Proof. Let us argue by contradiction. If the lemma is false, then there must have a sequence of complete properly immersed minimal submanifolds \(M_j\) satisfying that the total curvature \[\alpha_j := \int_{M_j} \lvert A_{M_j}\rvert^n d\mu_{M_j}\to 0\,,\] but \[\beta_j := \sup_{x\in M_j}\lvert A_{M_j}\rvert(x)d(\iota_j(x),\iota_j(\partial M_j)) \to \infty\,.\] Then the standard point picking argument(see [1] Lemma 2.2) by passing to a subsequence allows us to find \(\widetilde{q}_j \in M_j\) so that for \(\lambda_j := \lvert A_{M_j}\rvert(\widetilde{q}_j)\,,\) the rescaled minimal submanifold \[\widetilde{M}_j := \lambda_j(\iota_j(M_j) -\iota_j(\widetilde{q}_j))\] converges locally smoothly in \(\mathbb{R}^{n+m}\) to a complete immersed minimal submanifold \(\widetilde{M}_\infty\,.\) Moreover, \(\widetilde{M}_\infty\) has no boundary and the total curvature of \(\widetilde{M}_\infty\) equals 0. While \(\lvert A_{\widetilde{M}_\infty}\rvert(0) = 1,\) which derives a contradiction.

For the convenience of readers, we recall the point picking argument used above to construct \(\widetilde{M}_{\infty}\). Let \(\iota_j: M_j \to \mathbb{R}^{n+m}\) denote the immersion map. Choose \(\widetilde{p}_{j} \in M_{j}\) so that \[|A_{M_{j}}|(\widetilde{p}_{j}) d(\iota_j(\widetilde{p}_{j}),\iota_j(\partial M_{j})) > \frac{1}{2}\beta_{j} \to \infty\] and set \(r_{j} = |A_{M_{j}}|(\widetilde{p}_{j})^{-\frac{1}{2}} d(\iota_j(\widetilde{p}_{j}),\iota_j(\partial M_{j}))^{\frac{1}{2}}\,\). Then, we choose \(\widetilde{q}_{j} \in M_{j}\cap \iota_j^{-1}(B_{r_{j}}(\iota_j(\widetilde{p}_{j})))\) so that \[\label{eq:max46choose1} |A_{M_{j}}|(\widetilde{q}_{j}) d(\iota_j(\widetilde{q}_{j}),\partial B_{r_{j}}(\iota_j(\widetilde{p}_{j}))) = \max_{\iota_j(x) \in B_{r_{j}}(\iota_j(\widetilde{p}_{j}))}|A_{M_{j}}|(x) d(\iota_j(x),\partial B_{r_{j}}(\widetilde{p}_{j}))\, .\tag{1}\] Note that the right hand side is at least \(\left(|A_{M_{j}}|(\widetilde{p}_{j})d(\iota_j(\widetilde{p}_{j}),\iota_j(\partial M_{j}))\right)^{\frac{1}{2}}\) which is tending to infinity. Let \(R_{j} = d(\iota_j(\widetilde{q}_{j}),\partial B_{r_{j}}(\iota_j(\widetilde{p}_{j})))\,\). Because \(d(y,\partial B_{R_{j}}(\iota_j(\widetilde{q}_{j}))) \leqslant d(y,\partial B_{r_{j}}(\iota_j(\widetilde{p}_{j})))\) for any \(y \in B_{R_{j}}(\widetilde{q}_{j})\,\), we find that \[\label{eq:max46choose2} |A_{M_{j}}|(\widetilde{q}_{j}) d(\iota_j(\widetilde{q}_{j}),\partial B_{R_{j}}(\iota_j(\widetilde{q}_{j}))) = \max_{\iota_j(x) \in B_{R_{j}}(\iota_j(\widetilde{q}_{j}))}|A_{M_{j}}|(x) d(\iota_j(x),\partial B_{R_{j}}(\iota_j(\widetilde{q}_{j})))\, .\tag{2}\] Note that \(|A_{M_{j}}|(\widetilde{q}_{j})R_{j} \geqslant|A_{M_{j}}|(\widetilde{p}_{j})r_{j} \to \infty\,\).

As above, we set \(\lambda_{j} = |A_{M_{j}}|(\widetilde{q}_{j})\,\). Then, the rescaled submanifold \[\widetilde{M}_{j} =\lambda_{j}(\iota_j(M_{j}) - \iota_j(\widetilde{q}_{j}))\] with immersion map \(\widetilde{\iota}_j\) satisfies \[|A_{\widetilde{M}_{j}}|(x) d(\widetilde{\iota}_j(x)\,,\partial B_{\lambda_{j}R_{j}}(0)) \leqslant\lambda_{j}R_{j}\,,\] when \(\widetilde{\iota}_j(x) \in B_{\lambda_{j}R_{j}}(0)\,.\) If \(x\in\widetilde{M}_{j}\) and \(\widetilde{\iota}_j(x)\) lies in a given compact set of \(\mathbb{R}^{n+m}\,\), then \[|A_{\widetilde{M}_{j}}|(x) \leqslant\frac{\lambda_{j} R_{j}}{\lambda_{j}R_{j} - |\widetilde{\iota}_j(x)|} \to 1 = |A_{\widetilde{M}_{j}}|(0)\] as \(j\to\infty\,\). Then \(d(0,\widetilde{\iota}_j(\partial \widetilde{M}_j)) \to \infty\) due to \(\lambda_j R_j \to \infty\,\). After passing to a subsequence, we can take a smooth limit of \(\lambda_{j}(\iota_j(M_{j}) - \iota_j(\widetilde{q}_{j}))\) and find a complete, non-flat, immersed minimal submanifold \(\widetilde{M}_{\infty}\) in \(\mathbb{R}^{n+m}\) without boundary. ◻

Proof. We argue by contradiction. If the lemma is false, there must have a sequence of complete immersed minimal submanifolds \(M_j\) with the total curvature \[\alpha_j := \int_{M_j} \lvert A_{M_j}\rvert^n d\mu_{M_j}\to 0\,,\] but \[\beta_j := \sup_{x\in M_j}\lvert A_{M_j}\rvert(x)d^{M_j}(x,\partial M_j) \geqslant 2C_2>0\,.\] Then we can find \(\widetilde{q}_j \in M_j\) such that \(\lvert A_{M_j}\rvert(\widetilde{q}_j)d^{M_j}(\widetilde{q}_j,\partial M_j) > C_2\,.\) After rescaling, we can assume \(\lvert A_{M_j}\rvert(\widetilde{q}_j) = 1, \;d^{M_j}(\widetilde{q}_j,\partial M_j)> C_2\) and \(\iota_j(\widetilde{q}_j) = 0\,.\) If \(j\) is sufficiently large, then \(|A_{M_j}|(x)\) is uniformly bounded for any \(x \in B^{M_j}_{\frac{1}{2}C_2}(\widetilde{q}_j)\) by curvature estimates stated in Remark 3 similar to Lemma 2. We denote \(\widehat\iota_j\) as \(\iota_j\) restricted on \(M_j\cap B^{M_j}_{\theta C_2}(\widetilde{q}_j)\) for some \(\theta\) small. By taking \(\theta\) sufficiently small, we can assume \(\widehat\iota_j\) is an embedding and the image of \(\widehat\iota_j\) in \(\mathbb{R}^{n+m}\) is the graph of some function \(u_j\,.\) After passing to a subsequence, we can assume \(\widehat\iota_j\) converges locally smoothly to a minimal embedding \(\widehat\iota_\infty: \widehat M_\infty \to \mathbb{R}^{n+m}\) whose image in \(\mathbb{R}^{n+m}\) is also the graph of some function \(u_\infty\) and \(\iota_j(\widetilde{q}_j) = \iota_\infty( \widetilde{q}_\infty) = 0 \in \mathbb{R}^{n+m}\,.\) Hence \(\lvert A_{\widehat M_\infty}\rvert(\widetilde{q}_\infty) = 1\) but \[\int_{\widehat M_\infty} \lvert A_{\widehat M_\infty}\rvert^n d\mu_{M_\infty}= 0\,,\] which is a contradiction. ◻

Proof. The proof is based on the spectral decomposition of vector-valued functions on \(\mathbb{S}^{n-1}\).

