Give an example of smooth function $f:\mathbb{R}\to\mathbb{R}$ such that
supp $f=[0,1]$ and $|f'(x)|\leq1$ for all $x$.
Give an example of smooth function $f:\mathbb{R}\to\mathbb{R}$ such that supp $f=[0,1]$ and $|f'(x)|\leq1$ for all $x$.
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real-analysis
differential-topology
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0What does supp $f=[0,1]$ mean? – 2017-02-11
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2@Vikrant Desai The support of $f$ is the closure of the set $\{x : f(x) \neq 0\}$. – 2017-02-11
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0Am I missing something? or a function that only "lifts a little" from the x axis in $[0,1]$ does the job... – 2017-02-11
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0Try something of the form $e^{-1/t^2}$ and paste it with itself (I don't know if this makes sense to you as to me). – 2017-02-11
2 Answers
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Take your favorite smooth function $g$ which is supported on $[0,1]$.
$|g'|$, being continuous on $[0,1]$, will be bounded. Let $M$ be such a bound. Consider now $f:=\frac{g}{M}$.
If you want an explicit example, just pick a particular $g$ and follow the procedure.
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0how to get this $g$?? – 2017-02-11
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0@Eklavya Based on your question, I imagine you have already encountered the so-called *bump functions*, which are compactly supported smooth functions. If not, the construction is standard and found in any book of differential topology for example (and also in a lot of questions in this site). Having a compactly supported smooth function $h$ on $\mathbb{R}$, consider a interval $[-a,a]$ which contains its support. Now, you can for example define $g$ by $g(x)=f(2ax-a)$. – 2017-02-11
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This type of functions are to be constructed. for example $$f(x)=\left\{ \begin{array}{ll} e^\frac{-1}{1-x^2} , & \hbox{$ |x|\leq$ 1;} \\ 0 , & \hbox{otherwise} \end{array} \right.,$$ Using the fact that $e^\frac{-1}{x^2}$ approches $0$ when x approches $0.$ $\sin x, \cos x$ can also be used to construct this type of functions.