If $f:A\to \Bbb R^n$ is a continuous function whose support is compact then is it uniformly continuous?

Question

So let be $f$ a continuous real valued function defined in a open set $A$ of $\Bbb R^m$ and we suppose that its support $S$ is compact so that by a well know theorem we can state that $f$ is uniformly continuous on $S$. Moreover $f$ is zero outside $S$ and so surely uniformly continuous in $A\setminus S$ but unfortunately I did not able to conclude anything about the uniform continuity of $f$ in $A$. Anyway I here find a proof but unfortunately I did not understand it very well: in particular in the @md5's proof I did not understand why if $f$ is not uniformly continous there must exist two sequences $x_n\in S$ and $y_n\in A\setminus S$ such that $$ d(x_n,y_n)<2^{-n}\,\,\,\text{and}\,\,\,|f(x_n)-f(y_n)|>\epsilon $$ and moreover I do not really understand why $y_n$ converges to $x$ and finally why $f$ would not be continuous? So could someone help me, please?

Answer 1

I did not understand why if $f$ is not uniformly continuous there must exist two sequences $x_n\in S$ and $y_n\in A\setminus S$ such that $$ d(x_n,y_n)<2^{-n}\,\,\,\text{and}\,\,\,|f(x_n)-f(y_n)|>\epsilon $$

This is by negating the definition of uniform continuity. Uniform continuity says that for every $\epsilon > 0$, there exists $\delta > 0$ such that $|f(x_n) - f(y_n)| < \epsilon$ whenever $d(x, y) < \delta$.

Thus, if $f$ is not uniformly continuous, there exists some $\epsilon > 0$ for which no $\delta > 0$ does the job. So, fix any such $\epsilon$ and taking different values of $\delta$. More specifically, take $\delta = 2^{-n}$. Then, for each such $n$, there exists $x_n, y_n \in A$ such $d(x_n, y_n) < 2^{-n}$ and $|f(x_n) - f(y_n)| \geqslant \epsilon$.
However, note that $f$ is uniformly continuous when restricted to $S$ and when restricted to $A \setminus S$. Thus, we can conclude that we can find $x_n \in S$ and $y_n \in S \setminus A$ such that the above happens.

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and moreover I do not really understand why $y_n$ converges to $x$

They don't say that $(y_n)$ converges to $x$. There is a subsequence $(x_{\sigma(n)})$ which converges to $x$ and they are saying that the corresponding subsequence of $(y_n)$ will converge to $x$. That is, $$\lim_{n \to \infty} y_{\sigma(n)} = x.$$

This follows quite easily from the fact that $d(x_n, y_n) < 2^{-n}$. By the triangle inequality, we have $$d(x, y_{\sigma(n)}) \leqslant d(x, x_{\sigma(n)}) + d(x_{\sigma(n)}, y_{\sigma(n)}).$$

Let $n \to \infty$ above and use the fact that $d(x, x_{\sigma(n)}) \to 0$ and $2^{-\sigma(n)} \to 0$.

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and finally why $f$ would not be continuous?

You have two sequences $(x_{\sigma(n)})$ and $(y_{\sigma(n)})$ which converge to the same point $x$. However, $|f(x_{\sigma(n)}) - f(y_{\sigma(n)})| \geqslant \epsilon$ for all $n$. In particular, the difference $f(x_{\sigma(n)}) - f(y_{\sigma(n)})$ does not converge to $0$. However, if $f$ were continuous, then we would have $$\lim_{n \to \infty}[f(x_{\sigma(n)}) - f(y_{\sigma(n)})] = f(x) - f(x) = 0.$$

Answer 2

So if $f$ not uniformly continuous that for any $\epsilon>0$ there not exist $\delta>0$ such that holds the implication $$ d(x,y)<\delta\Rightarrow |f(x)-f(y)|<\epsilon $$ and so for this such $\epsilon$ there must exist $x,y\in A$ such that $$ d(x,y)<\delta\,\,\,\text{and}\,\,\,|f(x)-f(y)|\ge\epsilon $$ and in particular it must be $x\in S$ and $y\in A\setminus S$ because the function $f$ is uniformly continuous in this two sets. So putting $$ \delta_n:=\frac{1}{2^n} $$ for any $n\in\Bbb N$ there exist two sequences $(x_n)_{n\in\Bbb N}\in S$ and $(y_n)_{n\in\Bbb N}\in A\setminus S$ such that $$ d(x_n,y_n)<\delta_n\,\,\,\text{and}\,\,\,|f(x_n)-f(y_n)|\ge\epsilon $$ for any $n\in\Bbb N$. Now $S$ is closed and so the sequence $(x_n)_{n\in\Bbb N}$ converges to $x\in S$ and thus the sequence $(y_n)_{n\in\Bbb N}$ too as we show to follow. So let be $$ m:=\min\Big\{n\in\Bbb N:\varepsilon-\frac{1}{2^n}>0\Big\} $$ for any $\varepsilon>0$; now $x_n\rightarrow x$ then $$ d(x,x_n)<\varepsilon-\frac{1}{2^m} $$ for any $n\ge n_0$ and so for any such $n$ it must be $$ d(x,y_n)\le d(x,x_n)+d(x_n,y_n)<\frac{1}{2^n}+\Big(\varepsilon-\frac{1}{2^m}\Big)\le\frac{1}{2^m}+\Big(\varepsilon-\frac{1}{2^m}\Big)=\varepsilon $$ so that the statement follows. Finally we know that the function $f$ is continuous so that there must exsist $n_1$ and $n_2$ such that $$ |f(x)-f(x_n)|<\frac{\epsilon}2\,\,\,\text{and}\,\,\,|f(x)-f(y_n)|<\frac{\epsilon}2 $$ for any $n\ge\max\{n_1,n_2\}$ so that in this case it would be $$ \epsilon\le|f(x_n)-f(y_n)|\le|f(x_n)-f(x)|+|f(x)-f(y_n)|<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon $$ and clearly this is impossible.