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Let $f:mathbb R rightarrow mathbb R$ be continuous and compactly supported. How can I prove that $f$ is uniformly continuous ? I was trying to prove it by contradiction but get stuck. My attempt was as follows:

Let $E$ be the compact support of $f$. On $E$ we know that $f$ is uniformly continuous. If we assume that $f$ is not uniformly continuous we know
$$
exists epsilon > 0 forall delta > 0 exists x,y in mathbb R: |x-y|<delta wedge |f(x)-f(y)| geq epsilon
$$ Let $delta_n := frac 1n$ and fix this $epsilon > 0$. Compute a $delta > 0$ s.t. $forall x,y in E:|x-y| < delta rightarrow |f(x)-f(y)| <epsilon$. Let $n geq N$ s.t. $frac 1N < delta$. Then we may assume wlog that $x_n in E$ and $y_n in mathbb R setminus E$ wehere $x_n,y_n$ are the points corresponding with $delta_n$. This gives a sequence of points where $|x_n-y_n| rightarrow 0$ and $x_n in E$ and $y_n in mathbb R setminus E$.

I now want to use somehow the continuitiy of $f$. How can I do this ? Is this approach a good one ? Can it be more simple ?

New idea:
I know that $E$ is compact so $(y_n)_{n=0}^infty$ has a convergent subsequence $(y_{n_j})_{j=0}^infty$ whit limit say $y in E$. Now
$$
|x_{n_j}-y| leq |x_{n_j}-y_{n_j}|+|y_{n_j}+y| rightarrow 0
$$ So I can take $x_{n_j}$ close to $y$ to get $|f(x_{n_j})-f(y)| = |f(y)| <frac epsilon 2$ by the continuity. I can also take $y_{n_j}$ close to $y$ to get $|f(y_{n_j})-f(y)| <frac epsilon 2$. We further have
$$
|f(y_{n_j})| leq |f(y_{n_j})-f(y)| + |f(y)|
$$ s.t.
$$
|f(y)| geq |f(y_{n_j})| - |f(y_{n_j})-f(y)| geq frac epsilon 2
$$ because $|f(y_{n_j})| geq epsilon$ per construction. So we have $|f(y)| leq frac epsilon 2$ and $|f(y)| > frac epsilon 2$ which is a contradiction.

New solution: Let $epsilon > 0$. Let $delta_1$ for the uniform continuity on $E$. Further
$$
forall x in E, exists delta_x > 0 forall y in mathbb{R}: |x-y|< delta_x rightarrow |f(x)-f(y)| < epsilon
$$ Compute an open finite over of $E$
$$
E subseteq bigcup_{i=1}^N Bleft(x_i,frac{delta_{x_i}}2right)
$$ Write $delta_i := delta_{x_i}$. Let $delta_2 := min_{i=1,cdots,N} frac {delta_i} 2$ and $delta := min(delta_1,delta_2)$.

Assume $|x-y|< delta$. If $x,y in E$ or $x,y notin E$ we are done. Otherwise assume $x in E$ and $y notin E$. Then $x in Bleft(x_i,frac{delta_i}{2}right)$ for some $i$. Further
$$
|y-x_i| leq |y-x|+|x-x_i|leq delta_2 + frac{delta_i}2 leq delta_i
$$ thus $y in B(x_i,delta_i)$ which proves the claim.

Let us first see what is continuity. Given a positive $epsilon$, continuity of $f$ means that at each $x$, we can find a small $delta$ such that $x'in (x-delta,x+delta)$ implies $f(x')in (f(x)-epsilon,f(x)+epsilon)$. So around each $x$, we have a small open set that is mapped very close to $f(x)$. Also note that points in the same small open set are mapped to points close to each other, since they are both mapped to points close to $f(x)$. But the problem here is that this radius of the open sets $delta$ depends on the point $x$.

Now compactness means you can find a finite collection of such small open sets that they cover the support of the function, and this is good enough for us. Because among these finitely many open sets, you can determine the smallest radius, say $r$. If the distance between $x$ and $x'$ are less than $1/10r$, then you know they must lie in the same open set, then $f(x)$ and $f(x')$ must be close. Now this $r$ does not depend on $x$, you have uniform continuity.

Let $K$ be the support of $f$. Define
$$ M: K times [0,1]to {mathbb R}$$
by letting
$$M(x,delta) = max_{|y-x|le delta} |f(y)-f(x)|.$$
Observe
$$ |M(x,delta)- M(x',delta')| le |M(x,delta) - M(x,delta')| + |M(x,delta')-M(x',delta')|,$$
and a quick calculation shows that $M$ is continous. Let ${bar M}(delta) = max_{x} M(x,delta)=M(x_delta,delta)$, due to compactness. Note that $delta to{bar M}(delta)$ is nondecreasing. We show it tends to $0$ as $deltasearrow 0$. If $delta_n searrow 0$ we have a subsequence $x_{delta_{n_k}}to x_0in K$ and therefore by continuity, ${bar M}(delta_{n_k}) = M(x_{delta_{n_k}},delta_{n_k}) to M(x_0,0)=0$.