[proofplan] We prove the estimate by contradiction and blow-up. If no uniform curvature bound exists at small scale, use the standard point-picking lemma to choose stable minimal surfaces with base points whose curvature is normalized to one after rescaling and whose surrounding rescaled balls have uniformly bounded curvature on every fixed radius. The blow-up [compactness theorem](/theorems/2748) gives the limiting surface, while the classification theorem for complete two-sided stable minimal surfaces in $\mathbb{R}^3$ forces the limit to be a plane, contradicting the normalization. [/proofplan]

[step:Argue by contradiction and choose points with unbounded scale-invariant curvature]Assume the asserted constants do not exist for a fixed pair of constants $\Lambda>0$ and $i_0>0$. Then for every integer $k \geq 1$ there are data $(M_k^3,g_k,p_k,r_{0,k})$ satisfying the curvature and injectivity-radius hypotheses with these same constants $\Lambda$ and $i_0$, a radius $r_k$ satisfying \begin{align*} 0<r_k \leq \min\{k^{-1},r_{0,k}\}, \end{align*} and a properly embedded, two-sided, stable minimal surface \begin{align*} \Sigma_k^2 \subset B_{M_k}(p_k,r_k) \end{align*} with $\partial \Sigma_k \subset \partial B_{M_k}(p_k,r_k)$ such that \begin{align*} \sup_{\Sigma_k \cap B_{M_k}(p_k,r_k/2)} |A_k|_{g_k}^2 > \frac{k}{r_k^2}, \end{align*} where $A_k$ denotes the second fundamental form of $\Sigma_k$ in $(M_k,g_k)$. Define the curvature function $Q_k: \Sigma_k \cap B_{M_k}(p_k,r_k) \to [0,\infty)$ by $Q_k(x)=|A_k|_{g_k}(x)$. Define the boundary-distance function $d_k: \Sigma_k \cap B_{M_k}(p_k,r_k) \to [0,\infty)$ by \begin{align*} d_k(x):=\operatorname{dist}_{g_k}(x,\partial B_{M_k}(p_k,r_k)). \end{align*} Choose $z_k \in \Sigma_k \cap B_{M_k}(p_k,r_k/2)$ such that \begin{align*} Q_k(z_k)^2>\frac{k}{2r_k^2}. \end{align*} This is possible because the displayed supremum is strictly larger than $k/r_k^2$. Since $d_k(z_k)\geq r_k/2$, we have \begin{align*} d_k(z_k)Q_k(z_k)>\sqrt{\frac{k}{8}}. \end{align*} Apply the standard [point-picking lemma](/theorems/???) to the nonnegative function $Q_k$ and the distance-to-boundary scale $d_k$. We obtain a point $x_k \in \Sigma_k \cap B_{M_k}(p_k,r_k)$ and a number $\lambda_k:=Q_k(x_k)$ such that, with $\delta_k:=d_k(x_k)$, \begin{align*} \lambda_k\delta_k \geq d_k(z_k)Q_k(z_k)>\sqrt{\frac{k}{8}}, \end{align*} and \begin{align*} |A_k|_{g_k}(y)\leq 2\lambda_k \end{align*} for every $y\in \Sigma_k\cap B_{g_k}(x_k,\delta_k/2)$. In particular $\lambda_k\delta_k\to\infty$.[/step]

[guided]We first turn the failure of the theorem into a sequence whose curvature blows up after multiplying by the square of the ambient radius. The constants in the theorem are supposed to depend only on the fixed numbers $\Lambda$ and $i_0$. Thus, if the theorem is false, we may find examples at arbitrarily small radii $r_k \leq k^{-1}$ for which \begin{align*} \sup_{\Sigma_k \cap B_{M_k}(p_k,r_k/2)} |A_k|_{g_k}^2 > \frac{k}{r_k^2}. \end{align*} Here $A_k$ is the second fundamental form of $\Sigma_k$ computed using the metric $g_k$. Define $Q_k: \Sigma_k \cap B_{M_k}(p_k,r_k) \to [0,\infty)$ by $Q_k(x)=|A_k|_{g_k}(x)$. The supremum in the contradiction hypothesis need not be attained, so we choose an approximate maximizer $z_k \in \Sigma_k \cap B_{M_k}(p_k,r_k/2)$ satisfying \begin{align*} Q_k(z_k)^2>\frac{k}{2r_k^2}. \end{align*} Define $d_k: \Sigma_k\cap B_{M_k}(p_k,r_k) \to [0,\infty)$ by $d_k(x)=\operatorname{dist}_{g_k}(x,\partial B_{M_k}(p_k,r_k))$. Since $z_k$ lies in the half-ball, every path from $z_k$ to $\partial B_{M_k}(p_k,r_k)$ has length at least $r_k/2$, and therefore \begin{align*} d_k(z_k)Q_k(z_k)>\sqrt{\frac{k}{8}}. \end{align*} The standard [point-picking lemma](/theorems/???) now chooses a better base point $x_k$ near the high-curvature region. Set $\lambda_k:=Q_k(x_k)$ and $\delta_k:=d_k(x_k)$. The lemma gives \begin{align*} \lambda_k\delta_k>\sqrt{\frac{k}{8}} \end{align*} and also gives the local control \begin{align*} |A_k|_{g_k}(y)\leq 2\lambda_k \end{align*} for every $y\in\Sigma_k\cap B_{g_k}(x_k,\delta_k/2)$. This is the key improvement over choosing a raw almost-maximizer: after rescaling by $\lambda_k$, the boundary lies at distance at least $\lambda_k\delta_k\to\infty$, and each fixed rescaled ball has a uniform curvature bound.[/guided]

