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    Polynomials do not converge uniformly to $1/z$ on $S^1$

    P-4HP1O

    Area
    Complex Analysis
    Topics
    Cauchy Integral Formula, Counterexamples, Polynomials, Uniform Convergence
    Status
    draft

    Prove that there is no sequence of polynomials that uniformly converge to \(f(z) = {1\over z}\) on \(S^1\).

    Solution

    • By Cauchy’s integral formula, \(\int_{S^1} f = 2\pi i\)

    • If \(p_j\) is any polynomial, then \(p_j\) is holomorphic in \(\DD\), so \(\int_{S^1} p_j = 0\).

    • Contradiction: compact sets in \(\CC\) are bounded, so \[ \abs{\int f - \int p_j} &\leq \int \abs{p_j - f} \\ &\leq \int \norm{p_j - f}_\infty \\ &= \norm{p_j - f}_\infty \int_{S^1} 1 \,dz \\ &= \norm{p_j-f}_\infty \cdot 2\pi \\ &\to 0 \] which forces \(\int f = \int p_j = 0\).

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