Since \(|\nabla Y|_0 <\infty\) and Equation 4 is uniformly elliptic, the classical elliptic estimates give upper bounds on the derivatives of \(Y\): more precisely, for any \(\ell>0\,\), there is a constant \(C_\ell'\) independent of \(s\) such that for any \(s>1\,,\) \[\|\nabla^\ell Y \|_{C^0(\mathbb{S}^{n-1}\times [s,s+1])}\leqslant C_\ell'\|Y\|_{C^0(\mathbb{S}^{n-1}\times [s-1,s+2])}\,.\] This implies that for any \(\ell>0\,\), \(\|\nabla^\ell Y \|_\beta<\infty\,\). Since the term \(\mathcal{Q}(Y)\) in 4 gathers all the nonlinear terms consisting of \(Y\,, \nabla Y\, , \nabla^2Y\) at least cubic(see Appendix 5), we have \(\|\mathcal{Q}(Y)\|_{3\beta}<\infty\) and \(\|\nabla^\ell \mathcal{Q}(Y)\|_{3\beta}<\infty\,\).

In the preceding section, we have described the spectrum of the Laplace operator on the sphere. So let us denote \(\lambda_k:=k(k+n-2)\) and \(\Phi_{k,\alpha}\) the orthonormal basis of the eigenspace of \(\Delta_{\mathbb{S}^{n-1}}\) associated to \(-\lambda_k\,,\) that is \[\mathchoice {{\setbox 0=\displaystyle{\textstyle-}{\int}\vcenter{\textstyle- }\kern-.6\wd 0}} {{\setbox 0=\textstyle{\scriptstyle-}{\int}\vcenter{\scriptstyle- }\kern-.6\wd 0}} {{\setbox 0=\scriptstyle{\scriptscriptstyle-}{\int}\vcenter{\scriptscriptstyle- }\kern-.6\wd 0}} {{\setbox 0=\scriptscriptstyle{\scriptscriptstyle-}{\int}\vcenter{\scriptscriptstyle- }\kern-.6\wd 0}} \!\int_{\mathbb{S}^{n-1}}\langle \Phi_{k_1,\alpha_1}\,,\Phi_{k_2,\alpha_2} \rangle d\mu_{\mathbb{S}^{n-1}} = \delta_{k_1k_2}\delta_{\alpha_1\alpha_2}\,.\] The dimension of the eigenspace associated to \(-\lambda_k\) is bounded by \(c_1 (k^{n-1}+1)m\) with \(c_1\) only depending on \(n\,.\) Moreover, for \(k \geqslant 1\,,\) we have the following estimates for the \(L^\infty\) norm of the eigenfunctions (see [15]): \[\|\Phi_{k,\alpha}\|_\infty\leqslant c_2\lambda_k^{\frac{n-2}{4}} , c_2>1 \text{ only depending on } n\,.\]

Now let us define \[\begin{align} g_{k,\alpha}(t)&=\mathchoice {{\setbox 0=\displaystyle{\textstyle-}{\int}\vcenter{\textstyle- }\kern-.6\wd 0}} {{\setbox 0=\textstyle{\scriptstyle-}{\int}\vcenter{\scriptstyle- }\kern-.6\wd 0}} {{\setbox 0=\scriptstyle{\scriptscriptstyle-}{\int}\vcenter{\scriptscriptstyle- }\kern-.6\wd 0}} {{\setbox 0=\scriptscriptstyle{\scriptscriptstyle-}{\int}\vcenter{\scriptscriptstyle- }\kern-.6\wd 0}} \!\int_{\mathbb{S}^{n-1}} \langle Y(x,t), \Phi_{k,\alpha}(x) \rangle d\mu_{\mathbb{S}^{n-1}} \,,\\ f_{k,\alpha}(t)&=-\mathchoice {{\setbox 0=\displaystyle{\textstyle-}{\int}\vcenter{\textstyle- }\kern-.6\wd 0}} {{\setbox 0=\textstyle{\scriptstyle-}{\int}\vcenter{\scriptstyle- }\kern-.6\wd 0}} {{\setbox 0=\scriptstyle{\scriptscriptstyle-}{\int}\vcenter{\scriptscriptstyle- }\kern-.6\wd 0}} {{\setbox 0=\scriptscriptstyle{\scriptscriptstyle-}{\int}\vcenter{\scriptscriptstyle- }\kern-.6\wd 0}} \!\int_{\mathbb{S}^{n-1}} \langle \mathcal{Q}(Y)(x,t), \Phi_{k,\alpha}(x) \rangle d\mu_{\mathbb{S}^{n-1}}\,. \end{align}\]

Hence \(g_{k,\alpha}\) and \(f_{k,\alpha}\) are smooth functions on \(\mathbb{R}^+\,,\) and from 4 , they satisfy \[\begin{gather} g_{k,\alpha}''-(\lambda_{k,+}+\lambda_{k,-})g_{k,\alpha}'+ (\lambda_{k,+}\times\lambda_{k,-})g_{k,\alpha}= f_{k,\alpha}\,. \end{gather}\]

Using \(\Delta_{\mathbb{S}^{n-1}} \Phi_{k,\alpha}=-\lambda_k\Phi_{k,\alpha}\,\), and integration by parts, for \(k\geqslant 1\,,\) we get the following estimates for \(a,b\in\mathbb{Z}^+\): \[\begin{align} |g_{k,\alpha}(s)|&\leqslant\frac{\sup_{t=s}|\nabla^{2a}Y(x,t)|}{(1+\lambda_k)^a}\, ,\\ |f_{k,\alpha}(s)|&\leqslant\frac{\sup_{t=s}|\nabla^{2a}\mathcal{Q}(Y)(x,t)|}{(1+\lambda_k)^a}\,. \end{align}\] Thus we get \[\begin{align} \|g_{k,\alpha}\|_\beta&\leqslant\frac{\|\nabla^{2a}Y\|_\beta}{(1+\lambda_k)^a} \,,\\ \|f_{k,\alpha}\|_{3\beta}&\leqslant\frac{\|\nabla^{2a}\mathcal{Q}(Y)\|_{3\beta}}{(1+\lambda_k)^a}\,. \end{align}\]

For \(k=0\,,\) \[\|f_{k,\alpha}\|_{3\beta}\leqslant\| \mathcal{Q}(Y) \|_{3\beta}\,.\]

From the standard ordinary differential equation theory(see Lemma 10 in Appendix A of [16]), we can write \[g_{k,\alpha}(t)=a_{k,\alpha}e^{t\lambda_{k,+}} + b_{k,\alpha}e^{t\lambda_{k,-}}+ r_{k,\alpha}(t)\] with some estimates on the different terms. First we notice that \(|\lambda_{k,+}-\lambda_{k,-}| = |2k+n -2|\) and \(|3\beta+\lambda_{k,\pm}|\) are uniformly bounded from below from \(0\) and \(\frac{(2+|\lambda_{k,+}|^2+|\lambda_{k,-}|^2)^{1/2}}{|\lambda_{k,+}-\lambda_{k,-}|}\) is uniformly bounded. Hence, for \(k\geqslant 1\,,\) there is a uniform constant \(c_3\) independent of \(k\) such that \[\begin{align} \max(|a_{k,\alpha}|,|b_{k,\alpha}|)&\leqslant c_3 ( \|g_{k,\alpha}\|_\beta+ \|g'_{k,\alpha}\|_\beta+ \|f_{k,\alpha}\|_{3\beta})\\ &\leqslant c_3 \frac{\|\nabla^{2a}Y\|_\beta+ \|\nabla^{2a+1}Y\|_\beta+ \|\nabla^{2a}Q(Y)\|_{3\beta}}{(1+\lambda_k)^a} \end{align}\] and \[\begin{align} \|r_{k,\alpha}\|_{3\beta}&\leqslant c_3 \|f_{k,\alpha}\|_{3\beta}\\ &\leqslant c_3 \|\mathcal{Q}(Y)\|_{3\beta}\,. \end{align}\] For \(k=0\,,\) \[\max(|a_{k,\alpha}|,|b_{k,\alpha}|) \leqslant C'\, , \|r_{k,\alpha}\|_{3\beta} \leqslant C'\,.\] If \(\lambda_{k,+} > -\beta\), \(\left\lVert e^{t\lambda_{k,+}}\right\rVert_{\beta} = \infty\), so \(a_{k,\alpha} = 0\,.\) Also, if \(\lambda_{k,-} > - \beta\,\), \(b_{k,\alpha} = 0\,.\) If \(\lambda_{k,\pm}\leqslant-3\beta\,\), \(\left\lVert e^{t\lambda_{k,\pm}}\right\rVert_{3\beta} = 1\).

