# The derivative of harmonic function at origin is a bounded linear functional

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.

## 1 Answer

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 $u(x,y)=x$).

• This solution is nice.Thank you. – Ben Mar 24 '14 at 4:58