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From Melrose, Lecture notes on Microlocal Analysis, Chapter 1.

I was asked to show that the function $$ u(x)=e^{x}\cos[e^{x}] $$ is a tempered distribution. I tried to use the definition that there exist $k$ and $C_{k}$ such that $$ |\int uv|\le C_{k}|v|_{k} $$ where $$|v|_{k}=\sup_{\alpha+\beta\le k}|x^{\alpha}\partial_{x}^{\beta}v|$$ However I found I did know how to treat the oscillatory part $$ \int^{\infty}_{M}uv $$ where $M$ is a large enough number. Can someone give a hint?

Bombyx mori
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  • Try to express $u$ as the derivative of a function $v$ for which it is clear that $v$ defines a tempered distribution. You could then still use the definition (use partial integration). – PhoemueX Dec 24 '14 at 13:28
  • @PhoemueX: $u$ is a tempered distribution, $v$ is in Schwartz class. $u=D[\sin[e^{x}]]$, but this fact itself is not really useful. – Bombyx mori Dec 24 '14 at 13:30
  • For example integration by parts we have $\int^{\infty}{M}uv=\sin[e^{x}]v|^{\infty}{M}-\int_{M}^{\infty} \sin[e^{x}]v'(x)$. The second one is essentially bounded by $C|v|_3$, but it is not clear to me how to treat the first one rigorously, though the infinity part seems "intuitively" should be zero. – Bombyx mori Dec 24 '14 at 13:35
  • I guess one way to look at it is $\lim_{x\rightarrow \infty}|u|=0$ as it is in Schwartz class. But I feel $\sin[e^{x}]v$ evaluated at $\infty$ is still poorly defined and I do not know how to make it rigorous. – Bombyx mori Dec 24 '14 at 13:38
  • What is your notion of integration here? Riemann or Lebesgue? – Cameron Williams Dec 24 '14 at 17:31
  • @CameronWilliams: This is what confuses me, because if we use Lebesgue then it is obvious after change of variable that for some $v$ like $e^{-x/2}$ the integral $\int |uv|=\int |\frac{\cos[y]}{\sqrt{y}}|, y=e^x$ does not really converge. – Bombyx mori Dec 24 '14 at 20:04

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First note that $u$ is a ($C^\infty$) function and not a distribution. Hence, if we want to interpret $u$ as a distribution, we need to make some kind of identification.

Your book probably defines this identification as follows: If $f$ is a (locally integrable) function, we say that $f$ defines a tempered distribution if there is some tempered distribution $u$ such that

$$ u(\varphi)= \int f \cdot \varphi\, dx $$

for all $\varphi \in C_c^\infty$ (not for all Schwartz functions).

This definition is sensible, because it is often possible to write down a Schwartz function such that the integral $\int f \cdot \varphi$ does not exist (in the usual Lebesgue sense).

If you use that definition here, your problematic boundary terms will vanish and the distribution $u$ above will turn out to be $\frac{d}{dx}\sin(e^x)$.

PhoemueX
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  • Maybe I'm off base here but aren't tempered distributions a stricter subset of distributions that are in the dual space of the Schwartz class? Your argument then wouldn't work exactly since it would only give that it is a distribution, not a tempered distribution. – Cameron Williams Dec 24 '14 at 15:50
  • @CameronWilliams: Yes, you are correct. I should have written "tempered distribution" everywhere where I wrote "distribution". But still, one says that a (locally integrable) function $f$ defines a tempered distribution if there is some tempered distribution $u$ with $u(\varphi) = \int f \cdot \varphi$ for all $\varphi \in C_c^\infty$ (and not for all Schwartz functions). This determines $u$ uniquely (if it exists), because the test functions are dense in the Schwartz space. – PhoemueX Dec 24 '14 at 17:50
  • @PhoemueX: I have to say I am surprised to learn this, because in my notes I did not see this being mentioned explicitly. Let me look up some reference book on this. Of course this is works then the boundary term would go away. – Bombyx mori Dec 24 '14 at 20:18
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    @PhoemueX: So I guess one way to look at this is it is suffice to prove the inequality for $C^{\infty}_{c}$ functions, then by density arguments we can get the desired result. – Bombyx mori Dec 24 '14 at 20:24
  • @PhoemueX. How do you define tempered distribution here? – md2perpe May 17 '19 at 14:07
  • @md2perpe: A tempered distribution is a linear function $\varphi : \mathcal{S}(\mathbb{R}^n) \to \mathbb{C}$ which is continuous with respect to the usual topology on the Schwartz space $\mathcal{S}$. Specifically, this means that there are $M \in \Bbb{N}$ and $C > 0$ such that $|\varphi(f)| \leq C \cdot \max_{|\alpha| \leq M} \sup_{x \in \Bbb{R}^n} (1 + |x|)^M |\partial^\alpha f (x)|$. Of course this simplifies a little if we are only in one dimension. Specifically, instead of the multi-index $\alpha \in \Bbb{N}_0^n$, you would then just take all derivatives of $f$ up to order $M$. – PhoemueX May 17 '19 at 17:45
  • @PhoemueX. What you wrote about identification of functions with tempered distributions made me think that perhaps you had some different definition of tempered distributions. Obviously I was wrong. – md2perpe May 17 '19 at 18:42