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A global maximum of a real-valued function is a value $f(x_0)$ satisfying $f(x)\le f(x_0)$ throughout its domain.

Mathematical optimization studies extrema under constraints.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. We first use it to show that a continuous function on $[a,b]$ is bounded above. Otherwise, choose $x_n\in [a,b]$ with $f(x_n)>n$. A convergent subsequence $x_{n_k}\to c$ exists, with $c\in [a,b]$ because the interval is closed. Continuity would give $f(x_{n_k})\to f(c)$, contradicting $f(x_{n_k})>n_k\to \infty$.

Let $L=\sup \{f(x):x\in [a,b]\}$, which is finite. By the definition of supremum, there are $y_n\in [a,b]$ with $L-1/n<f(y_n)\le L$. Another application of the Bolzano-Weierstrass theorem gives a subsequence $y_{n_k}\to c\in [a,b]$. By continuity,Thus the supremum is actually attained:This proves the maximum part of the extreme value theorem from sequential compactness rather than assuming it.

Now suppose $f^{\prime \prime }<0$ on $\mathbb{R}$. Applying the mean value theorem to $f^{\prime }$ shows that $f^{\prime }$ is strictly decreasing. At a local maximum of a differentiable function, the left difference quotients are nonnegative and the right difference quotients are nonpositive; existence of the derivative forces $f^{\prime }=0$. A strictly decreasing function has at most one zero. ThereforeIf that zero exists, $f^{\prime }$ is positive to its left and negative to its right, so the local maximum is also a global maximum.

For the uniform bound by $K<0$, defineIts derivative is $h^{\prime }(x)=f^{\prime \prime }(x)-K<0$, so $h$ is strictly decreasing by the mean value theorem. Comparing with $h(0)=f^{\prime }(0)$ givesBecause $K<0$, this makes $f^{\prime }$ negative at sufficiently large positive $x$ and positive at sufficiently large negative $x$. The function $f^{\prime }$ is continuous, since it is differentiable; the intermediate value theorem supplies a zero $c$ between such points. Its strict decrease makes this zero unique. Again using the mean value theorem for $f$, the derivative signs imply that $f$ increases up to $c$ and decreases after $c$:This illustrates uniform negative curvature and global maximization.

Merely requiring $f^{\prime \prime }<0$ does not force existence. TakeThen $f^{\prime }(x)=e^{-x}>0$ and $f^{\prime \prime }(x)=-e^{-x}<0$, but $f$ is strictly increasing, approaches $0$ as $x\to +\infty$, and never equals $0$. The answer to the final question is no, even for a function bounded above:The stronger uniform curvature assumption prevents this behaviour. Here $f^{\prime \prime }(x)\to 0$ at the end where the supremum is approached, so no uniform negative bound of that kind is available.

The obstacle and wave pattern are stationary, so $\omega =\Omega =0$. Withthe dispersion relation becomesHence $\cos \varphi >0$, so $-\pi /2<\varphi <\pi /2$, and

Differentiating the dispersion relation with respect to the wavevector giveswhere the zero-frequency relation
$\alpha /\sqrt{\kappa }=V\cos \varphi$ was used. Its first component is always negative, so the ray angle $\psi$ lies inMoreover,

Put $q=|\tan \varphi |$. The magnitude of the slope relative to the negative $x_1$-axis isIts derivative isso its global maximum occurs at $q=1/\sqrt{2}$ and equalsTherefore all group-velocity rays lie in the stationary square-root-dispersion wakea wedge of semi-angle ${\tan }^{-1}(2^{-3/2})$ extending in the negative $x_1$-direction.

Use $x>0$ as the parameter of the smooth curve:Its velocity vector and speed areThe unit tangent vector is therefore

The parameter-independent formula for the curvature of a space curve givesDifferentiating,Hence $\kappa$ increases for $0<x<1/\sqrt{2}$ and decreases for $x>1/\sqrt{2}$. Its global maximum occurs at $x=1/\sqrt{2}$ and equalsThe corresponding point is

Write the base-state gradients asThe base heat flux and Stefan condition relations are

Under the quasi-stationary approximation, and with advective heat transport neglected, each temperature perturbation satisfies the Laplace equation. The decaying normal-mode solutions areKeeping the displaced ice interface at $T_m$ omits the curvature correction described by the Gibbs--Thomson relation.
The fixed melting temperature at the displaced ice interface givesExpanding temperature and conductive heat flux continuity at the displaced outer interface gives

Put $r=k_a/k_w$. Since $G_a=G_w/r$, the temperature condition in the limits $r\ll 1$ and $\alpha h\ll 1$ givesAlthough $r$ is small, the product $rC$ is therefore order one and cannot be discarded. The flux condition then givesThis is why the stated asymptotic warning matters.

The perturbed Stefan condition at $x=0$ isUsing $B\sim {\eta }_2{G}_w$ and $-k_w{G}_w={\rho }_i{L}_{f}V$ yieldsSubstitution of the interface-amplitude ratio from part i produces the thin-film icicle-ripple instability dispersion relation

Set $b=\Gamma h{\alpha }^3$. Rationalizing this complex number givesso its growth rate and imaginary part areDifferentiating the growth rate with respect to the wavenumber shows that its positive stationary point satisfiesIt is the unique global maximum, and henceAt this wavenumber $b=1/\sqrt{5}$. With the convention $e^{i\alpha z+\sigma t}$, a constant phase travels with phase velocityThereforeBecause $z$ increases downward, the negative sign means that the icicle ripples migrate upward with speed $\sqrt{5}V/6$.

If a twice differentiable real function on $\mathbb{R}$ satisfies $f^{\prime \prime }(x)\le -c$ for some fixed $c>0$, then it has a unique global maximum. The mean value theorem shows that $f^{\prime }$ is strictly decreasing, and that $f^{\prime }(x)+cx$ is nonincreasing. Hence $f^{\prime }(x)\le f^{\prime }(0)-cx$ for $x>0$ and $f^{\prime }(x)\ge f^{\prime }(0)-cx$ for $x<0$. These inequalities force opposite derivative signs at the two ends. The intermediate value theorem supplies a unique zero of $f^{\prime }$, at which $f$ changes from increasing to decreasing. Merely having $f^{\prime \prime }<0$ does not guarantee existence: $f(x)=-e^{-x}$ is increasing and approaches its unattained supremum zero.