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Study the uniform continuity of these functions:

  1. $f(x,y)=(\sin(2x-y), \cos(x-2y))$ in $\mathbb{R^2}$
  2. $f(x,y)=\frac{x^2+y^2}{x+y}$ if $x+y \neq 0$ and $f(x,y)=0$ if $x+y=0$ in $M=(0,1)\times(0,1)$

I think that the first part only need to prove that $\sin(x), \cos(x), 2x-y$ and $x-2y$ are uniformly continues. But I don't know how to do this.

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How can $x+y = 0$ in $M$? – copper.hat Dec 20 '12 at 6:31
up vote 1 down vote accepted

The mean value theorem for a real valued $C^1$ function gives $|\phi(x_1)-\phi(x_0)| \leq \max_{\xi \in [x_0,x_1]} |\phi'(\xi)|\|x_1-x_0\|$, so if $|\phi'(x)|$ is bounded by $L$ on some convex set, then $|\phi(x_1)-\phi(x_0)| \leq L \|x_1-x_0\|$ on this set and hence is uniformly continuous (given $\epsilon>0$, choose $\delta = \frac{\epsilon}{L}$).

If $\phi$ is a vector-valued function (ie, $\phi(x) = (\phi_1(x),...,\phi_n(x))^T$), and each of the derivatives $\phi_i'$ is bounded by some $L$ on some convex set, then $\|\phi(x_1)-\phi(x_0)\| \le \sqrt{\sum_i(\phi_i(x_1)-\phi_i(x_0))^2} \leq \sqrt{\sum_i L^2 \|x_1-x_0\|^2} = \sqrt{n} L \|x_1-x_0\|$ on this set, and hence is uniformly continuous.

For 1), it is easy to see that each element of $DF((x,y))$ is bounded by $2$, hence $f$ is uniformly continuous on $\mathbb{R}^2$.

For 2), $Df((x,y)) = \begin{bmatrix} \frac{y^2-2xy-x^2}{(x+y)^2} & \frac{y^2+2xy-x^2}{(x+y)^2} \end{bmatrix}$. For $(x,y) \in M$, $|y^2-2xy-x^2| \leq (x+y)^2$, and $|y^2+2xy-x^2| \leq (x+y)^2$, hence each entry is bounded by $1$. Hence $f$ is uniformly continuous on $M$.

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