Prove that there is no sequence of polynomials that converge uniformly to the function \(\textstyle f ( z ) = { \frac { 1 } { z } }\) on the unit circle.
Solution
Let \[ \Gamma=\{z\in\CC:\abs z=1\} \] with counterclockwise orientation.
1. For every polynomial \(p\),\[ \int_\Gamma p(z)\,dz=0. \]
Every polynomial has a polynomial antiderivative on all of \(\CC\). The integral of a derivative over a closed curve is zero, so the displayed integral vanishes.
2. If a sequence of polynomials \(p_n\) converged uniformly to \(1/z\) on\(\Gamma\), then\[ \int_\Gamma p_n(z)\,dz \longrightarrow \int_\Gamma \frac{dz}{z}. \]
Uniform convergence gives \[ \sup_{z\in\Gamma} \abs{p_n(z)-1/z} \longrightarrow0. \] Therefore \[ \begin{aligned} \left| \int_\Gamma \left(p_n(z)-\frac1z\right)\,dz \right| &\leq \operatorname{length}(\Gamma) \sup_{z\in\Gamma} \abs{p_n(z)-1/z}\\ &= 2\pi \sup_{z\in\Gamma} \abs{p_n(z)-1/z} \longrightarrow0. \end{aligned} \] This is exactly convergence of the contour integrals.
3. The contour integral of \(1/z\) around \(\Gamma\) is\[ \int_\Gamma\frac{dz}{z}=2\pi i. \]
Parametrize \[ z=e^{it}, \qquad 0\leq t\leq2\pi. \] Then \[ dz=ie^{it}\,dt, \] so \[ \int_\Gamma\frac{dz}{z} = \int_0^{2\pi}i\,dt = 2\pi i. \]
4. No sequence of polynomials can converge uniformly to \(1/z\) on theunit circle.
If such a sequence existed, step 1 would give \[ \int_\Gamma p_n(z)\,dz=0 \] for every \(n\). By step 2, these zero integrals would converge to \[ \int_\Gamma\frac{dz}{z}. \] Step 3 says that limit is \(2\pi i\neq0\), a contradiction.
5. Q.E.D.
Step 4 is the required nonexistence statement.