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The local maximum modulus principle says that if $|f|$ has a local maximum at an interior point of a connected domain, then $f$ is constant. On a small circle about the maximum, the mean-value property andforce equality everywhere. Equality in the triangle inequality makes the boundary values identical; Cauchy's formula then makes $f$ constant locally, and the identity theorem makes it constant on the domain.

Write $f=u+iv$. The hypothesis gives $u-v\le 0$. Thereforewith equality at zero. The local maximum modulus principle makes the exponential constant. Differentiation then gives $f^{\prime }=0$, so $f$ is constant; since $f(0)=0$, it is identically zero.

A contraction satisfies $d(Tx,Ty)\le qd(x,y)$ for some $0\le q<1$. Starting from $x_0$, define $x_{k+1}=T{x}_k$. The geometric bound on successive distances makes $(x_k)$ Cauchy, so completeness gives a limit $x$. Continuity of $T$ gives $Tx=x$. If $Ty=y$, then $d(x,y)\le qd(x,y)$, hence $x=y$. This is the contraction mapping theorem.

If $T^n$ is a contraction, it has a unique fixed point $x$. Since $T^n(Tx)=T(T^{n}x)=Tx$, the point $Tx$ is also fixed by $T^n$, so uniqueness gives $Tx=x$. Every fixed point of $T$ is fixed by $T^n$, so it too must equal $x$. Thus $T$ has exactly one fixed point.

The dual space $V^{\ast }$ is the vector space of linear maps $V\to \mathbb{F}$. If $B=(v_1,\dots ,v_n)$, its dual family satisfies $v_i^{\ast }(v_j)={\delta }_{ij}$. Every functional $f$ obeysso the family spans; evaluation on each $v_j$ proves linear independence. Thus it is a basis without any prior dimension argument.

For $p=a+bt+c{t}^2$,The basis dual to $(f_0,f_1,f_2)$ isas direct substitution gives $f_i(p_j)={\delta }_{ij}$. Hence the three functionals form a basis of $P_2^{\ast }$.

In the fluid region $y>\eta (x)$, incompressibility and potential flow giveNo penetration through the exact surface gives

The condition $hk\ll 1$ says the hill slope is small; $|\nabla \varphi |\ll U$ says the disturbance velocity is small compared with the background wind. Dropping the product ${\varphi }_x{\eta }_x$ and Taylor-shifting the boundary from $y=\eta$ to $y=0$ therefore givesThe decaying solution is

Bernoulli's equation isTo first order on the surface,ThusEquivalently, crest minus trough is the negative of this. For $k{U}^2/g\ll 1$, hydrostatic elevation dominates; for $k{U}^2/g\gg 1$, the Bernoulli pressure drop caused by faster crest flow dominates.

Spatial Fourier transformation givesHenceInverting and splitting the cosine into exponentials, or applying D'Alembert formula, yieldsThus the initial Gaussian separates into two half-amplitude Gaussian pulses travelling without distortion at speeds $c$ and $-c$.

The Euler--Lagrange equation isPutting $u=\cos (\varphi /2)$ and simplifying gives

Since $1-\cos \varphi =2(1-u^2)$, $1+\cos \varphi =2u^2$, and ${\dot{u}}^2=(1-u^2){\dot{\varphi }}^2/4$, the transformed functional isIts Euler--Lagrange equation is $\ddot{u}+(g/4a)u=0$, exactly the preceding equation.

For endpoint-vanishing $\eta$, the second variation isWriting $\eta ={\sum }_{n\ge 1}{c}_n{e}_n$ and using orthonormality givesIt is positive definite when $T<\pi /\omega$ and has a negative $e_1$ direction when $T>\pi /\omega$. Therefore