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


1 Answer 1


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.

  • $\begingroup$ wait I understand.... that works thanks so much $\endgroup$ Sep 15, 2019 at 23:39
  • $\begingroup$ Use the linearity of the inegral : $\int_{\mathbb{R}}\lambda f=\lambda\int_{\mathbb{R}}f$, applying this with $\lambda=\frac{1}{\int_{\mathbb{R}}f}$ you get $\int_{\mathbb{R}}{g}=\frac{\int_{\mathbb{R}}f}{\int_{\mathbb{R}}f}=1$. $\endgroup$
    – Tuvasbien
    Sep 15, 2019 at 23:41
  • $\begingroup$ yeah - it took me a few to catch on... thanks guys $\endgroup$ Sep 15, 2019 at 23:42

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