Finally we have the following equality \[\begin{align} \label{eq46pro46sum}Y(x,t)&=\sum_{-3\beta\leqslant\lambda_{k,+}\leqslant-\beta} a_{k,\alpha} e^{t\lambda_{k,+}} \Phi_{k,\alpha}(x) +\sum_{-3\beta\leqslant\lambda_{k,-} \leqslant-\beta} b_{k,\alpha} e^{t\lambda_{k,-}} \Phi_{k,\alpha}(x)\\ &\quad+\sum_{\lambda_{k,+} < -3\beta} a_{k,\alpha} e^{t\lambda_{k,+}} \Phi_{k,\alpha}(x) +\sum_{\lambda_{k,-}< -3\beta}b_{k,\alpha} e^{t\lambda_{k,-}} \Phi_{k,\alpha}(x)\\ &\quad+\sum_{k=0}^\infty r_{k,\alpha}(t) \Phi_{k,\alpha}(x)\, .\end{align}\tag{6}\] First we notice that the first two sums of 6 are finite and are elements of the kernel of \(L\), this is the expected function \(X\). In fact, we claim that the other sums converge and have finite \(3\beta\)-norms. Let \(A(x,t)\) be the sum of the term with \(\lambda_{k,-}< -3\beta\) . In the following computation, we use the expressions of \(\lambda_k\,,\) their multiplicities and the \(L^\infty\) estimates on \(\Phi_{k,\alpha}\,.\) \[\begin{align} \|A\|_{3\beta }&\leqslant c_2\sum_{\lambda_{k,-}< -3\beta,k \geqslant 1} |b_{k,\alpha}| \lambda_k^{\frac{n-2}{4}} + \sum_{\lambda_{0,-}<-3 \beta}|b_{0,\alpha}|\\ &\leqslant c_2c_3\sum_{k\geqslant 1,\alpha} \frac{\|\nabla^{2a}Y\|_\beta+ \|\nabla^{2a+1}Y\|_\beta+ \|\nabla^{2a}\mathcal{Q}(Y)\|_{3\beta}}{(1+\lambda_k)^a} \lambda_k^{\frac{n-2}{4}} + c_1C'm\\ &\leqslant 2c_1c_2c_3m(\|\nabla^{2a}Y\|_\beta+ \|\nabla^{2a+1}Y\|_\beta+ \|\nabla^{2a}\mathcal{Q}(Y)\|_{3\beta})\sum_{k=1}^\infty \frac{(1+k^{\frac{3n}{2}}) }{(1+k^2)^a }+c_1C'm\\ &<\infty \end{align}\] if \(a\) is chosen such that \(2a-\frac{3n}{2}\geqslant 2\,.\) We can prove the other two sums in the claim by the same method and we omit the details. ◻

Proof. Fix a point \(p \in M\,,\) up to a translation, we can assume \(\iota(p) = 0\in \mathbb{R}^{n+m}\,.\)

\((1)\Rightarrow(2):\) This has been proved in [6]. We have organized his proof as follows. We firstly prove that \(\iota\) is a proper immersion. Since the restriction of coordinate functions on \(M\) is harmonic on \(M\,,\) \(M\) is not a closed manifold. Let \(f(x)\) denote the function \(|\iota(x)|\) and \(X = \iota(x)\) as the position vector. By Lemma 4, we can choose \(R_0\) large such that for any \(q \in M\) if \(d^M(q,p) \geqslant R_0\,,\) then \(\lvert A\rvert(q)d^M(q,p)< \frac{1}{4}\,.\) Let \(t_0 := d^M(q,p)\) and \(\gamma(t)\) be the minimizing normal geodesic in \(M\) between \(p\) and \(q\) with \(\gamma(0) = p\) and \(\gamma(t_0) =q\,.\) Let \(V = \gamma'(t)\,,\) and if \(t \geqslant R_0\,,\) then we have \[V\langle V,X \rangle = \langle A(V,V),X \rangle + 1\geqslant 1-\lvert A\rvert|X| \geqslant\frac{3}{4}\,.\] At the point \(q\,,\) \[\begin{align}f(q) &\geqslant\langle X,V \rangle(t_0)\\ & = \langle X,V \rangle(R_0) +\int_{R_0}^{t_0} V\langle V,X \rangle(t) dt\\ & \geqslant\langle X,V \rangle(R_0) + \frac{3}{4}(t_0 - R_0) .\end{align}\] So \(\iota\) is a proper immersion. We compute \(\lvert\nabla f\rvert\) at \(\gamma (t_0) = q\,,\) \[\begin{align}\lvert\nabla f\rvert(q) &\geqslant\frac{\langle X,V \rangle(t_0)}{f(q)}\\ &\geqslant\frac{\langle X,V \rangle(t_0)}{t_0}\\ &\geqslant\frac{\langle X,V \rangle(R_0)-\frac{3}{4}R_0}{t_0} + \frac{3}{4}\\ &\geqslant-\frac{2R_0}{t_0} + \frac{3}{4} .\end{align}\] If \(t_0 \geqslant 8 R_0\,,\) then \(\frac{1}{2}\leqslant\lvert\nabla f\rvert(t_0) \leqslant 1\,.\) By the elementary Morse theory, \(M \setminus B^M_{8R_0}(p)\) is diffeomorphic to \(\left(M \cap \partial B^M_{8R_0}(p) \right)\times [0,\infty)\,.\) Since \(\iota\) is proper, \(M \cap \partial B^M_{8R_0}(p)\) is the union of finite \((n-1)\)-dimensional closed manifolds. So \(M\) is of finite ends. Fix an end \(E\) of \(M\,,\) and by the curvature estimate in Lemma 4, \(r_i^{-1}\iota(E)\) converges locally smoothly in \(\mathbb{R}^{n+m} \setminus \{0\}\) to an \(n\)-plane passing \(0\in \mathbb{R}^{n+m}\) as \(r_i\uparrow \infty\,.\) Since \(n\geqslant 3\) and \(\mathbb{S}^{n-1}\) is simply connected, the multiplicity of \(n\)-plane is 1. So in the sense of varifolds, \(|r_i^{-1}\iota(E)| \rightharpoonup|\psi(\mathbb{R}^n\times \{0^m\} )|\) in \(\mathbb{R}^{n+m} \setminus B_1(0)\) where \(\psi \in SO(n+m).\)

\((2)\Rightarrow(3):\) By using a result of Allard and Almgren [17] and Simon [18], outside a compact set, each end \(E\) up to a rotation can be described as the graph of a vector-valued function over the exterior of a bounded region in \(\mathbb{R}^n\times \{0^m\} \subset \mathbb{R}^{n+m}\) as 3 and \(F\) is defined on \(\mathbb{S}^{n-1} \times [t_1,+\infty)\) satisfying \(\|F\|_{2,\beta}<\infty\) for some \(\beta>0\). The result of Allard and Almgren can be applied since all Jacobi functions of the totally geodesic submanifold \(\mathbb{S}^{n-1} \subset \mathbb{S}^{n+m-1}\) are Killing vector fields of \(\mathbb{S}^{n-1}\) (see Theorem 5.1.1. in [19]). Decreasing slightly \(\beta\) if necessary, we can assume that \(-3\beta \neq \lambda_{k,\pm}\) and apply Proposition 5. So \(F = X+R\,,\) where \(X\) is in the kernel of \(L\) with decay between \(-\beta\) and \(-3\beta\) and \(\|R\|_{3\beta}<\infty\,\). If there are no elements in the kernel of \(L\) with decay between \(-\beta\) and \(-3\beta\,\), we get \(\|F\|_{3\beta}<\infty\,\); in that case we have improved the decay of \(F\,\). So we can iterate this argument until we get the first non-vanishing element in the kernel. The first decay of elements in the kernel is given by \(\lambda_{0,+}=-1\,\). Let \(\Phi_{k} :\mathbb{S}^{n-1} \mapsto \mathbb{R}^m\) denote the eigenfunction of \(\Delta_{\mathbb{S}^{n-1}}\) with eigenvalue \(-k(k+n-2)\,,\) and \(\Phi_k = (\Phi_k^1,\dots,\Phi_k^m)\,.\) Then \(\Phi_{0}\) is the constant vector-valued function and \(F\) can be written \[F(x,t)=e^{-t}b + R(x,t)\,, \text{ for some } b \in \mathbb{R}^m\,,\] with \(\|R\|_{1+\varepsilon}<\infty\) for some \(\varepsilon>0\,\). The first term can be interpreted as a translation. So the translated submanifold \(\iota(E)-b\) can be expressed as the graph of a vector-valued function \(G\) over the exterior of a bounded region in \(\mathbb{R}^{n} \times \{0^m\}\) with the estimate \(\|G\|_{1+\varepsilon}<\infty\,.\)

