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A harmonic polynomial is a polynomial annihilated by the Laplacian.

Let ${\wp }_{\Lambda }$ be the Weierstrass elliptic function forBecause $2\pi i$ is a period, the formulais independent of the branch of the complex logarithm. It is a nonconstant harmonic function onand its other period gives $u(qz)=u(z)$.

A bounded harmonic function on the complex plane is constant. More generally, an entire harmonic function bounded on either side is constant: write it as $\mathrm{Re}f$ by harmonic function as the real part of a holomorphic function, then apply the Liouville theorem to $e^f$ or $e^{-f}$.

A continuous function harmonic on a bounded domain attains its maximum and minimum on the boundary. Apply the strictly subharmonic result to $u+ϵ|x{|}^2$ and let $ϵ\downarrow 0$, then apply the same argument to $-u$.

A twice differentiable function is strictly subharmonic where $\Delta u>0$. It cannot have an interior local maximum because the Hessian at such a maximum is negative semidefinite.

A real-valued function $u$ on a simply connected domain is harmonic exactly whenfor some holomorphic function $f$. One construction observes that $u_x-i{u}_y$ is holomorphic and takes its holomorphic primitive.

If an entire harmonic function on ${\mathbb{R}}^n$ grows like $O(1+|x{|}^{k+\alpha })$ with $k\in \mathbb{N}$ and $0<\alpha <1$, then it is a harmonic polynomial of degree at most $k$. Apply the interior derivative estimate of order $k+1$ on expanding balls.

If $u$ is harmonic on $B_r(x_0)$, then for every integer $m\ge 1$,The first-order estimate follows by differentiating the ball mean-value formula and applying the divergence theorem; iteration gives higher orders.

The Weyl lemma says that every locally integrable distributional solution of $\Delta u=0$ agrees almost everywhere with a smooth harmonic function. Convolution with a mollifier and the harmonic mean value property provide a standard proof.

For every closed ball contained in the domain of a harmonic function, the value at its centre equals both its average over the ball and its average over the boundary sphere.

In harmonic gauge, the vacuum Einstein equations have principal part $-g^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }{g}_{\alpha \beta }/2$, forming a quasilinear wave system supplemented by constraints.

Harmonic slicing imposes the harmonic-coordinate condition only on time. In 3+1 form it obeys $({\partial }_t-{\beta }^i{\partial }_i)\alpha =-{\alpha }^{2}K$.

For canonical coordinates $(q_i,p_i)$ and Hamiltonian $H$,

A point outside a nonempty closed convex subset of a normed space can be strictly separated from it by the real part of a bounded linear functional.

Point evaluation at zero is bounded on $C([0,1])$ with the essential-supremum norm, and Hahn--Banach extends it to a bounded functional on $L^{\infty }([0,1])$. No $g\in L^1$ represents this extension: continuous functions supported in shrinking neighbourhoods of zero have value one at zero while their integrals against $g$ tend to zero. Hence $(L^{\infty }{)}^{\prime }$ is strictly larger than $L^1$.

For a normed space $X$, the map $J:X\to X^{\prime \prime }$ defined byis a linear isometry. The easy inequality is $\Vert Jx\Vert \le \Vert x\Vert$; Hahn--Banach extends the norm-one functional on $\mathrm{span}\{x\}$ that takes $x$ to $\Vert x\Vert$, proving the reverse inequality.

Adding a phase gate to the Hadamard test rotates the interference signal and measures the imaginary part of $⟨\psi |U|\psi ⟩$.

If $U|\psi ⟩=e^{2\pi i\theta }|\psi ⟩$, the ordinary Hadamard test returns ancilla zero with probability

A bias-free network with $p$ inputs, $m$ sign-activated hidden nodes, and a thresholded linear output satisfiesThe hidden units contribute $s(\mathcal{G},n{)}^m\le (n+1{)}^{pm}$ sample patterns, and the output weights contribute $(n+1{)}^m$.

For $N=4$ and one marked item, $\theta =\pi /6$. One Grover iteration gives marked amplitude $\sin (3\theta )=1$, so measurement returns the target with certainty.