For the following parts: $f(t)=\begin{cases}e^{-1/t^2}&t\neq0\\0&t=0\end{cases}$

$\quad(a)$ Show that $f\in C^\infty(\mathbb R)$; that is, $f$ is differentiable to all orders on $\mathbb R$.

$\quad(b)$ Use $f$ to define a function $g\in C^\infty(\mathbb R)$ whose support is $[a,b]$, where $a<b$.

$\quad(c)$ Show how $g$ can be used to define a function $h\in C^\infty(\mathbb R^n)$ such that $$h(x)\begin{cases}=1&||x||\leq1\\\in[0,1]&1<||x||\leq2\\=0&2<||x||.\end{cases}$$

The part I am struggling with is part c. I'm struggling to find a function which fits the given criteria for function values while also being infinitely differentiable with compact support. I used the bump function as my answer to part b. Any help would be greatly appreciated.

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    $\begingroup$ I edited the question into the body. Please do this for your future questions. $\endgroup$ Nov 15 '19 at 21:43

The function $h$ in part c is a bump function, so you aren’t intended to use it to do this problem. The directions also say to use $f$ to define $g$ in part b, so you should probably back up here. If you are like me, you will probably find it very helpful to graph $f$.

  • Graph everything (I use Desmos).
  • Let $g_1(x) = \frac{f(x-a)}{f(x-a)+f(x-b)}$ if $x\in[a,b]$, and 0 otherwise. Define $g_2$ to be a horizontally flipped version of $g_1$, with different constants $c$ and $d$. Stitch those together piecewise to make the function $g$ such that $g\equiv0$ if $x\not\in[a,d]$, $g\equiv1$ if $x\in[b,c]$ and $g$ is smooth everwhere.
  • Then just choose constants so $g$ is $h$.
  • $\begingroup$ I guess I don't know how to define g from f then. You say to mess around with 3 copies of f in a fraction and to define g' in such a way to get g. I'm honestly still lost. Am I adding scale/stretch factors to f? Am I adding an additional piecewise component to f? Is it a combination of the two? I'm sorry for not understanding but I just don't know what to do. The professor pretty much just reads us definitions and theorems and writes the proofs for them without any examples on how to apply them and pretty much just says "have at it". $\endgroup$ Nov 15 '19 at 22:29
  • $\begingroup$ That's alright. I don't think it's obvious at all if you've never seen it before. I'll give you some more to work with. $\endgroup$ Nov 16 '19 at 3:20
  • $\begingroup$ By "horizontally flipped" do you mean replacing x with -x in the definition of $g_1$? $\endgroup$ Nov 16 '19 at 5:37
  • $\begingroup$ Or could I just do $1-g_1$ with different constants? That's seeming to work on desmos, at least on the surface. $\endgroup$ Nov 16 '19 at 6:03
  • $\begingroup$ $g_2$ will definitely need to have different constants than $g_1$, so you can move them independently of each other. I changed the formula for $g_2$ slightly, so the other constant was on top. Subtracting from 1 simplifies to the same thing, so you're good. $\endgroup$ Nov 18 '19 at 2:17

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