Can $e^\frac{1+z}{z-1}$ be uniformly approximated in the disc by harmonic continuous functions

Question

Ie: For $f(z)=e^\frac{1+z}{z-1}$ is it true: $\inf\{||f-u||: \mbox{u is harmonic in the unit disc and continuous on the closed unit disc}\}=0$. Note the infinum is taken over the interior (ie: $|z|<1$)

Answer 1

No, in fact $\sup_{|z|<1}|f(z)-u(z)|\ge 1$ for all such $u.$ Proof: Observe that as $t \to 0^+,$ $f(e^{it})$ takes on every value in the unit circle infinitely many times. Using the continuity of $f$ at all boundary points save $1,$ we see there exists sequences $z_n,w_n$ in the open disc, $z_n, w_n \to 1,$ such that $f(z_n) \to 1, f(w_n)\to -1.$ Thus $f(z_n)-u(z_n)\to 1-u(1),$ while $f(w_n)-u(w_n)\to -1-u(0).$ Now one of $1-u(1),-1-u(1)$ has modulus at least $1,$ giving the result.