Monotonicity formula implies that the tangent map of harmonic map at singular point is homogeneous of degree zero

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It is Proposition 2.3.2 in page 27 in Lin's book, The analysis of harmonic maps and their heat flows.

Suppose $\Omega \subset \mathbb{R}^n$. Let $u \in H^1(\Omega ;\mathbb{R}^m)$ be a minimizing harmonic map and $0 \in \Omega$ is a singular point of $u$. Let $r_i$ be a sequence decreasing to $0$. We suppose $u_i(y) = u(r_iy)$ strongly converges to $\phi(y)$ in $H^1_{loc}(\mathbb{R}^n;\mathbb{R}^m)$ and is a minimizing harmonic map.

If we have the monotonicity formula for $u$, that is, for any $0<r<R$ and $B_R(x) \subset \Omega$ $$\dfrac{1}{R^{n-2}}\int_{B_R(x)}|\nabla u|^2 - \dfrac{1}{r^{n-2}}\int_{B_r(x)}|\nabla u|^2 \geq \int_{B_R(x) \setminus B_r(x)} \rho^{2-n} \left| \dfrac{\partial u}{\partial \rho} \right|^2.$$ Here $\rho=|x|$. Then how can we conclude that $\phi$ is homogeneous of degree zero? More precisely, $\phi(y) = \phi\left(\dfrac{y}{|y|}\right)$.

If i put $R=r_{i+1}$ and $r=r_i$, we have $$\int_{B_1(0)}|\nabla_y u_{i+1}|^2 - \int_{B_1(0)}|\nabla_y u_i|^2 \geq \int_{B_{r_{i+1}}(0) \setminus B_{r_i}(0)} \rho^{2-n} \left| \dfrac{\partial u}{\partial \rho} \right|^2.$$ Then what can we do next? Thanks!