Then, we study the asymptotic behavior of \(\iota(E)-b\,.\) By Proposition 5, we get the first non-vanishing element in the kernel. Then the first decay of elements in the kernel is given by \(\lambda_{0,-}=-n+1\,.\) Furthermore, \(\lambda_{1,-} = -n,\lambda_{2,-} = -(n+1)\) and \(\Phi_{1}^\ell(x) = \sum_{j=1}^nc_{j}^\ell x^j,1\leqslant\ell\leqslant m\,.\) Since \(n\geqslant 3\,,\) \(3(-n+1) <-(n+1)\,.\) By Proposition 5, \[G(x,t) = e^{-(n-1)t}a + e^{-nt}\Phi_1(x) + O(e^{-(n+1)t})\, , \text{ for some } a \in \mathbb{R}^m\,.\]

We have parametrization \[e^tF(x,t) = b+e^{-(n-2)t}a + e^{-(n-1)t}\Phi_1(x) + O(e^{-nt})\,.\] If \(x^1,\cdots ,x^n\) are coordinates in \(\mathbb{R}^n\times \{0^m\}\,,\) then the graph function \(Y=(Y^1,\cdots,Y^m)\) has the following asymptotic behavior for \(\lvert x\rvert\) large: \[Y^\ell(x)= b^\ell + a^\ell\lvert x\rvert^{2-n} + \sum_{j=1}^nc_{j}^\ell x^j\lvert x\rvert^{-n} +O(\lvert x\rvert^{-n})\, , 1\leqslant\ell \leqslant m\,.\]

\((3)\Rightarrow(1):\) If the immersed submanifold \(M\) is regular at infinity, then \(M\) is of finite embedded ends. So we only need to show that the total curvature of each end is finite. However, this is obvious from the asymptotic behavior. ◻

Proof. By Theorem 6, \(M\) is regular at infinity and \(\iota\) is a proper immersion. So for any fixed \(p \in M\,,\) there exists \(R_j \uparrow \infty\) such that \(\partial B_{R_j}(p)\) intersects \(M\) transversely. Then we have \(\lvert A_M\rvert(p) = 0\) by curvature estimates in Lemma 2, which implies \(M\) is a flat \(n\)-plane in \(\mathbb{R}^{n+m}\). ◻

Proof. Firstly, we assume the conclusion in item(1) holds for the fixed \(I\,,\) and prove the conclusion in item(2) holds for the same fixed \(I\,\). Since \(\iota_j(\mathcal{B}_j) \subset B_{\frac{3}{4}r_0}(0) \,,\) after passing to a subsequence, we can assume \(\iota_j(\mathcal{B}_j)\) converges to \(\widetilde{\mathcal{B}}_\infty \subset \mathbb{R}^{n+m}\) in the Hausdorff distance sense, i.e., \(d_{\mathcal{H}}(\iota_j(\mathcal{B}_j),\widetilde{\mathcal{B}}_\infty) \to 0\,\). Since \(\nu_j(\mathbb{R}^{n+m})=\nu_j(B_{R_0}(0)) = \int_{M_j} \lvert A_{M_j}\rvert^n d\mu_{M_j}< IK_0\,,\) after passing to a subsequence, we can assume \(\nu_j \to \nu_\infty\) in the Radon measure sense.

If there exists \(p_\infty \in \widetilde{\mathcal{B}}_\infty\) with \(\nu_\infty(p_\infty)< 2K_0\,,\) then there exists some \(\tau_0>0\) small such that \(\nu_j(B_{\tau_0}(p_\infty)) <2K_0\) and \(\iota_j(p_j) \subset B_{\frac{\tau_0}{4}}(p_\infty)\) with some \(p_j\in \mathcal{B}_j\) for all \(j\) large enough. We denote \(\kappa_j : = |A_{M_j}|(p_j)\) and fix \(R_1 > C_1\) where \(C_1\) is the constant in Lemma 2. By the definition of smooth blow-up sets, after passing to a subsequence, the boundary of \(M_j\) in \(B_{\frac{ R_1}{\kappa_j}}(\iota_j(p_j))\) is empty for all \(j\,\). Hence \(B_{\frac{ R_1}{\kappa_j}}(\iota_j(p_j)) \subset B_{\frac{\tau_0}{2}}(p_\infty)\) and \(\nu_j(B_{\frac{ R_1}{\kappa_j}}(\iota_j(p_j)))< 2K_0\) for all \(j\) large enough. By Lemma 2, \(R_1 =\lvert A_{M_j}\rvert(p_j) \frac{ R_1}{\kappa_j} \leqslant C_1\) for all \(j\) large enough, this is a contradiction. So \(\nu_\infty(p_\infty)\geqslant 2K_0\) for any \(p_\infty \in \widetilde{\mathcal{B}}_\infty\,.\)

We will prove ?? by induction on \(I\). When \(I = 1\), the lemma follows the definition of \(K_0\) and Lemma 2. Then we assume ?? holds for \(I-1(I>1)\,.\)

After passing to a subsequence, we may assume that \[\alpha_{j} : = \sup_{x\in M_{j}} |A_{M_{j}}|(x)d(x,\iota_j(\partial M_{j})) \to \infty \, .\] If we cannot find such a subsequence, it is easy to see that curvature estimates hold with \(\mathcal{B}_{j} = \emptyset\,\).

Then the standard point picking argument as in Lemma 2 by passing to a subsequence allows us to find \(\widetilde{p}_{j}\in M_{j}\) so that \(\lambda_{j}:=|A_{M_{j}}|(\widetilde{p}_{j})\to\infty\) and the rescaled submanifold \[\widehat M_{j} : = \lambda_{j}(\iota_j(M_{j})- \iota_j(\widetilde{p}_{j}))\] converges locally smoothly in \(\mathbb{R}^{n+m}\) to a complete, non-flat, immersed minimal submanifold \(\widehat M_{\infty}\) without boundary. Moreover \(\widehat M_{\infty}\) is of finite total curvature and for all \(x \in \widehat M_{\infty}\,,\) \[|A_{\widehat M_{\infty}}|(x) \leqslant|A_{\widehat M_{\infty}}|(0) = 1\, , \mathop{\mathrm{vol}}(\widehat M_{\infty}\cap B_R(0) ) \leqslant\Lambda R^n, \text{ for any } R>0\,.\]

Because \(\widehat M_{\infty}\) is non-flat, by Corollary 7, \[\int_{\widehat M_{\infty}}\lvert A_{\widehat M_{\infty}}\rvert^n d\mu_{\widehat M_{\infty}} \geqslant 2K_0\,.\] Then there is some radius \(\widehat R > 0\) such that \[\int_{\widehat M_{\infty}\cap B_{\widehat R}(0)} \lvert A_{\widehat M_{\infty}}\rvert^n d\mu_{\widehat M_{\infty}} > \frac{3}{2}K_0\,.\] By Theorem 6, \(\widehat M_{\infty}\) is regular at infinity. By taking \(\widehat R\) larger if necessary, we can assume \(\widehat M_{\infty}\) intersects \(\partial B_{\widehat{R}}(0)\) transversely and \[\label{eq:infty46cur46est} |A_{\widehat M_{\infty}}|(x) \leqslant\frac{1}{4}\tag{7}\] for any \(x \in \widehat M_{\infty}\setminus B_{\widehat R}(0)\,\). Then \(\int_{M_j \cap B_{\widehat R/\lambda_{j}}(\iota_j(\widetilde{p}_{j}))} \lvert A_{M_j}\rvert^n > K_0\) for all \(j\) large enough while \(\limsup \limits_{j\to \infty}\nu_j(B_{R_0}(0)\setminus B_{\frac{r_0}{2}}(0))< K_0\,.\) So after passing to a subsequence, we can assume \(\iota_j(\widetilde{p}_{j}) \in B_{\frac{3}{4}r_0}(0)\,.\)

