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
$$\|x\|\_A=\sqrt{x^TAx},\qquad \|x\|\_B=\sqrt{x^TBx}.$$



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> 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 $B$ ?
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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



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> 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 ?
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Interpolation of normed spaces *vs* geometrical mean of positive matrices
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 $Y$.


