[proofplan] We first view $N$ as an isometrically embedded compact submanifold of some Euclidean space and define the value of $u$ at the origin arbitrarily; the finite energy hypothesis then gives a function in $W^{1,2}(B(0,1);N)$. The main point is to show that the weak harmonic map equation survives across the puncture. This is done with logarithmic cutoffs whose gradients have vanishing $L^2$ norm, so the error produced by cutting off test functions near $0$ tends to zero. Once $u$ is weakly harmonic on the whole disk, interior regularity for weakly harmonic maps into compact smooth targets upgrades the extension to a smooth harmonic map. [/proofplan]

[step:Embed the target and extend the Sobolev map across the point] By the [Nash embedding theorem](/page/Nash%20Embedding%20Theorem), choose an integer $m \in \mathbb{N}$ and an isometric smooth embedding $\iota: N \to \mathbb{R}^m$. Replacing $u$ by $\iota \circ u$, we regard $N$ as a compact embedded submanifold of $\mathbb{R}^m$. Let $\nu$ denote the [second fundamental form](/page/Second%20Fundamental%20Form) of this embedding, with the sign convention for which the extrinsic harmonic map equation is $\Delta u+\nu_u(\nabla u,\nabla u)=0$. Thus, for each $p \in N$, $\nu_p:T_pN\times T_pN\to (T_pN)^\perp\subset\mathbb{R}^m$ is a symmetric bilinear map. The expression $\nu_u(\nabla u,\nabla u)$ denotes the map from $B(0,1)\setminus\{0\}$ to $\mathbb{R}^m$ whose value at $x$ is \begin{align*} \sum_{i=1}^2\nu_{u(x)}(\partial_{x_i}u(x),\partial_{x_i}u(x)). \end{align*} Choose a point $q \in N$ and define the extended map $\tilde u: B(0,1) \to N \subset \mathbb{R}^m$ by $\tilde u(0) := q$ and $\tilde u(x) := u(x)$ for $x \neq 0$. Since $N$ is compact, $\tilde u \in L^2(B(0,1);\mathbb{R}^m)$. The finite [Dirichlet energy](/page/Dirichlet%20Energy) hypothesis gives \begin{align*} \int_{B(0,1)\setminus\{0\}} |\nabla u(x)|^2\,d\mathcal{L}^2(x) < \infty. \end{align*} The logarithmic cutoff estimate constructed explicitly in the next step is the standard verification that the singleton $\{0\}\subset\mathbb{R}^2$ has zero $W^{1,2}$ capacity. By the [Sobolev removability theorem for zero-capacity sets](/page/Sobolev%20Capacity), applied to this singleton, the distributional first derivatives of $\tilde u$ on $B(0,1)$ are represented by the $L^2$ function $\nabla u$ on the punctured disk, extended arbitrarily on the null set $\{0\}$. Hence $\tilde u \in W^{1,2}(B(0,1);\mathbb{R}^m)$ and $\tilde u(x)\in N$ for $\mathcal{L}^2$-almost every $x\in B(0,1)$. We write $u$ for this extension from now on. [/step]

[step:Prepare logarithmic cutoffs around the puncture]For each $\varepsilon \in (0,e^{-2})$, define the radial cutoff $\eta_\varepsilon: B(0,1) \to [0,1]$ as follows. Set $\eta_\varepsilon(0):=0$; set $\eta_\varepsilon(x):=0$ when $0<|x|\leq\varepsilon$; set $\eta_\varepsilon(x):=\log(|x|/\varepsilon)/\log(\varepsilon^{-1/2})$ when $\varepsilon<|x|<\varepsilon^{1/2}$; and set $\eta_\varepsilon(x):=1$ when $\varepsilon^{1/2}\leq |x|<1$. Let $S^1:=\{\omega\in\mathbb{R}^2:|\omega|=1\}$, let $\mathcal{H}^1$ denote one-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) on $S^1$, and let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $(0,\infty)$. The notation $W^{1,\infty}(B(0,1))$ denotes the [Sobolev space](/page/Sobolev%20Space) of essentially bounded functions on $B(0,1)$ whose distributional first derivatives are essentially bounded. Then $\eta_\varepsilon \in W^{1,\infty}(B(0,1))$, $\eta_\varepsilon=0$ on $B(0,\varepsilon)$, $\eta_\varepsilon=1$ on $B(0,1)\setminus B(0,\varepsilon^{1/2})$, and \begin{align*} |\nabla \eta_\varepsilon(x)| \leq \frac{2}{|x|\log(1/\varepsilon)}\mathbb{1}_{B(0,\varepsilon^{1/2})\setminus B(0,\varepsilon)}(x) \end{align*} for $\mathcal{L}^2$-almost every $x \in B(0,1)$. Therefore, using the polar-coordinate formula for $\mathcal{L}^2$ under $x=r\omega$ with $d\mathcal{L}^2(x)=r\,d\mathcal{L}^1(r)\,d\mathcal{H}^1(\omega)$ on $(0,\infty)\times S^1$, and defining $I_\varepsilon:=\int_{B(0,1)} |\nabla \eta_\varepsilon(x)|^2\,d\mathcal{L}^2(x)$, we get \begin{align*} I_\varepsilon \leq \frac{4}{\log(1/\varepsilon)^2}\int_{S^1}\int_\varepsilon^{\varepsilon^{1/2}} \frac{1}{r^2}r\,d\mathcal{L}^1(r)\,d\mathcal{H}^1(\omega). \end{align*} Evaluating the $S^1$ integral gives \begin{align*} I_\varepsilon \leq \frac{4\mathcal{H}^1(S^1)}{\log(1/\varepsilon)^2}\int_\varepsilon^{\varepsilon^{1/2}} \frac{1}{r}\,d\mathcal{L}^1(r). \end{align*} Since $\int_\varepsilon^{\varepsilon^{1/2}} r^{-1}\,d\mathcal{L}^1(r)=\frac{1}{2}\log(1/\varepsilon)$, we obtain \begin{align*} I_\varepsilon \leq \frac{2\mathcal{H}^1(S^1)}{\log(1/\varepsilon)}. \end{align*} Thus $\|\nabla \eta_\varepsilon\|_{L^2(B(0,1))}\to 0$ as $\varepsilon\downarrow 0$.[/step]

[guided]The cutoff must vanish near the puncture so that we may use the weak equation on $B(0,1)\setminus\{0\}$, but its gradient must also become small enough not to create an error in the limit. A linear radial cutoff would have too much gradient in dimension two. The logarithmic cutoff is chosen because the point $\{0\}$ has zero $W^{1,2}$ capacity in two dimensions. For $\varepsilon \in (0,e^{-2})$, define $\eta_\varepsilon: B(0,1) \to [0,1]$ as follows. Set $\eta_\varepsilon(0):=0$; set $\eta_\varepsilon(x):=0$ when $0<|x|\leq\varepsilon$; set $\eta_\varepsilon(x):=\log(|x|/\varepsilon)/\log(\varepsilon^{-1/2})$ when $\varepsilon<|x|<\varepsilon^{1/2}$; and set $\eta_\varepsilon(x):=1$ when $\varepsilon^{1/2}\leq |x|<1$. Let $S^1:=\{\omega\in\mathbb{R}^2:|\omega|=1\}$, let $\mathcal{H}^1$ denote one-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) on $S^1$, and let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $(0,\infty)$. The notation $W^{1,\infty}(B(0,1))$ denotes the Sobolev space of essentially bounded functions on $B(0,1)$ whose distributional first derivatives are essentially bounded. The function is Lipschitz away from the two circles $|x|=\varepsilon$ and $|x|=\varepsilon^{1/2}$ and belongs to $W^{1,\infty}(B(0,1))$ because the radial pieces match continuously and have bounded derivative for fixed $\varepsilon$. On the annulus $B(0,\varepsilon^{1/2})\setminus B(0,\varepsilon)$, differentiating the radial expression gives \begin{align*} |\nabla \eta_\varepsilon(x)| = \frac{1}{|x|\log(\varepsilon^{-1/2})} = \frac{2}{|x|\log(1/\varepsilon)}. \end{align*} Outside this annulus the cutoff is constant, so its gradient vanishes. Hence \begin{align*} |\nabla \eta_\varepsilon(x)| \leq \frac{2}{|x|\log(1/\varepsilon)}\mathbb{1}_{B(0,\varepsilon^{1/2})\setminus B(0,\varepsilon)}(x) \end{align*} for $\mathcal{L}^2$-almost every $x \in B(0,1)$. Applying the polar-coordinate formula for $\mathcal{L}^2$, with substitution $x=r\omega$ and measure decomposition $d\mathcal{L}^2(x)=r\,d\mathcal{L}^1(r)\,d\mathcal{H}^1(\omega)$, and defining $I_\varepsilon:=\int_{B(0,1)} |\nabla \eta_\varepsilon(x)|^2\,d\mathcal{L}^2(x)$, yields \begin{align*} I_\varepsilon \leq \frac{4}{\log(1/\varepsilon)^2}\int_{S^1}\int_\varepsilon^{\varepsilon^{1/2}} \frac{1}{r^2}r\,d\mathcal{L}^1(r)\,d\mathcal{H}^1(\omega). \end{align*} Evaluating the angular integral gives \begin{align*} I_\varepsilon \leq \frac{4\mathcal{H}^1(S^1)}{\log(1/\varepsilon)^2}\int_\varepsilon^{\varepsilon^{1/2}} \frac{1}{r}\,d\mathcal{L}^1(r). \end{align*} The radial integral is $\frac{1}{2}\log(1/\varepsilon)$, so \begin{align*} I_\varepsilon \leq \frac{2\mathcal{H}^1(S^1)}{\log(1/\varepsilon)}. \end{align*} The right-hand side tends to $0$ as $\varepsilon \downarrow 0$, so $\|\nabla \eta_\varepsilon\|_{L^2(B(0,1))}\to 0$. This computation is the explicit zero-capacity cutoff estimate used in the Sobolev extension step and in the passage of the weak equation through the puncture.[/guided]

[step:Pass the weak harmonic map equation through the puncture]Let $\varphi \in C_c^\infty(B(0,1);\mathbb{R}^m)$ be an arbitrary compactly supported smooth [test function](/page/Test%20Function), where $C_c^\infty(B(0,1);\mathbb{R}^m)$ denotes the space of smooth maps from $B(0,1)$ to $\mathbb{R}^m$ with compact support. Let $\operatorname{supp}\varphi$ denote the closed support of $\varphi$. For each $\varepsilon \in (0,e^{-2})$, the map $\psi_\varepsilon: B(0,1)\setminus\{0\}\to\mathbb{R}^m$ defined by \begin{align*} \psi_\varepsilon(x) := \eta_\varepsilon(x)\varphi(x) \end{align*} has compact support in $B(0,1)\setminus\{0\}$. Since $u$ is a [weakly harmonic map](/page/Harmonic%20Map) on the punctured disk and $\nu$ was defined with the convention $\Delta u+\nu_u(\nabla u,\nabla u)=0$, the extrinsic weak formulation says that every $\mathbb{R}^m$-valued test function compactly supported in the punctured disk satisfies \begin{align*} \int_{B(0,1)} \nabla u(x)\cdot \nabla \psi_\varepsilon(x)\,d\mathcal{L}^2(x) = \int_{B(0,1)} \nu_{u(x)}(\nabla u(x),\nabla u(x))\cdot \psi_\varepsilon(x)\,d\mathcal{L}^2(x). \end{align*} This is the Euclidean form of the intrinsic weak harmonic map equation after the isometric embedding $N\subset\mathbb{R}^m$: the tangential variations give the intrinsic equation, while the normal part is represented by the second fundamental form term. Here the integrals over $B(0,1)$ equal the integrals over the punctured disk because $\psi_\varepsilon$ vanishes near $0$ and $\mathcal{L}^2(\{0\})=0$. We pass to the limit in the left-hand side. Expanding $\nabla(\eta_\varepsilon\varphi)$ gives \begin{align*} \int_{B(0,1)} \nabla u\cdot \nabla \psi_\varepsilon\,d\mathcal{L}^2 &= \int_{B(0,1)} \eta_\varepsilon\nabla u\cdot \nabla\varphi\,d\mathcal{L}^2 + \int_{B(0,1)} \nabla u\cdot (\varphi\otimes \nabla\eta_\varepsilon)\,d\mathcal{L}^2. \end{align*} The first term converges to $\int_{B(0,1)}\nabla u\cdot\nabla\varphi\,d\mathcal{L}^2$ by dominated convergence, using $|\eta_\varepsilon\nabla u\cdot\nabla\varphi|\leq |\nabla u|\,\|\nabla\varphi\|_\infty$ and $\nabla u\in L^2(B(0,1))\subset L^1(\operatorname{supp}\varphi)$. The second term satisfies, by Cauchy-Schwarz, \begin{align*} \left|\int_{B(0,1)} \nabla u\cdot (\varphi\otimes \nabla\eta_\varepsilon)\,d\mathcal{L}^2\right| &\leq \|\varphi\|_\infty\left(\int_{B(0,1)}|\nabla u|^2\,d\mathcal{L}^2\right)^{1/2} \left(\int_{B(0,1)}|\nabla\eta_\varepsilon|^2\,d\mathcal{L}^2\right)^{1/2}, \end{align*} which tends to $0$ by the cutoff estimate. For the right-hand side, define \begin{align*} \|\nu\|_\infty := \sup\{ |\nu_p(v,w)| : p\in N,\ v,w\in T_pN,\ |v|\leq 1,\ |w|\leq 1\}. \end{align*} Compactness of $N$ and smoothness of $\nu$ imply $\|\nu\|_\infty<\infty$. Since $|\nabla u|^2\in L^1(B(0,1))$, $\eta_\varepsilon\varphi\to\varphi$ pointwise and \begin{align*} |\nu_{u(x)}(\nabla u(x),\nabla u(x))\cdot\eta_\varepsilon(x)\varphi(x)| \leq \|\nu\|_\infty\|\varphi\|_\infty |\nabla u(x)|^2, \end{align*} dominated convergence gives \begin{align*} \int_{B(0,1)} \nu_{u}(\nabla u,\nabla u)\cdot \eta_\varepsilon\varphi\,d\mathcal{L}^2 \to \int_{B(0,1)} \nu_{u}(\nabla u,\nabla u)\cdot \varphi\,d\mathcal{L}^2. \end{align*} Therefore \begin{align*} \int_{B(0,1)} \nabla u\cdot\nabla\varphi\,d\mathcal{L}^2 = \int_{B(0,1)} \nu_{u}(\nabla u,\nabla u)\cdot\varphi\,d\mathcal{L}^2 \end{align*} for every $\varphi\in C_c^\infty(B(0,1);\mathbb{R}^m)$. Thus $u$ is weakly harmonic on all of $B(0,1)$.[/step]

[guided]We want to test the weak harmonic map equation with an arbitrary function $\varphi\in C_c^\infty(B(0,1);\mathbb{R}^m)$, where $C_c^\infty(B(0,1);\mathbb{R}^m)$ denotes the space of smooth maps from $B(0,1)$ to $\mathbb{R}^m$ with compact support. Let $\operatorname{supp}\varphi$ denote the closed support of $\varphi$. The original equation is only known on the punctured disk. The cutoff $\eta_\varepsilon$ solves this mismatch: the product test function $\psi_\varepsilon: B(0,1)\setminus\{0\}\to\mathbb{R}^m$ defined by \begin{align*} \psi_\varepsilon(x) := \eta_\varepsilon(x)\varphi(x) \end{align*} has compact support in $B(0,1)\setminus\{0\}$ because $\eta_\varepsilon=0$ on $B(0,\varepsilon)$. Since $u$ is weakly harmonic on the punctured disk and the embedding convention gives the extrinsic equation $\Delta u+\nu_u(\nabla u,\nabla u)=0$, the weak formulation gives \begin{align*} \int_{B(0,1)} \nabla u(x)\cdot \nabla \psi_\varepsilon(x)\,d\mathcal{L}^2(x) = \int_{B(0,1)} \nu_{u(x)}(\nabla u(x),\nabla u(x))\cdot \psi_\varepsilon(x)\,d\mathcal{L}^2(x). \end{align*} The integrals may be written over $B(0,1)$ because $\psi_\varepsilon$ vanishes near $0$ and $\mathcal{L}^2(\{0\})=0$. We now pass to the limit. Expanding the gradient of the product gives \begin{align*} \int_{B(0,1)} \nabla u\cdot \nabla \psi_\varepsilon\,d\mathcal{L}^2 &= \int_{B(0,1)} \eta_\varepsilon\nabla u\cdot \nabla\varphi\,d\mathcal{L}^2 + \int_{B(0,1)} \nabla u\cdot (\varphi\otimes \nabla\eta_\varepsilon)\,d\mathcal{L}^2. \end{align*} The first integral converges to $\int_{B(0,1)}\nabla u\cdot\nabla\varphi\,d\mathcal{L}^2$ by dominated convergence: $\eta_\varepsilon\to 1$ pointwise away from $0$, the point $0$ is $\mathcal{L}^2$-null, and the integrand is dominated by $|\nabla u|\,\|\nabla\varphi\|_\infty$, which is integrable on $\operatorname{supp}\varphi$ because $\nabla u\in L^2(B(0,1))$ and $\operatorname{supp}\varphi$ has finite $\mathcal{L}^2$ measure. For the cutoff-error term, Cauchy-Schwarz gives \begin{align*} \left|\int_{B(0,1)} \nabla u\cdot (\varphi\otimes \nabla\eta_\varepsilon)\,d\mathcal{L}^2\right| &\leq \|\varphi\|_\infty\left(\int_{B(0,1)}|\nabla u|^2\,d\mathcal{L}^2\right)^{1/2} \left(\int_{B(0,1)}|\nabla\eta_\varepsilon|^2\,d\mathcal{L}^2\right)^{1/2}. \end{align*} The first factor is finite by the finite-energy hypothesis and the last factor tends to $0$ by the logarithmic cutoff estimate. Hence this error term vanishes. For the nonlinear term, define \begin{align*} \|\nu\|_\infty := \sup\{ |\nu_p(v,w)| : p\in N,\ v,w\in T_pN,\ |v|\leq 1,\ |w|\leq 1\}. \end{align*} Because $N$ is compact and $\nu$ is smooth, this supremum is finite. Therefore \begin{align*} |\nu_{u(x)}(\nabla u(x),\nabla u(x))\cdot\eta_\varepsilon(x)\varphi(x)| \leq \|\nu\|_\infty\|\varphi\|_\infty |\nabla u(x)|^2. \end{align*} The dominating function on the right belongs to $L^1(B(0,1))$, and $\eta_\varepsilon\varphi\to\varphi$ pointwise. Dominated convergence gives \begin{align*} \int_{B(0,1)} \nu_{u}(\nabla u,\nabla u)\cdot \eta_\varepsilon\varphi\,d\mathcal{L}^2 \to \int_{B(0,1)} \nu_{u}(\nabla u,\nabla u)\cdot \varphi\,d\mathcal{L}^2. \end{align*} Taking limits in the weak equation yields \begin{align*} \int_{B(0,1)} \nabla u\cdot\nabla\varphi\,d\mathcal{L}^2 = \int_{B(0,1)} \nu_{u}(\nabla u,\nabla u)\cdot\varphi\,d\mathcal{L}^2. \end{align*} Because $\varphi\in C_c^\infty(B(0,1);\mathbb{R}^m)$ was arbitrary, this is exactly the weak harmonic map equation on the whole disk.[/guided]

[step:Apply interior regularity to obtain a smooth harmonic extension] The extended map satisfies $u\in W^{1,2}(B(0,1);N)$ and is weakly harmonic on $B(0,1)$ in the extrinsic sense. The target $N$ is compact and smooth, so Hélein's interior [regularity theorem](/theorems/2750) for two-dimensional weakly harmonic maps into compact targets, stated in the [regularity theory for harmonic maps](/page/Harmonic%20Map), applies: every weakly harmonic $W^{1,2}$ map from a two-dimensional domain into such a target is smooth and satisfies the harmonic map equation classically. Hence $u$ has a representative, still denoted $u$, which is a smooth harmonic map $u:B(0,1)\to N$. This representative agrees with the original smooth harmonic map on $B(0,1)\setminus\{0\}$ because Sobolev representatives agree $\mathcal{L}^2$-almost everywhere there and both are smooth solutions of the same elliptic system. Therefore the original finite-energy punctured harmonic map extends to a smooth harmonic map on the whole disk. [/step]