We define \(\widetilde{M}_{j} : = M_{j} \setminus \iota_j^{-1}\left(B_{\widehat R/\lambda_{j}}(\iota_j(\widetilde{p}_{j}))\right)\,\). For all \(j\) large, due to the choice of \(\widetilde{p}_{j}\) and \(\alpha_{j}\to\infty\,,\) \(B_{\widehat R/\lambda_{j}}(\iota_j(\widetilde{p}_{j})) \cap \partial M_j = \emptyset\,.\) Since \(\widehat M_{\infty}\) intersects \(\partial B_{\widehat{R}}(0)\) transversely, \(M_{j}\) intersects \(\partial B_{\widehat R/\lambda_{j}}(\iota_j(\widetilde{p}_{j}))\) transversely. Thus, \(\widetilde{M}_{j}\) is a smooth compact minimal submanifold with smooth, compact boundary \[\partial\widetilde{M}_{j} = \partial M_{j} \cup (\partial B_{\widehat R/\lambda_{j}}(\widetilde{\iota}_j(p_{j}))\cap M_{j})\, .\] For all \(j\) large, \[\int_{\widetilde{M}_{j}}\lvert A_{\widetilde{M}_{j}}\rvert^n d\mu_{\widetilde{M}_{j}}< (I-1)K_0\,.\] By the inductive hypothesis, after passing to a subsequence, there is a sequence of smooth blow-up sets \(\widetilde{\mathcal{B}}_{j} \subset \widetilde{M}_{j}\) with \(|\widetilde{\mathcal{B}}_{j}|< I-1\,, \iota_j(\widetilde{\mathcal{B}}_{j} ) \subset B_{\frac{3r_0}{4}}(0)\) and a constant \(\widetilde{C}\) (independent of \(j\)) so that \[\label{eq:curv-est-induct-hyp} |A_{\widetilde{M}_{j}}|(x)d(\iota_j(x),\iota_j(\widetilde{\mathcal{B}}_{j} \cup \partial\widetilde{M}_{j})) \leqslant\widetilde{C}\tag{8}\] for all \(x\in \widetilde{M}_j\,.\)

We claim that \(\mathcal{B}_{j} : = \widetilde{\mathcal{B}}_{j}\cup \{\widetilde{p}_{j}\}\) is a sequence of smooth blow-up sets for \(M_j\). We only need to check that none of the points in \(\widetilde{\mathcal{B}}_{j}\) can appear in the blow-up at \(\widetilde{p}_{j}\) and \(\widetilde{p}_{j}\) cannot appear in the blow-up at any point in \(\widetilde{\mathcal{B}}_{j}\) (which implies that rescaling \(M_{j}\) around points in \(\widetilde{\mathcal{B}}_{j}\) still yields a smooth limit).

We will prove the above claim by contradiction. If \[\begin{gather} \label{lemm46fin46contra} \liminf_{j\to\infty}\min_{\widetilde{q} \in \widetilde{\mathcal{B}}_{j}} \lambda_{j} |\iota_j(\widetilde{q})-\iota_j(\widetilde{p}_{j})| < \infty\, , \end{gather}\tag{9}\] then we can assume that the minimum is attained at \(\widetilde{q}_{j}\in\widetilde{\mathcal{B}}_{j}\). Since \(\widetilde{q}_{j} \in \widetilde{M}_{j}= M_{j} \setminus \iota_j^{-1}\left( B_{\widehat R/\lambda_{j}}(\iota_j(\widetilde{p}_{j}))\right)\) and 7 , 9 hold, after passing to a subsequence, \[\beta_{j} : = |A_{M_{j}}|(\widetilde{q}_{j}) \leqslant\frac{1}{2} |A_{M_{j}}|(\widetilde{p}_{j}) = \frac{1}{2} \lambda_{j}\, .\] Hence, we have \[\label{eq46contra1} \liminf_{j\to\infty} \beta_{j}|\iota_j(\widetilde{q}_{j})-\iota_j(\widetilde{p}_{j})| < \infty\, .\tag{10}\]

So if the claim is not hold, 10 must hold. However, the blow-up of \(\widetilde{M}_{j}\) around \(\widetilde{q}_{j}\) has no boundary due to the definition of smooth blow-up set, which will contradict with 10 . So \(\mathcal{B}_j\) is a sequence of smooth blow-up sets.

Now, we prove that curvature estimates in ?? hold. We argue by contradiction. Suppose that there is \(y_{j} \in M_{j}\) such that \[\label{eq:cur46contra} \limsup_{j\to\infty}|A_{M_{j}}|(y_{j})d(\iota_j(y_j),\iota_j(\mathcal{B}_{j}\cup\partial M_{j})) = \infty\, .\tag{11}\] Combined 8 and the choice of \(\widetilde{p}_{j}\), after passing to a subsequence, we can assume that \(y_{j} \in \widetilde{M}_{j}\) and \[\label{eq46lem46finite46dist} \Xi_j : = d(\iota_j(y_{j}) \,,\iota_j(\mathcal{B}_{j}\cup\partial M_{j})) = |\iota_j(y_{j})-\iota_j(\widetilde{p}_{j})|\, .\tag{12}\] Then, we have \[\label{eq46omega} \Omega_j := d(\iota_j(y_{j})\, , \iota_j(\widetilde{\mathcal{B}}_{j}\cup \partial\widetilde{M}_{j})) = |\iota_j(y_{j})-\iota_j(\widetilde{p}_{j})| - \frac{\widehat R}{\lambda_{j}}\,.\tag{13}\] Hence, after passing to a subsequence, we can assume \(\Xi_j \to 0\,.\) Otherwise \(\Omega_j/\Xi_j \to 1\,,\) then 11 will contradict with 8 . Because \(\widehat M_{\infty}\) has bounded curvature, so \(y_{j}\) cannot appear in the blow-up at \(\widetilde{p}_{j}\,\), i.e., \[\label{eq46q46blow} \liminf_{j\to\infty} \lambda_{j}\Xi_j = \infty\, .\tag{14}\] Hence, combined 8 , 13 and 14 , we have \[\label{eq46lem46cur46small} \limsup_{j\to\infty} |A_{M_{j}}|(y_{j}) \frac{\widehat R}{\lambda_{j}} \leqslant\limsup_{j\to\infty} \frac{\widetilde{C} \widehat R}{\lambda_{j}d(\iota_j(y_{j}),\iota_j(\widetilde{\mathcal{B}}_{j}\cup\partial\widetilde{M}_{j}))} = 0\, .\tag{15}\] Combined 8 , 13 , 14 and 15 , we have \[\begin{align} \widetilde{C} &\geqslant\limsup_{j\to\infty} |A_{M_{j}}|(y_{j})d(\iota_j(y_{j}) ,\iota_j(\widetilde{\mathcal{B}}_{j}\cup\partial\widetilde{M}_{j})) \\ &= \limsup_{j\to\infty} |A_{M_{j}}|(y_{j}) \left(|\iota_j(y_{j})-\iota_j(\widetilde{p}_{j})| - \frac{\widehat R}{\lambda_{j}}\right)= \infty\, , \end{align}\] which is a contradiction. So we complete the proof. ◻

