Answer: The Harmonic Function That Knows Its Boundary

Key Idea / Intuition

The value of a harmonic function at the center of the disk is the average of its boundary values — this is the mean value property. So $u(0)$ is simply the average of $u$ over the unit circle. Since $u$ is zero on the upper half and non-negative on the lower half, the average is automatically non-negative. The exact formula drops out immediately from the mean value theorem.

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Formal Proof / Solution

Step 1: The Mean Value Property.

For any function $u$ that is continuous on $\overline{\mathbb{D}}$ and harmonic on $\mathbb{D}$, the mean value property states:

$u(0) = \frac{1}{2\pi} \int_0^{2\pi} u(e^{i\theta})\, d\theta.$

This is one of the most fundamental facts in complex analysis: the value at the center equals the average over any centered circle.

Step 2: Split the integral by the two semicircles.

$u(0) = \frac{1}{2\pi} \int_0^{\pi} u(e^{i\theta})\, d\theta + \frac{1}{2\pi} \int_{\pi}^{2\pi} u(e^{i\theta})\, d\theta.$

The first integral vanishes (since $u = 0$ on the upper semicircle), giving:

$u(0) = \frac{1}{2\pi} \int_{\pi}^{2\pi} u(e^{i\theta})\, d\theta.$

Step 3: Sign.

Since $u \geq 0$ on the lower semicircle $\{e^{i\theta} : \pi \leq \theta \leq 2\pi\}$, the integrand is non-negative, hence:

$u(0) = \frac{1}{2\pi} \int_{\pi}^{2\pi} u(e^{i\theta})\, d\theta \geq 0.$

Conclusion.

Yes, $u(0) \geq 0$ is guaranteed, with equality if and only if $u \equiv 0$ on the lower semicircle as well (which by the maximum principle would force $u \equiv 0$ everywhere).

The exact value is:

$\boxed{u(0) = \frac{1}{2\pi} \int_{\pi}^{2\pi} u(e^{i\theta})\, d\theta.}$

Why this is beautiful: The mean value property gives a global conclusion ($u(0) \geq 0$) from local boundary information, with no computation needed — just the realization that the center "sees" the boundary democratically and equally in every direction.