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Let $S = [0,1]\times [0,1].$ For $(x,y) \in S,$ define $f(x,y) = x^2 + 3y.$ The problem is to show $f$ is uniformly continuous on $S.$

Let $s > 0$. Suppose $(x,y),(x_0,y_0) \in S$ and $\lvert x - x_0 \rvert < s, \lvert y - y_0 \rvert < s.$ Then

$$\begin{align} |f(x,y) - f(x_0,y_0)| &\leq \lvert x - x_0\rvert \lvert x + x_0\rvert + 3\lvert y - y_0\rvert \\ &\leq (4 + 2\lvert x_0\rvert)s. \end{align}$$ Let $e$ be epsilon in which I'm trying to create a neighbourhood around $f(x,y) - f(x_0,y_0) $ such that $f(x,y)$ is continuous. I know that since I found a variation or an upper bound for $f(x,y)$ in regards of $x_0$ and $s$ let me find epsilon $e$ by

\begin{align} 4 + 2 \lvert x_0\rvert s \leq e, \end{align}

so I end up with

\begin{align} s < \frac{e}{4 + 2\lvert x_0\rvert}, \end{align}

in which $s$ is dependent on both $e$ and $x_0$. It seems like it is not uniformly convergent, however I can choose a very very small $s$ so that it is satisfied for all $x_0$ (since as $x_0$ increases $s$ decreases from the equation obtained) Can we say that $f$ is uniformly continuous in this case?

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  • $\begingroup$ The first sentence makes no sense, which expression defines $f$? $\endgroup$ – copper.hat Oct 17 '16 at 14:17
  • $\begingroup$ Fixed sorry about that $\endgroup$ – Xenidia Oct 17 '16 at 14:24
  • $\begingroup$ It still makes no sense. Read the comment. $\endgroup$ – copper.hat Oct 17 '16 at 14:26
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    $\begingroup$ Since $f$ is continuous and $S$ is compact, then we can show that $f$ is uniformly continuous on $S$. Alternatively, you could show that the derivative is bounded and use the mean value theorem. $\endgroup$ – copper.hat Oct 17 '16 at 14:58
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    $\begingroup$ The question is sloppily written. I still don't know which of the two $f$s you intend. $\endgroup$ – copper.hat Oct 17 '16 at 15:48
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since $x\le1$ $$s<\frac e 6\le\frac{e}{4+2|x_0|}$$ So s doesn't really depend on $x_0$, meaning $f$ is uniformly continuous

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You're definitely on the right track. But when you write

$$\begin{align} f(x,y) - f(x_0,y_0) &\leq \lvert x - x_0\rvert \lvert x + x_0\rvert + 3\lvert y - y_0\rvert \\ &\leq (4 + 2\lvert x_0\rvert)s. \end{align}$$

it's a little strange. First, you need absolute values on the left. We then have

$$ |f(x,y) - f(x_0,y_0)| \le |x+x_0||x-x_0|+ 3|y-y_0| \le (|x+x_0| + 3)s.$$

For some reason you must be thinking $|x+x_0|\le 1 + 2|x_0|.$ This is true, but weird. Why not just use $|x+x_0| \le 1 + 1 = 2?$ Then we get $|f(x,y) - f(x_0,y_0)| \le 5s.$ You're then ready to go. Let $\epsilon>0.$ Choose $s<\epsilon/5.$ With this $s$ you'll get $|f(x,y) - f(x_0,y_0)| < \epsilon$ as desired.

Your question needs some rewriting. You seem to have ignored @copper.hat's comment. When you do that, people get a bit frustrated and sometimes move on.

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