Proof. Choose a smooth cutoff function \(\eta \in C^{\infty}_{c}([0,1))\) so that \(\eta(r) \in [0,1]\,\), \(\eta(r) = 1\) for \(r \leqslant\frac{1}{4}\) and \(\eta(r) = 0\) for \(r\geqslant\frac{81}{100}\,.\) We will take \(\sigma_0>0\) sufficiently small based on this fixed cutoff function. Let \(\phi(x):=\eta(|\iota(x)|^{2}).\) Consider the function \[f(x) = |\iota(x)-p|^{2}\phi(x) + |\iota(x)|^{2}(1-\phi(x))\, .\] We see that \(f(x) = |\iota(x)-p|^{2}\) if \(|\iota(x)|\leqslant\frac{1}{2}\) and \(f(x) = |\iota(x)|^{2}\) if \(|\iota(x)| \geqslant\frac{9}{10}\,.\) For any point \(q \in \mathbb{R}^{n+m}\) and \(\xi \in \mathfrak{X}(M)\,,\) we have \[\begin{align} \label{lem46top461} \nabla(|\iota(x)-q|^{2}) & = 2(\iota(x)-q)^T, \end{align}\tag{16}\] \[\begin{align} \label{lem46top462}\nabla^2 (|\iota(x)-q|^{2}) (\xi,\xi) & = \xi(\xi(|\iota(x)-q|^{2})) - \nabla_\xi \xi (|\iota(x)-q|^{2}) \\ &= \overline{\nabla}^2(|\iota(x)-q|^{2})(\xi,\xi) + A(\xi,\xi)|\iota(x)-q|^{2} \\ & = 2\left(\lvert\xi\rvert^2 + \langle A(\xi,\xi),\iota(x)-q \rangle\right)\, ,\end{align}\tag{17}\] \[\begin{align} \label{lem46top463} \nabla\phi & = \eta'\nabla|\iota(x)|^2 = 2 \eta' (\iota(x))^T\, , \end{align}\tag{18}\] \[\begin{align} \label{lem46top464}\nabla^2 \phi(\xi,\xi) &= \eta' \nabla^2|\iota(x)|^2(\xi,\xi) + 4\eta''\langle \iota(x),\xi \rangle^2\\ & = 2\eta'\left(\lvert\xi\rvert^2 + \langle A(\xi,\xi),\iota(x) \rangle\right)+ 4\eta''\langle \iota(x),\xi \rangle^2\, .\end{align}\tag{19}\] Combined 16 , 17 , 18 and 19 , we compute \[\begin{align} \nabla^{2} f (\xi,\xi) = & \phi\nabla^2 |\iota(x)-p|^{2}(\xi,\xi) +(1-\phi) \nabla^2|\iota(x)|^{2}(\xi,\xi)\\ &+ 2\langle \nabla\phi ,\xi \rangle\langle \nabla(|\iota(x)-p|^2),\xi \rangle -2 \langle \nabla\phi ,\xi \rangle\langle \nabla(|\iota(x)|^2),\xi \rangle\\ &+ |\iota(x)-p|^{2} \nabla^2\phi(\xi,\xi) - |\iota(x)|^{2}\nabla^2\phi(\xi,\xi)\\ =&2\phi\left(|\xi|^2+ \langle A(\xi,\xi),\iota(x)-p \rangle\right) + 2(1-\phi)\left(|\xi|^2+ \langle A(\xi,\xi),\iota(x) \rangle\right)\\ & + 8 \eta' \langle \iota(x)-p,\xi \rangle \langle \iota(x),\xi \rangle - 8 \eta' \langle \iota(x),\xi \rangle^{2}\\ & + 2\eta'(|\iota(x)-p|^{2}-|\iota(x)|^{2})\left(|\xi|^2+ \langle A(\xi,\xi),\iota(x) \rangle\right)\\ & + 4\eta''\left(|\iota(x)-p|^{2}-|\iota(x)|^{2}\right) \langle \iota(x),\xi \rangle^{2}\\ =& 2\left(|\xi|^2 + \langle A(\xi,\xi),\iota(x)-p \rangle\right) +2(1-\phi)\langle A(\xi,\xi),p \rangle\\ & - 8 \eta' \langle p,\xi \rangle \left\langle \iota(x),\xi \right\rangle \\ & + 2\eta'\left(|\iota(x)-p|^{2}-|\iota(x)|^{2}\right)\left(|\xi|^2+\langle A(\xi,\xi),\iota(x) \rangle\right)\\& + 4\eta''\left(|\iota(x)-p|^{2}-|\iota(x)|^{2}\right)\left\langle \iota(x),\xi \right\rangle^{2}\, . \end{align}\] If \(|\iota(x)| < \frac{1}{2}\,,\) then \(x\) is not in the supports of \(1-\phi\,\), \(\eta'(\lvert\iota(x)\rvert^2)\) and \(\eta''(\lvert\iota(x)\rvert^2)\,\). Hence, by taking \(\sigma_0\) sufficiently small, \(|A|(x)\leqslant\frac{5}{8}\) on the supports of \(1-\phi\,\), \(\eta'(\lvert\iota(x)\rvert^2)\) and \(\eta''(\lvert\iota(x)\rvert^2)\,.\) It is easy to see that on \(M \cap \left(\overline{B_{1}(0)}\setminus {B_{\sigma}(p)}\right)\,\), \[\nabla^{2}f(\xi,\xi) \geqslant 2\left(|\xi|^2+ \langle A(\xi,\xi),\iota(x)-p \rangle\right) - C\lvert p\rvert |\xi|^{2}\, ,\] for some \(C>0\) only depending on \(|\eta|\, ,|\eta'|\) and \(|\eta''|\,.\) Since \(|A|(x)|\iota(x)-p| \leqslant\frac{1}{4}\) for all \(x \in M \cap \left( \overline{B_{1}(0)} \setminus B_{\sigma}(p) \right)\,,\) we have that \[\nabla^{2}f(\xi,\xi) \geqslant 2\left(\frac{3}{4} - C\lvert p\rvert\right) |\xi|^{2}\, .\] Thus, as long as \(\sigma_0 > \lvert p\rvert\) is sufficiently small, \(\nabla^{2}f\) is strictly positive.

Choosing such a \(\sigma_0\), then any critical point of \(f\) in \(M \cap \left(\overline{B_{1}(0)}\setminus {B_{\sigma}(p)}\right)\) must be a strict local minimum. Suppose \(f\) has critical points in \(M \cap \left(\overline{B_{1}(0)}\setminus {B_{\sigma}(p)}\right)\,\). Since \(M\) is properly immersed, \(f\) has finite critical points on \(M \cap \left(\overline{B_{1}(0)}\setminus {B_{\sigma}(p)}\right)\,.\) We fix a component \(M'\) of \(M\,.\) By Morse theory[22], if \(f\) has critical points on \(M'\), then \(M'\) must have only one critical point \(x'\) and \(M'\cap f^{-1}(\{x\in M'|f(x)<f(x')\}) = \emptyset\,.\) Then \(M' \cap B_{\sigma}(p) = \emptyset\,,\) which is a contradiction. Hence \(f\) cannot have any critical points in \(M \cap \left(\overline{B_{1}(0)}\setminus {B_{\sigma}(p)}\right)\,\). The lemma follows from the standard Morse theory. ◻

Proof. Let \(f(x):=|\iota(x)|\,.\) We claim that after rescaling \(M\,,\) we can assume that \(|\widehat\iota(x)|\) has no critical points in \(B_2(0)\) and \(M\) intersects \(\partial B_1(0)\) transversely.

We may rescale \(M\) and still denote the immersed submanifold \(M\) after rescaling, so that any critical point of \(f\) on \(M\cap B_{2}(0)\) is a non-degenerate local minimum and \(M\) satisfies the assumption in Theorem 10. Let \(M'\) be a component of \(M\) and \(f\) has at least a critical point on \(M'\,.\) Then by Morse theory(as argued in Lemma 9) \(M'\) is a smooth immersed minimal submanifold in \(B_2(0)\) while \(0 \not \in \iota(M')\,.\) Hence, by the volume bound and monotonicity formula, there exists \(\delta_2\) small such that every component \(M'\) of \(M \cap B_{ 2}(0)\) with \(M' \cap B_{\delta_2}(0) \not = \emptyset\) will have no critical points. Hence we can rescale \(M\) so that \(f\) has no critical points and \(M\) intersects \(\partial B_1(0)\) transversely.

The above arguments are similar to arguments in [23]. At below, we use several classical results in Geometric Measure Theory to address the problems we consider. Due to the volume bound and monotonicity formula, \(M\) has finite components in \(B_2(0)\,.\) Let \(M'\) be a component of \(M\) and let \(\eta_j \uparrow \infty \,\). Let \(M'_j := \eta_j\iota(M')\,.\) By curvature estimates in Lemma 4, after passing to a subsequence, \(M_j'\) converges locally smoothly in \(\mathbb{R}^{n+m} \setminus\{0\}\) to an union of \(n\)-planes \(M_\infty'\) . Since \(M'_j\) has no critical points, we can argue as Lemma 11 so that all \(n\)-planes in \(M_\infty'\) are passing the point \(0 \in \mathbb{R}^{n+m}\,.\) Since \(M'\) is connected, \(f^{-1}(t)\cap M'\) is also connected by the standard Morse theory. Then \(M'_\infty\) is an \(n\)-plane passing \(0\in \mathbb{R}^{n+m}\) and \(f^{-1}(1)\cap M'_j\) converges smoothly to \(M'_\infty\cap \partial B_1(0) = \mathbb{S}^{n-1}\,.\) So for all \(j\) large enough, \(f^{-1}(1)\cap M'_j\) is a \(K_j\)-sheeted covering space of \(\mathbb{S}^{n-1}\,.\) It is well known that \(\mathbb{S}^{n-1}(n\geqslant 3)\) is simply connected, so \(K_j = 1\) due to the standard covering space theory. Hence \(M'_{\infty}\) is of multiplicity one. Then by Theorem 5 and corresponding Corollary of [24], the varifold \(|M'_\infty|\) is the unique tangent cone of the varifold \(|M'|\) at \(0 \in \mathbb{R}^{n+m}\)(\(|M'_\infty|\) and \(|M'|\) denote the the corresponding varifolds of \(M'_\infty\) and \(M'\)). So we get that \(M'\) is a smooth minimal graph passing \(0 \in \mathbb{R}^{n+m}\) in a neighborhood of \(0 \in \mathbb{R}^{n+m}\) due to Allard’s regularity theorem[25]. So there exists \(\widehat\iota\) which extends \(\iota\) to a smooth minimal immersion in \(B_2(0)\,\). ◻

Proof. We prove the theorem by induction on \(I\). For \(I=1\) the theorem trivially follows from curvature estimates in Lemma 2, the volume bound and Lemma 13.

