0
$\begingroup$

Let $\phi \in C_{c}^{\infty}\mathbb{(R)}$ . I need to show that $\int_{\mathbb{R}} \phi (x) dx = 0$ iff there exists a function $\psi \in C_{c}^{\infty}\mathbb{(R)}$ such that $\phi(x) = \psi ' (x)$.

I have absolutely no clue how to begin. Any help with this is appreciated!

$\endgroup$
11
  • $\begingroup$ What is $C_c^\infty$ the Riemann Sphere? $\endgroup$ – caverac Apr 23 '17 at 16:41
  • $\begingroup$ Is it right that $\;\int_{\Bbb R}=\int_{-\infty}^\infty\;$ , an improper integral? $\endgroup$ – DonAntonio Apr 23 '17 at 16:45
  • $\begingroup$ I was thinking it was the extended complex plane. In that case if $\phi(x) = \phi'(x)$ then the integral only depends on the end points, which are the same on this space $\endgroup$ – caverac Apr 23 '17 at 16:46
  • 1
    $\begingroup$ @Dark_Knight Even if this is Lebesgue Integral we have a problem here as improper Lebesgue integrals are defined by means of improper Riemann integrals, and this last doesn't exist in this case. $\endgroup$ – DonAntonio Apr 23 '17 at 17:00
  • 1
    $\begingroup$ If nothing is mentioned then I suppose they follow the standard use for $C_c^\infty$, then we can define $\psi(x) = \int _{-\infty}^x \phi(u) du$ and since $ \int _{-\infty}^\infty \phi(u) du=0$ we see that $\psi(x) \in C_c^\infty$ $\endgroup$ – clark Apr 23 '17 at 17:29
1
$\begingroup$

Hint:

Say $\operatorname{supp}\psi \subseteq [-M, M]$. Then, it follows $$ \int_{\mathbb R} \phi = \int_{\mathbb R} \psi' = \int_{-M}^M \psi' = \psi(M) - \psi(-M). $$

On other hand, assume $\operatorname{supp} \phi \subseteq [-M, M]$. Then, $$ \psi(x) = \int_{-M}^x \phi $$ has compact support if and only if $$ \psi(M) = \int_{\mathbb R} \phi = 0. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.