Proof of uniqueness for semisimple groups
Mostow (1955) also related the general problem for semisimple groups to the case of GL(n, R). The corresponding symmetric space is the space of positive symmetric matrices. A direct proof of uniqueness relying on elementary properties of this space is given in Hilgert & Neeb (2012).
Let
be a real semisimple Lie algebra with Cartan involution σ. Thus the fixed point subgroup of σ is the maximal compact subgroup K and there is an eigenspace decomposition

where
, the Lie algebra of K, is the +1 eigenspace. The Cartan decomposition gives

If B is the Killing form on
given by B(X,Y) = Tr (ad X)(ad Y), then

is a real inner product on
. Under the adjoint representation, K is the subgroup of G that preserves this inner product.
If H is another compact subgroup of G, then averaging the inner product over H with respect to the Haar measure gives an inner product invariant under H. The operators Ad p with p in P are positive symmetric operators. This new inner produst can be written as

where S is a positive symmetric operator on
such that
Ad(h)tS Ad h = S for h in H (with the transposes computed with respect to the inner product). Moreover, for x in G,

So for h in H,

For X in
define

If ei is an orthonormal basis of eigenvectors for S with Sei = λi ei, then

so that f is strictly positive and tends to ∞ as |X| tends to ∞. In fact this norm is equivalent to the operator norm on the symmetric operators ad X and each non-zero eigenvalue occurs with its negative, since i ad X is a skew-adjoint operator on the compact real form
.
So f has a global minimum at Y say. This minimum is unique, because if Z were another then

where X in
is defined by the Cartan decomposition

If fi is an orthonormal basis of eigenvectors of ad X with corresponding real eigenvalues μi, then

Since the right hand side is a positive combination of exponentials, the real-valued function g is strictly convex if X ≠ 0, so has a unique minimum. On the other hand, it has local minima at t = 0 and t = 1, hence X = 0 and p = exp Y is the unique global minimum. By construction
f(x) = f(σ(h)xh−1) for h in H, so that p = σ(h)ph−1 for h in H. Hence σ(h)= php−1. Consequently, if g = exp Y/2, gHg−1 is fixed by σ and therefore lies in K.