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Abstract

We prove that a $K$-quasiconformal mapping $f:\mathbb R^2\rightarrow\mathbb R^2$ which maps the unit disk $\mathbb D$ onto itself preserves the space ${\rm EXP}(\mathbb D)$ of exponentially integrable functions over $\mathbb D$, in the sense that $u \in {\rm EXP}(\mathbb D)$ if and only if $u \circ f^{-1} \in {\rm EXP}(\mathbb D)$. Moreover, if $f$ is assumed to be conformal outside the unit disk and principal, we provide the estimate $$ \frac 1{1+K\log K}\le \frac{\|u \circ f^{-1}\|_{{\rm EXP}(\mathbb D)}}{\|u\|_{\rm{EXP}(\mathbb D)} } \le 1+K\log K $$ for every $ u \in {\rm EXP}(\mathbb{D})$. Similarly, we consider the distance from $L^\infty$ in $\rm EXP$ and we prove that if $f:{\mit\Omega} \rightarrow {\mit\Omega}^\prime$ is a $K$-quasiconformal mapping and $G \subset \subset \mit\Omega$, then $$ \frac 1 K \le \frac{{\rm dist}_{{\rm EXP}(f(G))} (u \circ f^{-1},L^\infty(f(G)))}{ {\rm dist}_{{\rm EXP}(f(G))} (u,L^\infty(G ))}\le K $$ for every $ u \in{\rm EXP}(\mathbb G)$. We also prove that the last estimate is sharp, in the sense that there exist a quasiconformal mapping $f:\mathbb D \rightarrow \mathbb D$, a domain $G \subset \subset \mathbb D$ and a function $u\in {\rm EXP}(G)$ such that $$ {\rm dist}_{{\rm EXP}(f(G))} (u \circ f^{-1},L^\infty(f(G)))= K\,{\rm dist}_{{\rm EXP}(f(G))} (u,L^\infty(G )). $$