Consider $G/K$ a symmetric space of compact type. We have $\mathfrak{g}=\mathfrak{k}+\mathfrak{m}$. Let $\mathfrak{a}$ be a maximal abelian subspace of $\mathfrak{m}$. Then we have a root system $\Sigma$ and we chose positive roots $\Sigma^+$. If we denote $$\mathfrak{a}_r=\left\lbrace H\in \mathfrak{a}\vert \forall \lambda\in \Sigma,\lambda(H)\not\in i\pi\mathbb{Z}\right\rbrace,$$an $Q$ the connected component of $\mathfrak{a}_r$ containend in the Weyl chamber associated to the choice of positive roots and containing $0$ in its closure.
In Helgason, Differential Geometry, Lie Groups and Symmetric Spaces, Theorem 8.6 p.323, it is shown that for any $g\in G$, there are $k_1,k_2\in G$ and a unique $H\in \overline{Q}$ such that $g=k_1\exp(H)k_2$.
My question is, is the map $g\mapsto H(g)$ smooth ? And $k_i$ are not unique, but is there a choice such that $g\mapsto k_i(g)$ is also smooth ?
Consider the Lie subgroup $D_+$ of $SL(2, {\mathbb R})$ consisting of diagonal matrices with positive diagonal entries, $A=diag(a, a^{-1}), a>0$. Then $$ H(A)= \log(a),\ a\ge 1,$$ $$ H(A)=\log(a^{-1})= -\log(a),\ 0< a\le 1 $$ (up to a normalizing multiplicative factor). In other words, $H(A)=|\log(a)|$.
I will leave it to you to verify that this function is not even differentiable at $a=1$, let alone smooth. Since $D_+$ is a smooth submanifold of $SL(2, {\mathbb R})$, it follows that $H: SL(2, {\mathbb R})\to {\mathfrak a}^+=\bar{Q}$ is not smooth.
The situation will be similar for higher rank Lie groups: The Cartan projection $H$ will not be smooth over the boundary of ${\mathfrak a}^+$ (but smooth over ${\mathfrak a}^+_r$).