I am asked to sketch the proof seen below in a graph.
We state that the function $f: \mathbb{R} \to \mathbb{R}$ given by
$f(x)=e^{-x^2}$
Is uniformly continuous
Proof: \begin{eqnarray*} |f(x)|<\epsilon/2&\Longleftrightarrow& e^{-x^2}<\epsilon/2\\&\Longleftrightarrow& \log\left(e^{-x^2}\right)<\log (\epsilon/ 2)\\&\Longleftrightarrow& -x^2<\log \epsilon-\log 2\\&\Longleftrightarrow& x^2>\log 2-\log \epsilon\\&\Longleftrightarrow& |x|>\sqrt{\log 2-\log \epsilon} \end{eqnarray*}
By setting $K=\sqrt{\log 2-\log \epsilon}$ we get $|x|,|y|>K\Longrightarrow |f(x)-f(y)|\leq |f(x)|+|f(y)|<\frac\epsilon2+\frac\epsilon2=\epsilon.$ (1)
We can now use our third theoreom ($F: A->\mathbb{R}$ on the set $A \subset \mathbb{R}^{k}$ is uniformly continuous if $\forall\epsilon>0 \exists\delta>0: ||f(y)-f(x)||<\epsilon$ for all $x,y\in$A with $||y-x||<\delta$ on f in the interval $[-K-1; K+1]$ and conclude that f is uniformly continuous on this interval. We can therefore chose a $\delta'>0$ with the proberty
$|f(x)-f(y)|<\epsilon,$ when $|x-y|<\delta'$ and $x,y\in I$ (2)
By setting $\delta=min({\delta',1})$ we can parry off (not sure if this is the right term at all...) the chosen $\epsilon$ for all $x,y\in \mathbb{R}$. Set $|x-y|<\delta$ - if both x,y $\in$ I, then (2) says that $|f(x)-f(y)|<\epsilon$.
If $x\notin I$ we have $x>K+1$ or $x<-K-1$. If $x>K-1$ we get that $y>K$ because of $|x-y|<1$ whereas (1) gives that $|f(x)-f(y)|<\epsilon$. And if $x<-K-1$ we have that $y<-K$, which gives us that (1) is $|f(x)-f(y)|<\epsilon$
I am having trouble understanding the proof completely - I got the main idea, that something is uniformly continuous if the function behaves nicely. However, sketching this is giving me loads of troubles. Can anyone help out? We have to sketch it with the values $\epsilon,\delta,K$ on the figgure.
Your proof is not yet printable, but basically correct. The main thing I missed is an enouncement of "our third theorem", namely that a function which is continuous on a compact set $C$ is uniformly continuous on $C$.
Some minor improvements: It is not necessary to exhibit an explicit $K$ in terms of square roots and logs. Just use that $f$ is continuous on all of ${\Bbb R}$ and that $\lim_{|x|\to\infty} f(x)=0$. So you can say:
Given an $\epsilon>0$ there is a $K>0$ with $$\bigl|f(x)\bigr|<{\epsilon\over2}\qquad (|x|>K)\ .\tag{1}$$ By "our third theorem" the function $f$ is uniformly continuous on the compact interval $I:=[-K-1,K+1]$. This implies that there is a $\delta'>0$ with $$x, \> y\in I, \quad |x-y|<\delta'\qquad\Rightarrow\qquad |f(x)-f(y)|<\epsilon\ .\tag{2}$$
I claim that $\delta:=\min\{\delta',1\}$ does the job for the given $\epsilon$. Proof: Given two real numbers $x$, $y$ with $|x-y|<\delta$ and one of them, say $x$, is not in $I$ then $|x|> K+1$, and it then follows that $|y|> K$ as well. This implies $|f(x)-f(y)|<\epsilon$ by $(1)$. If both $x$ and $y$ are in $I$ then $|f(x)-f(y)|<\epsilon$ by $(2)$.