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Let $f:B^2(0,1)-\bar{B^2}(0,\frac{1}{2}) \to B^2(0,1)-{0}$ be a homeomorphism. To show that $f$ is not quasiconformal.

I can construct one such homeomorphism explicitly. However, I am not getting why such homeomorphism would not be quasiconformal. Any hints or idea regarding how to proceed ,would be very much helpful.

  • Please include what the map is that you've constructed, and why you think it is quasiconformal – MathTrain Apr 15 '22 at 13:49
  • The homeomorphism is $ h(x)= \frac{ 2||x||-1}{||x||} x $ from the annulas to the punctured ball. –  Apr 17 '22 at 13:49

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Quasiconformal mappings almost preserve modulus of annuli, they can distorce it by an amount between $K$ and $1/K$, where $K$ is it qc constant: $$ \frac{1}{K}\mbox{mod}(A)\le \mbox{mod}f(A)\le K \mbox{mod}(A) $$ Modulus of a puncture ball is infinity and the modulus of $B^{2}(0,1)\backslash B^{2}(0,\frac{1}{2})$ is $\frac{1}{2\pi}\log(2)$; so, your homeomorphism cannot be qc.

Lucas Oliveira
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