Then we suppose the theorem 12 holds for \(I-1(I>1)\,.\) We choose \(r_0\) small enough so that for any \(\widetilde{p}_\infty \in \widetilde{\mathcal{B}}_\infty\,,\) \(M_\infty\) is sufficiently smoothly close to the union of \(n\)-planes passing \(\widetilde{p}_\infty\) in \(B_{r_0}(\widetilde{p}_\infty)\,,\) and \(\min \limits_{\substack{\widetilde{p}_\infty,\widetilde{q}_\infty \in \widetilde{\mathcal{B}}_\infty \\ \widetilde{p}_\infty \not=\widetilde{q}_\infty}}|\widetilde{p}_\infty- \widetilde{q}_\infty|>4r_0\, .\) Since \(M_j\) converges locally smoothly in \(B_{\frac{3}{2}}(0) \setminus \widetilde{\mathcal{B}}_\infty\) to \(M_\infty\,,\) fixing \(r_0\) small enough, for all \(j\) large enough, \(M_j \cap B_{1}(0) \setminus B_{r_0}(\widetilde{\mathcal{B}}_\infty)\) is a smooth covering space of \(M_\infty \cap B_{1}(0) \setminus B_{r_0}(\widetilde{\mathcal{B}}_\infty)\) with the numbers of sheets on different components of \(M_\infty \cap B_{1}(0) \setminus B_{r_0}(\widetilde{\mathcal{B}}_\infty)\) uniformly bounded due to the monotonicity formula and the uniform volume bound. So after passing to a subsequence, we can assume that all of \(M_j\cap B_{1}(0) \setminus B_{r_0}(\widetilde{\mathcal{B}}_\infty)\) are diffeomorphic by Lemma 13. From Theorem 7.6.2 and Proposition 7.6.4 of [26], the smooth structure is independent of gluing maps. So there are only finitely many ways to connect the regions of \(M_j\cap B_{1}(0) \setminus B_{r_0}(\widetilde{\mathcal{B}}_\infty)\) to the regions of \(M_j \cap B_{r_0}(\widetilde{\mathcal{B}}_\infty)\), and we only need to prove that after passing to a subsequence, \(M_j \cap B_{r_0}(\widetilde{p}_\infty)\) are diffeomorphic for every \(\widetilde{p}_\infty \in \widetilde{\mathcal{B}}_\infty\,.\) After rescaling, \[\check M_j := r_0^{-1}( \iota_j(M_j) - \widetilde{p}_\infty)\,.\] By the volume bound and monotonicity formula, we can choose a component of \(\check M_j\) in \(B_2(0)\,,\) and we get a sequence of minimal submanifolds in \(\mathbb{R}^{n+m}\) satisfying \((\beth)\) with limit submanifold in item (6) is smoothly sufficiently close to the union of \(n\)-planes passing \(0 \in \mathbb{R}^{n+m}\,.\) Abusing notation slightly, we will still denote this sequence of submanifolds \(M_j\,.\) Hence we only need to prove the theorem for this sequence of minimal submanifolds.

If \(|\widetilde{\mathcal{B}}_\infty| \geqslant 2\,,\) we can argue as the proof of Lemma 8, and get that the new sequence \(M_j\) as defined above has total curvature \[\int_{M_j} \lvert A_{M_j}\rvert^n d \mu_{M_j}< (I-1)K_0\,.\] So the theorem holds by the induction hypothesis.

If \(|\widetilde{\mathcal{B}}_\infty| = 1\) and \(\liminf \limits_{j\to \infty}|\mathcal{B}_{j}| = 1\,,\) after passing to a subsequence, we can write \(\mathcal{B}_{j} = \{p_{j}\}\,,\) \(\widetilde{\mathcal{B}}_{\infty}=\{\widetilde{p}_{\infty}\}\) and \(\lambda_{j} : = |A_{M_{j}}|(p_{j})\,\). By passing to a subsequence, we have that \[\breve M_{j} : = \lambda_{j}(\iota_j(M_{j})-\iota_j(p_{j}))\] converges to a complete, non-flat, properly immersed minimal submanifold \(\breve M_{\infty}\subset \mathbb{R}^{n+m}\) without boundary satisfying that the total curvature of \(\breve M_{\infty}\) \(\leqslant\) \(IK_0\) and \(\mathop{\mathrm{vol}}(\breve{M}_{\infty} \cap B_{r}(0))\leqslant\Lambda r^{n}\) for any \(r>0\) due to the monotonicity formula. So by Theorem 6, \(\breve{M}_{\infty}\) is regular at infinity. In particular, we can choose \(R>0\) so that \(\breve{M}_{\infty}\) intersects \(\partial B_{R}(0)\) transversely and \[|A_{\breve{M}_{\infty}}|(x) |\iota_\infty(x)| < \frac{1}{4}\] for all \(x \in\breve{M}_{\infty}\setminus B_{R}(0)\), where \(|\iota_\infty(x)|\) denotes the Euclidean distance between \(0 \in \mathbb{R}^{n+m}\) and the image of \(x \in \breve M_\infty\) in \(\mathbb{R}^{n+m}\,.\)

Case I: after passing to a subsequence, if for all \(j\,,\) \[\label{eq:prop46cur1} |A_{M_{j}}|(x)|\iota_j(x)-\iota_j(p_{j})| < \frac{1}{4}\tag{20}\] for all \(x \in M_{j} \cap \left( B_{2}(0) \setminus B_{R/\lambda_{j}}(\iota_j(p_{j}))\right)\,\). Then Lemma 13 and Lemma 9 imply that after passing to a subsequence, all of the submanifolds \(M_{j}\cap B_{1}(0)\) are diffeomorphic (here, we have used the fact that the ends of \(\breve M_\infty\) are diffeomorphic to \(\mathbb{S}^{n-1}\times (0,1)\) with the standard smooth structure and \(\breve M_j\cap B_{R}(0)\) is a smooth covering space of \(\breve M_\infty \cap B_{R}(0)\,\)).

Case II: on the other hand, if 20 does not hold, we may choose \(\delta_{j}\) to be the smallest radius greater than \(R/\lambda_{j}\) so that \[|A_{M_{j}}|(x)|\iota_j(x)- \iota_j(p_{j})| < \frac{1}{4}\] for all \(x \in M_{j} \cap \left( B_{2}(0) \setminus B_{\delta_{j}}(\iota_j(p_{j}))\right)\,\). For all \(j\) sufficiently large, such a \(\delta_{j}\) exists with \(\delta_{j}\to 0\). This follows from the fact that \(M_{j}\) converges smoothly away from \(\widetilde{p}_{\infty}\) to a submanifold sufficiently close to the union of \(n\)-planes. Furthermore, we may assume \(\liminf \limits_{j\to \infty }\lambda_j\delta_j = \infty\,,\) otherwise we can take a subsequence of \(M_j\) and take \(R\) larger so that \(R > \liminf \limits_{j\to \infty }\lambda_j\delta_j\,,\) then 20 holds for all \(x \in M_{j} \cap \left( B_{2}(0) \setminus B_{R/\lambda_{j}}(\iota_j(p_{j}))\right)\) with all \(j\) large enough, and the theorem follows the same arguments in Case I. At below, we define \[\widehat M_{j} := \delta_{j}^{-1} (\iota_j(M_{j})-\iota_j(p_{j}))\, .\] After passing to a subsequence, there is an immersed minimal submanifold \(\widehat M_{\infty}\) in \(\mathbb{R}^{n+m} \setminus \{0\}\) so that \(\widehat M_{j}\) converges locally smoothly in \(\mathbb{R}^{n+m}\setminus \{0\}\) to \(\widehat M_{\infty}\) with finite multiplicity(the multiplicity may be different for distinct components of \(\widehat M_\infty\)) by the curvature estimates in item (5). Moreover, after passing to a subsequence, \(|\widehat M_{j}| \rightharpoonup\lvert\widehat M_\infty\rvert\) in the sense of varifolds in \(B_1(0)\,.\)

