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I came across the problem which says:

The map $L\colon \mathbb R^{2}\rightarrow \mathbb R^{2}$ given by $L(x,y)= (x,-y)$ is
(a) differentiable everywhere in $\mathbb R^{2}$,
(b) differentiable only at $(0,0)$,
(c) $DL(0,0)=L$,
(d) $DL(x,y)=L$ for all $(x,y) \in \mathbb R^2$.

I do not know how to approach the problem.Any kind of hints will be helpful. Thanks everyone in advance for your time .

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    In g$e$neral an approach could be to treat each question in turn, and answer by yes or no. Hint: this ia a linear map.2012-11-21

2 Answers 2

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The correct answer is (a), since $\frac{\partial L}{\partial x}=(1, 0)$, $\frac{\partial L}{\partial y}=(0,-1)$ wich are continuous in all $\mathbb{R}^2.$

Another way is the follow, we can write $L(x, y)=(L_1(x,y), L_2(x,y))$, where $L_1(x,y)=x, L_2(x,y)=-y$, it's clear that $L_1, L_2$ are differentiabl scalar field, in fact $DL_1(x,y)=(1,0), DL_2(x,y)=(0,-1)$ therefore $L$ is differentiable and $DL(x,y)=\left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right). $

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    thank you sir. But what about options (c) and (d)? Since {(1,0),(0,1)} is the usual basis of R^{2},we see L(1,0)=(1,0),L(0,-1)=(0,-1).So the matrix representation of L with respect to the usual basis is given by \begin{pmatrix} 1 &0 \\ 0 & -1 \end{pmatrix} which equals to DL(x,y). so options (c) and (d) should be correct. Am i right,sir?2012-11-22
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Hint: For a map $f \colon \mathbb R^2 \to \mathbb R^2$ to be differentiable at $(x,y)$ you need to find a linear map $Df(x,y) \colon \mathbb R^2 \to \mathbb R^2$ such that $ \frac{f(x+ h, y+k) - f(x,y) - Df(x,y)(h,k)}{\|(h,k)\|} \to 0, \qquad (h,k) \to 0 $ Your $f = L$ is linear. Can you spot a map $Df(x,y)$ that makes the numerator "small"? Compute $L(x+h, y+k) - L(x,y)$ and try to identify something that is linear in $h$ and $k$.