Can every continuous function on \(\bar \DD\) be uniformly approximated by polynomials in the variable \(z\)?
Hint: compare to Weierstrass for the real interval.
Solution
1. If a sequence of polynomials \(p_n\) converges uniformly on \(\overline\DD\) to a function \(g\), then \(g\) is holomorphic on \(\DD\).
Each \(p_n\) is holomorphic on \(\DD\), and a locally uniform limit of holomorphic functions on an open set is holomorphic there.
2. The function \(f(z)\da\bar z\) is continuous on \(\overline\DD\) and not holomorphic on \(\DD\).
Continuity is clear. Writing \(f=u+iv\) with \(u(x,y)=x\) and \(v(x,y)=-y\) gives \(u_x=1\neq-1=v_y\), so the Cauchy–Riemann equations fail at every point.
3. Q.E.D.
By steps 1 and 2, no sequence of polynomials in \(z\) converges uniformly to \(\bar z\) on \(\overline\DD\). The answer is \(\boxed{\text{no}}\).