I'm having difficulty answering this question:
Questions statement:
Consider the function:
$
f(x) = \begin{cases}
0 & \text{if $x\leq 0$,}\\
e^{-\frac{1}{x}} & \text{otherwise.}
\end{cases}
$
Show that there exists a positive function $g(x)\in C^{\infty}(\mathbb{R})$ that cancels outside the interval $ ]-1,1[ $ and that satisfies:
$ 1=\int_{\mathbb{R}}^{} g(x) dx $.
Thoughts on what to do
1) I can show that $f$ is infinitely differentiable and define the function $g$ by displacing $f$ over the interval $]-1,1[$ and define $g$ such that it's zero outside the interval $]-1,1[$ – this effectively solves the first part of the question
2) My problem is that I don't know how to guaranty the condition $ 1=\int_{\mathbb{R}}^{} g(x) dx $ without evaluating directly $ \int_{\mathbb{R}}^{} g(x) dx $ which in my mind doesn't have an analytical solution. I think it's the equality here that's got me stumped.
Any help would be appreciated
Best Answer
If we define $f$ by $$ f(x)=\left\{\begin{aligned} &0 &\text{if } x\notin ]-1,1[ \\ &e^{-\frac{1}{1-x^2}} &\text{if } x\in ]-1,1[ \end{aligned}\right. $$ (this is kinda your function $f$) then $f$ is $\mathcal{C}^{\infty}$ and you can take $$g=\frac{f}{\int_{\mathbb{R}}f}$$ I think this is the best example and the one expected.