Why is this function uniformly continuous?
$f:\mathbb{R}^2 \rightarrow\mathbb{R}$,
$f(x,y)=e^{-(x^2+y^2)}$.
I tried using the mean value theorem for several variables, but I am not able to complete it. Here is my attempt:
If we have two points in $\mathbb{R}^2$, a and b. Then:
$|f(\textbf{a})-f(\textbf{b})|=|\triangledown f(\textbf{c}) \cdot(\textbf{b}-\textbf{a})|$, where c is a point on the line between a and b.
I get that the gradient is:
$-2*e^{-(c_1^2+c_2^2)}[c_1,c_2]$. I get that the fight HS then is:
$|\triangledown f(\textbf{c}) \cdot(\textbf{b}-\textbf{a})|=|-2e^{-(c_1^2+c_2^2)}[c_1,c_2]\cdot(\textbf{b}-\textbf{a})|\le |2[c_1,c_2]\cdot(\textbf{b}-\textbf{a})|$.
But how do I proceed? I need the c's to disappear and to have $|\textbf{b}-\textbf{a}|$,ont the right side. If I could get something with $(\textbf{b}-\textbf{a})\cdot(\textbf{b}-\textbf{a})$ that would be ok, but I don't see how to get it?
Daniel Fischer offered a fix for the proof in the comments: instead of the wasteful $$e^{-c_1^2-c_2^2}\sqrt{c_1^2+c_2^2}\le \sqrt{c_1^2+c_2^2}$$ is the estimate $$e^{-c_1^2-c_2^2}\sqrt{c_1^2+c_2^2}\le \sup_{r>0} re^{-r}=e^{-1}$$ (since the supremum is attained at $r=1$).
For completeness, a more general result: if $f$ is continuous on $\mathbb R^n$ and tends to $0$ as $|x|\to\infty$, then $f$ is uniformly continuous. This can be proved by contradiction: suppose there is $\epsilon>0$ and two sequences $x_n,y_n$ such that $|x_n-y_n|\to 0$ and $|f(x_n)-f(y_n)|\ge \epsilon$. Then these sequences must be bounded (else $f$ would be too small), so we extract a convergent subsequence $x_n\to a$ and get a contradiction with the continuity of $f$ at $a$.