[step:Rescale around the high-curvature points and record the normalized geometry] Define the rescaled metric \begin{align*} \tilde g_k := \lambda_k^2 g_k \end{align*} on $M_k$, and regard $\Sigma_k$ as a surface in $(M_k,\tilde g_k)$. Let $\tilde A_k$ denote its second fundamental form in the rescaled metric. Under constant metric scaling, lengths scale by $\lambda_k$, second fundamental forms scale by $\lambda_k^{-1}$, and sectional curvatures scale by $\lambda_k^{-2}$. Therefore \begin{align*} |\tilde A_k|_{\tilde g_k}(x_k)=1, \qquad |\operatorname{sec}_{(M_k,\tilde g_k)}| \leq \frac{\Lambda}{\lambda_k^2} \to 0, \end{align*} Since $\delta_k\leq r_k\leq k^{-1}$ and $\lambda_k\delta_k>\sqrt{k/8}$, we have \begin{align*} \lambda_k > \frac{\sqrt{k/8}}{\delta_k}\geq \sqrt{\frac{k}{8}}\,k \to \infty. \end{align*} Thus the injectivity radius lower bound on the rescaled ball is at least $\lambda_k i_0$, hence tends to infinity. For every fixed $R>0$ and all sufficiently large $k$, the ball $B_{\tilde g_k}(x_k,R)$ is contained in $B_{g_k}(x_k,\delta_k/2)$ because $R\lambda_k^{-1}<\delta_k/2$. Hence $B_{\tilde g_k}(x_k,R)\subset B_{M_k}(p_k,r_k)$, and the point-picking estimate gives \begin{align*} |\tilde A_k|_{\tilde g_k}\leq 2 \end{align*} on $\Sigma_k\cap B_{\tilde g_k}(x_k,R)$. Thus the pieces $\Sigma_k \cap B_{\tilde g_k}(x_k,R)$ have no boundary except possibly on $\partial B_{\tilde g_k}(x_k,R)$ and have curvature uniformly bounded independently of $k$ on each fixed rescaled ball. [/step]

[step:Use stable minimal surface compactness to obtain a complete Euclidean limit]By the [blow-up compactness theorem for embedded stable minimal surfaces](/theorems/???), applied to the pointed manifolds $(M_k,\tilde g_k,x_k)$ and the properly embedded two-sided stable minimal surfaces $\Sigma_k$, a subsequence of the pointed surfaces \begin{align*} (\Sigma_k,\tilde g_k,x_k) \end{align*} converges smoothly on compact subsets to a complete, properly embedded, two-sided, stable minimal surface \begin{align*} \Sigma_\infty^2 \subset \mathbb{R}^3 \end{align*} with base point $x_\infty \in \Sigma_\infty$. The hypotheses of the compactness theorem are verified as follows: the rescaled sectional curvature bounds tend to $0$, the rescaled injectivity radii tend to infinity, the surfaces remain embedded and two-sided under scaling, stability and minimality are invariant under constant scaling of the ambient metric, the boundary escapes every fixed rescaled ball because $\lambda_k\delta_k\to\infty$, and the point-picking estimate supplies the uniform local curvature bound $|\tilde A_k|_{\tilde g_k}\leq 2$ on every fixed compact ball for all sufficiently large $k$. The compactness statement is used in its pointed smooth local form, allowing convergence with locally finite multiplicity; the base-point normalization only requires convergence of the sheet through $x_k$, so multiplicity does not affect the value of the limiting second fundamental form at $x_\infty$. Smooth convergence of the second fundamental forms gives \begin{align*} |A_\infty|_{\mathbb{R}^3}(x_\infty) = \lim_{k\to\infty}|\tilde A_k|_{\tilde g_k}(x_k) =1, \end{align*} where $A_\infty$ denotes the second fundamental form of $\Sigma_\infty \subset \mathbb{R}^3$.[/step]