Since \(\widehat M_{\infty}\) has finite total curvature and satisfies the volume growth condition of Theorem 10 by the varifold convergence and monotonicity formula, the possible singularity at \(\{0\}\) is removable. Hence \(\widehat M_{\infty}\) is an immersed minimal submanifold in \(\mathbb{R}^{n+m}\) with total curvature \(\int_{\widehat M_{\infty}} \lvert A_{\widehat M_{\infty}}\rvert^n d\mu_{\widehat M_{\infty}}\leqslant I K_0\) and \(\mathop{\mathrm{vol}}(\widehat M_{\infty}\cap B_{r}(0)) \leqslant\Lambda r^{n}\) for any \(r>0\,.\) Moreover, \(\widehat M_{\infty}\) has bounded number of components due to the volume bound. By Theorem 6, every component of \(\widehat M_{\infty}\) is regular at infinity. Then we can choose \(\gamma \geqslant 1\) large enough so that \(\partial B_{\gamma}(0)\) intersects each component of \(\widehat M_{\infty}\) transversely, and each component of \(\widehat M_{\infty} \cap \partial B_{\gamma}(0)\) is diffeomorphic to \(\mathbb{S}^{n-1}\) with the standard smooth structure. Moreover the number of components of \(\widehat M_{\infty} \cap \partial B_{\gamma}(0)\) is no more than \(\Lambda\,.\) By the choice of \(\delta_{j}\,\), the curvature estimates 20 hold for all \(x \in M_{j} \cap \left( B_{2}(0) \setminus B_{\gamma \delta_{j}}(\iota_j(p_{j}))\right)\,\). Then by applying Lemma 9, we see that \(M_{j} \cap \left( B_{2}(0) \setminus B_{\gamma \delta_{j}}(\iota_j(p_{j}))\right)\) is diffeomorphic to the union of annular regions. In particular, \(M_{j}\cap B_{\gamma\delta_{j}}(\iota_j(p_{j}))\) must be connected (because we have assumed that \(M_{j}\) is connected in \((\beth)\)). Then we only need to prove \(\widehat M_j \cap B_\gamma(0)\) are diffeomorphic to each other after passing to a subsequence.

By the choice of \(\delta_j\,,\) there exists at least a non-flat component of \(\widehat M_{\infty}\,.\) Applying Corollary 7, we can take \(\gamma\) sufficiently large and \(\kappa\) sufficiently small so that \[\int_{ \widehat M_{\infty}\cap B_\gamma(0)\setminus B_{3\kappa}(0)} \lvert A_{\widehat M_{\infty}}\rvert^n d\mu_{\widehat M_{\infty}}\geqslant\frac{3}{2}K_0\,.\] Then after passing to a subsequence, \[\int_{\widehat M_j \cap B_{2\kappa}(0) }\lvert A_{\widehat M_j}\rvert^n d\mu_{\widehat M_j} < (I-1)K_0\] for all \(j\,.\) After rescaling \(\widetilde{M}_j := \kappa^{-1} \widehat\iota_j(\widehat M_j)\,.\) We can choose some component \(\widetilde{M}_j'\) of \(\widetilde{M}_j\,,\) then we have the fact that the sequence \(\widetilde{M}_j'\) will satisfy \((\beth)\) with total curvature \[\int_{\widetilde{M}_j'\cap B_2(0)} \lvert A_{\widetilde{M}_j'}\rvert^n d\mu_{\widetilde{M}'_j} <(I-1)K_0\,.\] By the induction hypothesis, \(\widetilde{M}_j'\cap B_1(0)\) are diffeomorphic. Then after passing to a subsequence, \(\widehat M_j\cap B_\kappa(0)\) are diffeomorphic. By the locally smooth convergence of \(\widehat M_j\,,\) we have \(\widehat M_j \cap B_\gamma(0)\) are diffeomorphic to each other after passing to a subsequence. This completes the proof in the case that \(|\widetilde{\mathcal{B}}_\infty| = 1\) and \(\liminf \limits_{j\to \infty}|\mathcal{B}_{j}| = 1\,.\)

If \(|\widetilde{\mathcal{B}}_\infty| =1\, ,\liminf\limits_{j\to \infty}|\mathcal{B}_j| \geqslant 2\,\). Then \(\varepsilon_j : = \max \limits_{\substack{p_i,q_j \in \mathcal{B}_j \\ p_j \not = q_j}}d (\iota_j(p_j),\iota_j(q_j)) \to 0\,.\) By the definition of smooth blow-up sets, \(\lim \limits_{j\to \infty} \varepsilon_j |A_{M_j}|(p_j) = \infty\) for all \(p_j \in \widetilde{\mathcal{B}}_j\,.\) Then we fix \(p_j\,,q_j \in \mathcal{B}_j\) satisfying \(\varepsilon_j = |\iota_j(p_j)-\iota_j(q_j)|\,.\) Let \(M^*_j: = \frac{\sigma_0}{4\varepsilon_j}(\iota_j(M_j)-\iota_j(p_j))\,,\) and we can choose a component \(M^{*'}_j\) of \(M^*_j\) in \(B_2(0)\,.\) After passing to a subsequence, this sequence will satisfy \((\beth)\) with \(|\widetilde{\mathcal{B}}_{\infty}| \geqslant 2\) or \[\int_{M^{*'}_j\cap B_2(0)} \lvert A_{M^{*'}_j}\rvert^nd\mu_{M^{*'}_j} <(I-1)K_0\,.\]

Hence after passing to a subsequence, all of the \(B_{\frac{4\varepsilon_j}{\sigma_0}}(\iota_j(p_j)) \cap M_j\) are diffeomorphic. As the situation of \(|\widetilde{\mathcal{B}}_\infty| = 1\) and \(\liminf \limits_{j\to \infty}|\mathcal{B}_{j}| = 1\,,\) then we can argue in two cases and prove the theorem. ◻

Proof of Theorem 1. We will prove Theorem 1 by contradiction. Since the volume bound and monotonicity formula imply that there exist at most finite number of components for any minimal submanifold satisfying the assumption of Theorem 1, without loss of generality, we can assume the minimal submanifolds satisfying the assumption of Theorem 1 are connected. If \(M_j^n\) is a sequence of pairwise non-diffeomorphic complete connected, immersed minimal submanifold in \(\mathbb{R}^{n+m}\) with \(\mathop{\mathrm{vol}}(M_j\cap B_{R}(0)) \leqslant\Lambda R^{n}\) for any \(R>0\) and \[\int_{M_j} \lvert A_{M_j}\rvert^n d \mu_{M_j}\leqslant\Gamma < IK_0\,\,.\] By rescaling \(M_j,\) we can assume \[\int_{M_j\setminus B_{\frac{1}{j}}(0)}\lvert A_{M_j}\rvert^n d \mu_{M_j} <\frac{1}{j}\,,\] and \(M_j\) intersects \(\partial B_1(0)\) transversely. By Theorem 6, \(M_j\) is properly immersed and regular at infinity. By rescaling \(M_j\,,\) we can assume \(M_j \setminus B_{\frac{1}{2}}(0)\) is the union of minimal graph and each minimal graph is defined over the exterior of a bounded region in an \(n\)-plane passing \(0 \in \mathbb{R}^{n+m}\,.\) So after passing to a subsequence, we can assume the \(M_j\cap B_1(0)\) are pairwise non-diffeomorphic. By Lemma 8, the sequence \(M_j\cap B_2(0)\) satisfies \((\beth)\). Then by Theorem 12, after passing to a subsequence, all of the \(M_j\cap B_1(0)\) are diffeomorphic. This is a contradiction. ◻

Proof. Since \(M\) is a smooth compact \(n\)-manifold, \(M\) and some CW-complex with finite cells are homotopy equivalent(see [27] for more details about CW-complex, fundamental group and covering space). Hence \(\pi_1(M)\) is finitely generated. From [27], we know that \(k\)-sheeted covering spaces of \(M\) are classified by equivalence classes of homomorphisms \(\pi_1(M) \to \Sigma_k\,,\) where \(\Sigma_k\) is the symmetric group on \(k\) symbols and the equivalence relation identifies a homomorphism \(\rho\) with each of its conjugates \(h^{-1}\rho h\) by elements \(h \in \Sigma_k\,\). Since \(\pi_1(M)\) is finitely generated, a homomorphism is determined by the image of the \(\ell(\in \mathbb{N})\) generators of \(\pi_1(M)\, \,.\) Hence there are at most \((k!)^\ell\) such homomorphisms. It implies that there are at most \((k!)^\ell\) equivalence classes of homomorphisms \(\pi_1(M) \to \Sigma_k\,,\) and at most \((k!)^\ell\) pairwise non-diffeomorphic smooth \(k\)-sheeted covering spaces of \(M.\) ◻