Define $\pi_\mu(z)=N(z; \mu, \sigma^2 I)$. Define $\phi(z)$ to be the pdf of a gaussian mixture where every gaussian also has covariance $\sigma^2 I$. I wonder what is the solution for $\hat\mu = \arg\max_{\|\mu\|_2\le r} \mathrm{TV}(\pi_\mu||\phi)$?
The definition of total variation divergence is $\mathrm{TV}(\pi_\mu||\phi) = \frac{1}{2}\int|\pi_\mu(z) - \phi(z)|dz = \int\max(\pi_\mu(z) - \phi(z),0)dz$.
A natural guess is that $\hat\mu$ is at the opposite direction of the barycenter $z_c$ of $\phi$, which means $\hat\mu = \frac{-r}{\|z_c\|_2}z_c$ and $z_c = \int \phi(z)z \mathrm{d}z$. But I doubt its correctness.