1} X_1 = B_1$. The prior step used the identity $\lVert T^* \rVert = \lVert T \rVert$, accounting for the fact that $\lVert f \rVert_{Y^{-1}}$ is the norm dual to $\lVert u \rVert_Y$ for SPD $Y$.