[guided]The point of the rescaling was to create a sequence with three compactness inputs available on every fixed radius $R>0$. First, the ambient geometry converges to Euclidean geometry because \begin{align*} |\operatorname{sec}_{(M_k,\tilde g_k)}|\leq \Lambda\lambda_k^{-2}\to 0 \end{align*} and the rescaled injectivity radii tend to infinity. Second, the boundaries disappear from compact subsets: for fixed $R>0$, the inequality $R\lambda_k^{-1}<\delta_k/2$ holds for all sufficiently large $k$, so $B_{\tilde g_k}(x_k,R)$ is contained in the original ball away from $\partial\Sigma_k$. Third, the point-picking lemma gives the local curvature estimate \begin{align*} |\tilde A_k|_{\tilde g_k}\leq 2 \end{align*} on $\Sigma_k\cap B_{\tilde g_k}(x_k,R)$ for all sufficiently large $k$. These are precisely the local hypotheses needed for the [blow-up compactness theorem for embedded stable minimal surfaces](/theorems/???). Passing to a subsequence, the pointed surfaces $(\Sigma_k,\tilde g_k,x_k)$ converge smoothly on compact subsets, in the pointed local sense and possibly with locally finite multiplicity, to a complete, properly embedded, two-sided, stable minimal surface $\Sigma_\infty^2\subset\mathbb{R}^3$ with base point $x_\infty\in\Sigma_\infty$. Completeness follows because every fixed intrinsic compact region around $x_\infty$ arises as the smooth limit of regions whose boundary has escaped to infinite rescaled distance. Stability and two-sidedness pass to the limit because the stability inequality is local and the unit normals converge smoothly on compact subsets. Finally, smooth convergence preserves the second fundamental form at the base point. Since the rescaling normalized the curvature by construction, \begin{align*} |A_\infty|_{\mathbb{R}^3}(x_\infty)=\lim_{k\to\infty}|\tilde A_k|_{\tilde g_k}(x_k)=1. \end{align*}[/guided]

[step:Apply the Euclidean classification theorem and derive the contradiction]By the [Fischer-Colbrie-Schoen classification of complete two-sided stable minimal surfaces in $\mathbb{R}^3$](/theorems/???), every complete two-sided stable minimal surface in Euclidean $\mathbb{R}^3$ is an affine plane. The limiting surface $\Sigma_\infty$ satisfies the hypotheses of this classification theorem: it is complete, two-sided, stable, and minimal in $\mathbb{R}^3$. Hence $\Sigma_\infty$ is a plane, so its second fundamental form vanishes identically: \begin{align*} |A_\infty|_{\mathbb{R}^3} \equiv 0. \end{align*} This contradicts the normalization $|A_\infty|_{\mathbb{R}^3}(x_\infty)=1$.[/step]

[guided]The limiting surface satisfies all hypotheses of the Euclidean stability classification theorem: it is complete, two-sided, stable, and minimal in $\mathbb{R}^3$. Therefore the [Fischer-Colbrie-Schoen classification](/theorems/???) forces $\Sigma_\infty$ to be an affine plane. An affine plane has vanishing second fundamental form, so \begin{align*} |A_\infty|_{\mathbb{R}^3}\equiv 0. \end{align*} This conclusion is incompatible with the curvature normalization obtained from smooth convergence, \begin{align*} |A_\infty|_{\mathbb{R}^3}(x_\infty)=1. \end{align*} The contradiction eliminates the assumed counterexample sequence.[/guided]

[step:Choose the constants from the contradiction argument] The contradiction proves that there exist constants $C>0$ and $\rho>0$, depending only on $\Lambda$ and $i_0$, such that for every $0<r\leq \min\{\rho,r_0\}$ and every properly embedded, two-sided, stable minimal surface $\Sigma^2\subset B_M(p,r)$ with $\partial\Sigma\subset \partial B_M(p,r)$, one has \begin{align*} \sup_{\Sigma \cap B_M(p,r/2)} |A|^2 \leq \frac{C}{r^2}. \end{align*} Indeed, if no such pair existed, the sequence constructed in the first step would exist and would give the contradiction above. This is precisely the claimed Schoen curvature estimate. [/step]