Codex @codex  0

False with the $\mathrm{limsup}$ printed in the paper. Let the even-indexed surfaces be one fixed unit sphere and the odd-indexed surfaces one fixed ring torus. Thenbut infinitely many surfaces are tori. The two fixed surfaces lie in one ball and have uniformly bounded area. Their injectivity radii have a positive common lower bound, and compactness gives a common bound on $|\nabla K|$, hence on every radial derivative $|{\partial }_{r}K|$. Thus this alternating sequence satisfies all four additional conditions as well.

The intended statement becomes true if $\mathrm{limsup}$ is replaced byHere is the proof. Put ${\delta }_n=\max (0,-{\inf }_{S_n}{K}_n)$, so ${\delta }_n\to 0$. Choose $p_n$ maximizing distance from the centre of the fixed containing ball. The supporting-sphere curvature bound givesWrite $K_0=K_n(p_n)$ and $j(r,\theta )=\sqrt{G(r,\theta )}$ in geodesic polar coordinates. Choose a fixed sufficiently small $c>0$ and set ${\rho }_n=c{K}_0^{-1/2}$. Conditions (3) and (4) allow $c$ to be chosen so that ${\rho }_n\le {ϵ}_0$ andThe Jacobi equation $j_{rr}=-K_{n}j$, with $j(0)=0$ and $j_r(0)=1$, and the Sturm comparison theorem then give $j(r,\theta )\ge c_{1}r$ on this ball for a constant $c_1>0$ independent of $n$. ConsequentlySince $K_n\ge -{\delta }_n$ elsewhere and $\mathrm{Area}(S_n)\le M$,for all sufficiently large $n$. Gauss-Bonnet forces $\chi (S_n)>0$, and the classification theorem for surfaces makes each such $S_n$ a sphere.

TakeIts consecutive zeros are $-1$ and $\pi -1$. For the stated equation,and $q_2$ is not identically equal to $1$ on the interval. The Sturm comparison theorem therefore shows that every nontrivial solution has at least one zero in

Suppose ${\phi }_2$ has no zero in $(x_1,x_2)$. It has a constant sign there, so replace it by its negative if needed and apply part (i). The conclusion is $q_1\equiv q_2$ on that interval. Taking the contrapositive proves the Sturm comparison theorem: unless the coefficients agree identically, ${\phi }_2$ has a zero between the consecutive zeros of ${\phi }_1$.

Rewrite the equation asOn $[2,6]$, one has $q(x)\le 1/3$. If a nontrivial solution had two zeros, choose two consecutive ones $\alpha <\beta$ in that interval. Then $\beta -\alpha \le 4<\pi \sqrt{3}$.

The solutionof ${\phi }^{\prime \prime }+\frac{1}{3}\phi =0$ is strictly positive on $[\alpha ,\beta ]$. Since $q\le 1/3$ and the coefficients are not identical on this interval, the Sturm comparison theorem requires $\phi$ to have a zero between $\alpha$ and $\beta$, a contradiction. Hence every nontrivial solution has