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Spring 2013 - Problem 5

harmonic functions, Hilbert spaces

Let D={(x,y)∣x2+y2<1}\D = \set{\p{x, y} \mid x^2 + y^2 < 1} and let us define a Hilbert space

H={u ⁣:D→R | u is harmonic and ∫D∣u(x,y)∣2 dx dy<∞}H = \set{\func{u}{\D}{\R} \st u \text{ is harmonic and } \int_\D \abs{u\p{x,y}}^2 \,\diff{x} \,\diff{y} < \infty}

with inner product ⟨f,g⟩=∫Dfg dx dy\inner{f, g} = \int_\D fg \,\diff{x} \,\diff{y}.

  1. Show that f↦∂f∂x(0,0)f \mapsto \pder{f}{x}\p{0, 0} is a bounded linear functional on HH.
  2. Compute the norm of this linear functional.
Solution.

  1. Let u∈Hu \in H. Since D\D is simply connected, u=Re⁡fu = \Re{f} for some holomorphic function on the disk. Write f(z)=∑n=0∞anznf\p{z} = \sum_{n=0}^\infty a_nz^n so that ux(0)=Re⁡a1u_x\p{0} = \Re{a_1}. Thus, for r∈(0,1)r \in \p{0, 1},

    f(reiθ)=∑n=0∞anrneinθ=∑n=0∞anrn(cos⁡nθ+isin⁡nθ)  ⟹  u(reiθ)=∑n=0∞rn(Re⁡ (an)cos⁡nθ−Im⁡ (an)sin⁡nθ).\begin{gathered} f\p{re^{i\theta}} = \sum_{n=0}^\infty a_nr^ne^{in\theta} = \sum_{n=0}^\infty a_nr^n\p{\cos{n\theta} + i\sin{n\theta}} \\ \implies u\p{re^{i\theta}} = \sum_{n=0}^\infty r^n\p{\Re\,\p{a_n}\cos{n\theta} - \Im\,\p{a_n}\sin{n\theta}}. \end{gathered}

    By a change of variables, we have

    ∫D∣u(x,y)∣2 dA=∫01∫02πru2(reiθ) dθ dr=∫01∫02πr(∑n=0∞rn(Re⁡ (an)cos⁡nθ−Im⁡ (an)sin⁡nθ))2 dθ dr=∫01∫02πr∑n,m=0∞rnrm((Re⁡ (an)cos⁡nθ−Im⁡ (an)sin⁡nθ))((Re⁡ (am)cos⁡mθ−Im⁡ (am)sin⁡mθ)) dθ dr=∫01∑n=0∞r2n+1∫02π(Re⁡ (an)2cos⁡2nθ+Im⁡ (an)2sin⁡2nθ) dθ dr,\begin{aligned} \int_\D \abs{u\p{x, y}}^2 \,\diff{A} &= \int_0^1 \int_0^{2\pi} ru^2\p{re^{i\theta}} \,\diff\theta \,\diff{r} \\ &= \int_0^1 \int_0^{2\pi} r\p{\sum_{n=0}^\infty r^n \p{\Re\,\p{a_n}\cos{n\theta} - \Im\,\p{a_n}\sin{n\theta}}}^2 \,\diff\theta \,\diff{r} \\ &= \int_0^1 \int_0^{2\pi} r\sum_{n,m=0}^\infty r^n r^m \p{\p{\Re\,\p{a_n}\cos{n\theta} - \Im\,\p{a_n}\sin{n\theta}}}\p{\p{\Re\,\p{a_m}\cos{m\theta} - \Im\,\p{a_m}\sin{m\theta}}} \,\diff\theta \,\diff{r} \\ &= \int_0^1 \sum_{n=0}^\infty r^{2n+1} \int_0^{2\pi} \p{\Re\,\p{a_n}^2\cos^2{n\theta} + \Im\,\p{a_n}^2\sin^2{n\theta}} \,\diff\theta \,\diff{r}, \end{aligned}

    by orthogonality of the trigonometric terms and uniform convergence of the power series for uu on compact sets. Continuing,

    ∫01∑n=0∞r2n+1∫02π(Re⁡ (an)2cos⁡2nθ+Im⁡ (an)2sin⁡2nθ) dθ dr=∫01π∑n=0∞r2n+1(Re⁡ (an)2+Im⁡ (an)2) dr≥∫01πr3Re⁡ (a1)2 dr=π4Re⁡ (a1)2.\begin{aligned} &\int_0^1 \sum_{n=0}^\infty r^{2n+1} \int_0^{2\pi} \p{\Re\,\p{a_n}^2\cos^2{n\theta} + \Im\,\p{a_n}^2\sin^2{n\theta}} \,\diff\theta \,\diff{r} \\ ={}& \int_0^1 \pi \sum_{n=0}^\infty r^{2n+1} \p{\Re\,\p{a_n}^2 + \Im\,\p{a_n}^2} \,\diff{r} \\ \geq{}& \int_0^1 \pi r^3 \Re\,\p{a_1}^2 \,\diff{r} \\ ={}& \frac{\pi}{4} \Re\,\p{a_1}^2. \end{aligned}

    Hence, ux(0)≤2π∥u∥L2u_x\p{0} \leq \frac{2}{\sqrt{\pi}}\norm{u}_{L^2}, so the functional is bounded.

  2. Notice that if u(x,y)=xu\p{x, y} = x, then

    ∫D∣u(x,y)∣2 dA=∫01∫02πr3cos⁡2θ dθ dr=π4,\int_\D \abs{u\p{x, y}}^2 \,\diff{A} = \int_0^1 \int_0^{2\pi} r^3\cos^2\theta \,\diff\theta \,\diff{r} = \frac{\pi}{4},

    so v(x,y)=2xπv\p{x, y} = \frac{2x}{\sqrt{\pi}} satisfies ∥v∥L2=1\norm{v}_{L^2} = 1. Moreover, vx(0)=2πv_x\p{0} = \frac{2}{\sqrt{\pi}}, so the functional attains the upper bound 2π\frac{2}{\sqrt{\pi}}, i.e., its norm is 2π\frac{2}{\sqrt{\pi}}.