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Let $\mathbb{R}^2$ be equipped with the norm \begin{align*} \|x\|=\max\{|x_1|,|x_2|\},\quad x=(x_1,x_2)\in\mathbb{R}^2 \end{align*} Let $A:\mathbb{R}^2\to\mathbb{R}$ be given by $A(x_1,x_2)=|x_1|+|x_2|$. Show that $A$ is uniformly continuous w.r.t. the norm defined above.

Here is how I went so far, Let $A:\mathbb{R}^2\to\mathbb{R}$ be the linear map $A(x_1,x_2)=|x_1|+|x_2|$. We must show that given any $\epsilon>0$ we can find $\delta>0$ such that \begin{align*} \|\mathbf{x}-\mathbf{y}\|<\delta\,\,\Longrightarrow\,\,|A(\mathbf{x})-A(\mathbf{y})|<\epsilon \end{align*} in other words, \begin{align*} |x_1-y_1|+|x_2-y_2|<\delta\,\,\Longrightarrow\,\,\left.|A(y_1,y_2)-A(x_1,x_2)\right.| \end{align*} If $\mathbf{x}$ and $\mathbf{y}$ are any two points in $\mathbb{R}^2$, then \begin{align*} \left.|A(x_1,x_2)-A(y_1,y_2)\right.|&=\left.||x_1|+|x_2|-(|y_1|+|y_2|)\right.|\\ &=\bigl\lvert|x_1|-|y_1|+|x_2|-|y_2|\bigr\rvert\\ &\le \bigl\lvert|x_1|-|y_1|\bigr\rvert+\bigl\lvert|x_2|-|y_2|\bigr\rvert\\ &\le \bigl\lvert|x_1-y_1|\bigr\rvert+\bigl\lvert|x_2-y_2|\bigr\rvert\\ &\le |x_1-y_1|+|x_2-y_2|<\epsilon\quad\text{iff}\quad\delta=??\epsilon \end{align*}

Now I am lost, I am not sure how to use the "max"... Please help...

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  • $\begingroup$ \begin{align*} \|x\|=\max\{|x_1|,|x_2|\},\quad x=(x_1,x_2)\in\mathbb{R}^2 \end{align*} why have you used $|x_1-y_1|+|x_2-y_2|<\delta$ $\endgroup$ – R.N Oct 8 '15 at 12:08
  • $\begingroup$ Are $|x_1 -y_1|$ and $|x_2 -y_2|$ smaller or bigger than $||\bf{x} - \bf{y}||$? $\endgroup$ – s.harp Oct 8 '15 at 12:08
  • $\begingroup$ $\delta\leq \frac{\epsilon}{2}$ is good choice. but you should write \begin{align*} \left.|A(x_1,x_2)-A(y_1,y_2)\right.|&=\left.||x_1|+|x_2|-(|y_1|+|y_2|)\right.|\\ &=\bigl\lvert|x_1|-|y_1|+|x_2|-|y_2|\bigr\rvert\\ &\le \bigl\lvert|x_1|-|y_1|\bigr\rvert+\bigl\lvert|x_2|-|y_2|\bigr\rvert\\ &\le \bigl\lvert|x_1-y_1|\bigr\rvert+\bigl\lvert|x_2-y_2|\bigr\rvert\\ &\le |x_1-y_1|+|x_2-y_2|<2\delta<\epsilon\quad\text{iff}\quad\delta=??\epsilon \end{align*} $\endgroup$ – R.N Oct 8 '15 at 12:14
  • $\begingroup$ It may be helpful to note that if $a$ and $b$ are nonnegative then $\max(a,b)\le a+b\le2\max(a,b)$. $\endgroup$ – John Dawkins Oct 8 '15 at 13:07
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Let $\varepsilon > 0$; if $x_{1},x_{2},x_{1}',x_{2}' \in \mathbb{R}$, then $$ \big| |x_{1}| + |x_{2}| - |x'_{1}| - |x'_{2}| \big| \leq \big| |x_{1}| - |x_{1}'| \big| + \big| |x_{2}| - |x'_{2}| \big| \leq |x_{1}-x_{1}'| + |x_{2} - x'_{2}| \leq\\ 2\max \{ |x_{1}-x_{1}'|, |x_{2}-x_{2}'| \} = 2|| (x_{1},x_{2}) - (x_{1}', x_{2}') ||, $$ which is $< \varepsilon$ if in addition we have $|| (x_{1},x_{2}) - (x_{1}',x_{2}') || < \varepsilon/2$.

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