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Fall 2017 - Problem 6

Lp spaces, Minkowski's inequality

Let f∈L2(C)f \in L^2\p{\C}. For z∈Cz \in \C we define

g(z)=∫{w∈C∣∣w−z∣≤1}∣f(w)∣∣w−z∣ dA(w),g\p{z} = \int_{\set{w \in \C \mid \abs{w - z} \leq 1}} \frac{\abs{f\p{w}}}{\abs{w - z}} \,\diff{A\p{w}},

where dA\diff{A} denotes integration with respect to Lebesgue measure on C≃R2\C \simeq \R^2. Show that then ∣g(z)∣<∞\abs{g\p{z}} < \infty for almost every z∈Cz \in \C and that g∈L2(C)g \in L^2\p{\C}.
Solution.

First, by a change of variables,

g(z)=∫01∫02πr∣f(z+reiθ)∣r dθ dr=∫01∫02π∣f(z+reiθ)∣ dθ dr.g\p{z} = \int_0^1 \int_0^{2\pi} r \frac{\abs{f\p{z + re^{i\theta}}}}{r} \,\diff\theta \,\diff{r} = \int_0^1 \int_0^{2\pi} \abs{f\p{z + re^{i\theta}}} \,\diff\theta \,\diff{r}.

By Minkowski's inequality

(∫C∣g(z)∣2 dA(z))1/2=(∫C(∫01∫02π∣f(z+reiθ)∣ dθ dr)2 dA(z))1/2≤∫01∫02π(∫C∣f(z+reiθ)∣2 dA(z))1/2 dθ dr=∫01∫02π∥f∥L2 dθ dr=2π∥f∥L2<∞.\begin{aligned} \p{\int_\C \abs{g\p{z}}^2 \,\diff{A\p{z}}}^{1/2} &= \p{\int_\C \p{\int_0^1 \int_0^{2\pi} \abs{f\p{z + re^{i\theta}}} \,\diff\theta \,\diff{r}}^2 \,\diff{A}\p{z}}^{1/2} \\ &\leq \int_0^1 \int_0^{2\pi} \p{\int_\C \abs{f\p{z + re^{i\theta}}}^2 \,\diff{A\p{z}}}^{1/2} \,\diff\theta \,\diff{r} \\ &= \int_0^1 \int_0^{2\pi} \norm{f}_{L^2} \,\diff\theta \,\diff{r} \\ &= 2\pi \norm{f}_{L^2} \\ &< \infty. \end{aligned}

Thus, g∈L2(C)g \in L^2\p{\C} and ∣g(z)∣2<∞\abs{g\p{z}}^2 < \infty almost everywhere, which implies ∣g(z)∣<∞\abs{g\p{z}} < \infty almost everywhere as well.