3
$\begingroup$

The following problem is the 5th problem in the qualifying exam of UCLA (spring 2013).

Let $\mathbb{D}=\{(x,y):x^2+y^2<1\}$ and let us define a Hilbert space $$H:=\{u:\mathbb{D} \rightarrow \mathbb{R}:u \text{ is harmonic and }\int_{\mathbb{D}}u^2 dxdy< \infty\}$$ with inner product $<u,v>=\int_\mathbb{D} uvdxdy$.

(a)Show that $l(u)=\frac{\partial u}{\partial x}(0,0)$ is a bounded linear functional on $H$.

(b) Find the norm of $l$.

I have some ideas in proving part (a).The first idea:

For all $\delta\in (0,1)$, $$\frac{u(\delta,0)-u(-\delta,0)}{2\delta}=\frac{\int_{D_1\Delta D_2}u}{2\delta(1-\delta)^2 \pi}$$,where $D_1$,$D_2$ are the disc of radius $1-\delta$ centred at $(-\delta,0)$ and $(\delta,0)$ respectively.If I can show $\frac{\sqrt{m(D_1\Delta D_2)}}{\delta}=O(1)$then I can prove part (a).However,it turns out that the value is quite hard to calculate and it seems that it does not provide a nice value for part (b) so I think my idea is wrong.

I have a second idea: Let $v$ be the harmonic conjugate of $u$ which vanishes at origin.Let $f=u+iv$.Then by Cauchy integral formula,$$f'(0)=\frac{1}{2\pi i}\int_{C_r}\frac{f(w)}{(w-0)^2} dw=\frac{1}{2\pi}\int_0^{2\pi} f(re^{i \theta})(re^{i \theta})^{-1} d \theta.$$So we get $$f'(0)=\frac{1}{\pi}\int_{\mathbb{D}}\frac{f(z)}{z} dA.$$Hence,$$\frac{1}{\pi}\int_D \frac{ux+vy}{x^2+y^2}dxdy=\frac{\partial u}{\partial x}(0,0)$$.If I can bound the integral involving $v$ by the norm of $u$,then we are done. However,I don't have a way to bound $v$ at this moment.

Can anyone give me some hints on this problem? I greatly appreciate it.

$\endgroup$
1
2
$\begingroup$

The second idea is better, but instead of Cauchy integral formula you should have used the power series representation for $f$. Both the norm and the functional can be expressed in terms of the Taylor coefficients, reducing the problem to elementary inequalities for complex numbers.

Indeed, in the expansion $$u(z)= \sum_{n=0}^\infty \operatorname{Re}(c_n z^n)$$ the terms in the sum are mutually orthogonal (in $L^2$ sense), which is easy to see by integrating in polar coordinates. It follows that $$\int_{\mathbb D}u^2 = \int_0^1 r\,dr \int_0^{2\pi} u^2\,d\theta = \int_0^1 \left(2\pi (\operatorname{Re} c_0)^2 + \pi \sum_{n=1}^\infty |c_n|^2 r^{2n}\right)r\,dr $$ hence $$\int_{\mathbb D}u^2 = \pi (\operatorname{Re} c_0)^2 + \pi \sum_{n=1}^\infty \frac{|c_n|^2}{2n+2} \tag{1}$$

Since $l(u) = \operatorname{Re} c_1$, it follows that $$(l(u))^2 \le \frac{4}{\pi} \int u^2$$ and this is sharp (attained by $u(x,y)=x$).

$\endgroup$
1
  • $\begingroup$ This solution is nice.Thank you. $\endgroup$ – Ben Mar 24 '14 at 4:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.