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Consider $f(x,y) = x$ We must have: \[ D = \left( \frac{\|f(\vec{u}) - f(\vec{v}) - L(\vec{u}-\vec{v})\|}{\|\vec{u}-\vec{v}\|}\right)^2 = \frac{\|(u_1 - v_1, 0) - (L_1, L_2)\cdot (u_1 - v_1, u_2 - v_2)\|^2}{\|\vec{u}-\vec{v}\|^2} \to 0 \] \[ D = \frac{(u_1-v_1 - L_1(u_1-v_1))^2 + (L_2(u_2-v_2))^2}{(u_1-v_1)^2 + (u_2-v_2)^2} = \frac{(u_1-v_1 - L_1(u_1-v_1))^2}{(u_1-v_1)^2 + (u_2-v_2)^2} + \frac{(L_2(u_2-v_2))^2}{(u_1-v_1)^2 + (u_2-v_2)^2} \] Both terms are nonnegative, so they must approach 0 individually. \[ \frac{(u_1-v_1 - L_1(u_1-v_1))^2}{(u_1-v_1)^2 + (u_2-v_2)^2} \to 0 \implies L_1 = 1 \] \[ \frac{(L_2(u_2-v_2))^2}{(u_1-v_1)^2 + (u_2-v_2)^2} \to 0 \implies L_2 = 0 \] Thus $L = (1,0)$.
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Fri, 17 Aug 2018 01:01 GMT